Topology generated by a set

• Metric spaces
• Continuity
• Compactness
• Topological spaces
• Examples
• Closed subsets
• Continuous functions
• Examples
• Homeomorphisms
• Counter Example
• Separation axioms
• TnT_n spaces
• TnT_n reflection
• Sober spaces
• Frames of opens
• Sober reflection
• Universal constructions
• Limits and colimits
• Examples
• Subspaces
• Connected components
• Embeddings
• Compact spaces
• Nonexample
• Compact Hausdorff spaces
• Locally compact spaces
• Nonexample
• Paracompact spaces
• Examples
• Partitions of unity
• Vector bundles
• Transition functions
• Properties
• Manifolds
• Nonexample
• Counter-Example
• Tangent bundles
• Embeddings
• Nonexample
• References
• General
• Special topics
• Index

(classical logic)

The proofs in the following freely use the principle of excluded middle, hence proof by contradiction, and in a few places they also use the axiom of choice/Zorn's lemma.

Hence we discuss topology in its traditional form with classical logic.

We do however highlight the role of frame homomorphisms (def. below) and that of sober topological spaces (def. below). These concepts pave the way to a constructive formulation of topology in terms not of topological spaces but in terms of locales (remark below). For further reading along these lines see Johnstone 83.

(set theory)

Apart from classical logic, we assume the usual informal concept of sets. The reader (only) needs to know the concepts of

1. subsetsSXS \subset X;

2. complementsXSX \setminus S of subsets;

3. image sets f(X)f(X) and pre-image sets f1(Y)f^{-1}(Y) under a function

   f:XYf \colon X \to Y;

4. unionsiISi\underset{i \in I}{\cup} S_i and intersections iISi\underset{i \in I}{\cap} S_i of indexed sets of subsets {SiX}iI\{S_i \subset X\}_{i \in I}.

The only rules of set theory that we use are the

1. interactions of images and pre-images with unions and intersections

2. de Morgan duality.

For reference, we recall these:

(metric space)

A metric space is

1. a set XX (the underlying set);

2. a function d:X×X[0,)d \;\colon\; X \times X \to [0,\infty) (the distance function) from the Cartesian product of the set with itself to the non-negative real numbers

such that for all x,y,zXx,y,z \in X:

1. (symmetry) d(x,y)=d(y,x)d(x,y) = d(y,x)

2. (triangle inequality) d(x,z)d(x,y)+d(y,z)d(x,z) \leq d(x,y)+ d(y,z).

3. (non-degeneracy) d(x,y)=0x=yd(x,y) = 0 \;\;\Leftrightarrow\;\; x = y

(open balls)

Let (X,d)(X,d), be a metric space. Then for every element xXx \in X and every ϵ+\epsilon \in \mathbb{R}_+ a positive real number, we write

for the open ball of radius ϵ\epsilon around xx. Similarly we write

for the closed ball of radius ϵ\epsilon around xx. Finally we write

for the sphere of radius ϵ\epsilon around xx.

For ϵ=1\epsilon = 1 we also speak of the unit open/closed ball and the unit sphere.

For (X,d)(X,d) a metric space (def. ) then a subset SXS \subset X is called a bounded subset if SS is contained in some open ball (def. )

around some xXx \in X of some radius rr \in \mathbb{R}.

(normed vector space)

A normed vector space is

1. a real vector space VV;

2. a function (the norm)

   :V⟶0{\Vert {-} \Vert} \;\colon\; V \longrightarrow \mathbb{R}_{\geq 0}

   from the underlying set of VV to the non-negative real numbers,

such that for all cc \in \mathbb{R} with absolute value |c|{\vert c \vert} and all v,wVv , w \in V it holds true that

1. (linearity) cv=|c|v{\Vert c v \Vert} = {\vert c \vert} {\Vert v \Vert };

2. (triangle inequality) v+wv+w{\Vert v+w \Vert} \leq {\Vert v \Vert } + {\Vert w \Vert};

3. (non-degeneracy) if v=0{\Vert v \Vert} = 0 then v=0v = 0.

Every normed vector space (V,)(V, {\Vert {-} \Vert}) becomes a metric space according to def. by setting

(Euclidean space)

For nn \in \mathbb{N}, the Cartesian space

carries a norm (the Euclidean norm ) given by the square root of the sum of the squares of the components:

Via prop. this gives n\mathbb{R}^n the structure of a metric space, and as such it is called the Euclidean space of dimension nn.

More generally, for nn \in \mathbb{N}, and pp \in \mathbb{R}, p1p \geq 1, then the Cartesian space n\mathbb{R}^n carries the p-norm

One also sets

and calls this the supremum norm.

The graphics on the right (grabbed from Wikipedia) shows unit circles (def. ) in 2\mathbb{R}^2 with respect to various p-norms.

By the Minkowski inequality, the p-norm generalizes to non-finite dimensional vector spaces such as sequence spaces and Lebesgue spaces.

(epsilontic definition of continuity)

For (X,dX)(X,d_X) and (Y,dY)(Y,d_Y) two metric spaces (def. ), then a function

is said to be continuous at a point xXx \in Xif for every positive real number ϵ\epsilon there exists a positive real number δ\delta such that for all xXx' \in X that are a distance smaller than δ\delta from xx then their image f(x)f(x') is a distance smaller than ϵ\epsilon from f(x)f(x):

The function ff is said to be continuous if it is continuous at every point xXx \in X.

(distance function from a subset is continuous)

Let (X,d)(X,d) be a metric space (def. ) and let SXS \subset X be a subset of the underlying set. Define then the function

from the underlying set XX to the real numbers by assigning to a point xXx \in X the infimum of the distances from xx to ss, as ss ranges over the elements of SS:

This is a continuous function, with \mathbb{R} regarded as a metric space via its Euclidean norm (example ).

In particular the original distance function d(x,)=d({x},)d(x,-) = d(\{x\},-) is continuous in both its arguments.

Let xXx \in X and let ϵ\epsilon be a positive real number. We need to find a positive real number δ\delta such that for yXy \in X with d(x,y)<δd(x,y) \lt \delta then |d(S,x)d(S,y)|<ϵ{\vert d(S,x) - d(S,y) \vert} \lt \epsilon.

For sSs \in S and yXy \in X, consider the triangle inequalities

Forming the infimum over sSs \in S of all terms appearing here yields

which implies

This means that we may take for instance δϵ\delta \coloneqq \epsilon.

(rational functions are continuous)

Consider the real line \mathbb{R} regarded as the 1-dimensional Euclidean space \mathbb{R} from example .

For P[X]P \in \mathbb{R}[X] a polynomial, then the function

is a continuous function in the sense of def. . Hence polynomials are continuous functions.

Similarly rational functions are continuous on their domain of definition: for P,Q[X]P,Q \in \mathbb{R}[X] two polynomials, then fPfQ:{x|fQ(x)=0}\frac{f_P}{f_Q} \colon \mathbb{R} \setminus \{x | f_Q(x) = 0\} \to \mathbb{R} is a continuous function.

Also for instance forming the square root is a continuous function ():00\sqrt(-) \colon \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}.

(neighbourhood and open set)

Let (X,d)(X,d) be a metric space (def. ). Say that:

1. A neighbourhood of a point xXx \in X is a subset UxXU_x \subset X which contains some open ball Bx(ϵ)UxB_x^\circ(\epsilon) \subset U_x around xx (def. ).

2. An open subset of XX is a subset UXU \subset X such that for every xUx \in U it also contains an open ball Bx(ϵ)B^\circ_x(\epsilon) around xx (def. ).

3. An open neighbourhood of a point xXx \in X is a neighbourhood UxU_x of xx which is also an open subset, hence equivalently this is any open subset of XX that contains xx.

(the empty subset is open)

Notice that for (X,d)(X,d) a metric space, then the empty subset X\emptyset \subset X is always an open subset of (X,d)(X,d) according to def. . This is because the clause for open subsets UXU \subset X says that for every point xUx \in U there exists, but since there is no xx in U=U = \emptyset, this clause is always satisfied in this case.

Conversely, the entire set XX is always an open subset of (X,d)(X,d).

(open/closed intervals)

Regard the real numbers \mathbb{R} as the 1-dimensional Euclidean space (example ).

For a<ba \lt b \in \mathbb{R} consider the following subsets:

1. (a,b){x|a<x<b}(a,b) \coloneqq \left\{ x \in \mathbb{R} \vert a \lt x \lt b \right\} AA\phantom{AA} (open interval)

2. (a,b]{x|a<xb}(a,b] \coloneqq \left\{ x \in \mathbb{R} \vert a \lt x \leq b \right\} AA\phantom{AA} (half-open interval)

3. [a,b){x|ax<b}[a,b) \coloneqq \left\{ x \in \mathbb{R} \vert a \leq x \lt b \right\} AA\phantom{AA} (half-open interval)

4. [a,b]{x|axb}[a,b] \coloneqq \left\{ x \in \mathbb{R} \vert a \leq x \leq b \right\} AA\phantom{AA} (closed interval)

The first of these is an open subset according to def. , the other three are not. The first one is called an open interval, the last one a closed interval and the middle two are called half-open intervals.

Similarly for a,ba,b \in \mathbb{R} one considers

1. (,b){x|x<b}(-\infty,b) \coloneqq \left\{ x \in \mathbb{R} \vert x \lt b \right\} AA\phantom{AA} (unbounded open interval)

2. (a,){x|a<x}(a,\infty) \coloneqq \left\{ x \in \mathbb{R} \vert a \lt x \right\} AA\phantom{AA} \,\, (unbounded open interval)

3. (,b]{x|xb}(-\infty,b] \coloneqq \left\{ x \in \mathbb{R} \vert x \leq b \right\} AA\phantom{AA} (unbounded half-open interval)

4. [a,){x|ax}[a,\infty) \coloneqq \left\{ x \in \mathbb{R} \vert a \leq x \right\} AA\phantom{AA} \,\, (unbounded half-open interval)

The first two of these are open subsets, the last two are not.

For completeness we may also consider

• (,)=(-\infty , \infty) = \mathbb{R}

• (a,a)=(a,a) = \emptyset

which are both open, according to def. .

(rephrasing continuity in terms of open sets)

Let (X,dX)(X,d_X) and (Y,dY)(Y,d_Y) be two metric spaces (def. ). Then a function f:XYf \colon X \to Y is continuous in the epsilontic sense of def. precisely if it has the property that its pre-images of open subsets of YY (in the sense of def. ) are open subsets of XX:

principle of continuity

\, \, Continuous pre-Images of open subsets are open.

Observe, by direct unwinding the definitions, that the epsilontic definition of continuity (def. ) says equivalently in terms of open balls (def. ) that ff is continous at xx precisely if for every open ball Bf(x)(ϵ)B^\circ_{f(x)}(\epsilon) around an image point, there exists an open ball Bx(δ)B^\circ_x(\delta) around the corresponding pre-image point which maps into it:

With this observation the proof immediate. For the record, we spell it out:

First assume that ff is continuous in the epsilontic sense. Then for OYYO_Y \subset Y any open subset and xf1(OY)x \in f^{-1}(O_Y) any point in the pre-image, we need to show that there exists an open neighbourhood of xx in f1(OY)f^{-1}(O_Y).

That OYO_Y is open in YY means by definition that there exists an open ball Bf(x)(ϵ)B^\circ_{f(x)}(\epsilon) in OYO_Y around f(x)f(x) for some radius ϵ\epsilon. By the assumption that ff is continuous and using the above observation, this implies that there exists an open ball Bx(δ)B^\circ_x(\delta) in XX such that f(Bx(δ))Bf(x)(ϵ)Yf(B^\circ_x(\delta)) \subset B^\circ_{f(x)}(\epsilon) \subset Y, hence such that Bx(δ)f1(Bf(x)(ϵ))f1(OY)B^\circ_x(\delta) \subset f^{-1}\left(B^{\circ}_{f(x)}(\epsilon)\right) \subset f^{-1}(O_Y). Hence this is an open ball of the required kind.

Conversely, assume that the pre-image function f1f^{-1} takes open subsets to open subsets. Then for every xXx \in X and Bf(x)(ϵ)YB_{f(x)}^\circ(\epsilon) \subset Y an open ball around its image, we need to produce an open ball Bx(δ)XB_x^\circ(\delta) \subset X around xx such that f(Bx(δ))Bf(x)(ϵ)f(B_x^\circ(\delta)) \subset B^\circ_{f(x)}(\epsilon).

But by definition of open subsets, Bf(x)(ϵ)YB^\circ_{f(x)}(\epsilon) \subset Y is open, and therefore by assumption on ff its pre-image f1(Bf(x)(ϵ))Xf^{-1}(B^\circ_{f(x)}(\epsilon)) \subset X is also an open subset of XX. Again by definition of open subsets, this implies that it contains an open ball as required.

(step function)

Consider \mathbb{R} as the 1-dimensional Euclidean space (example ) and consider the step function

graphics grabbed from Vickers 89

Consider then for a<ba \lt b \in \mathbb{R} the open interval (a,b)(a,b) \subset \mathbb{R}, an open subset according to example . The preimage H1(a,b)H^{-1}(a,b) of this open subset is

By example , all except the last of these pre-images listed are open subsets.

The failure of the last of the pre-images to be open witnesses that the step function is not continuous at x=0x = 0.

(sequence)

Given a set XX, then a sequence of elements in XX is a function

from the natural numbers to XX.

A sub-sequence of such a sequence is a sequence of the form

for some injection ι\iota.

(convergence to limit of a sequence)

Let (X,d)(X,d) be a metric space (def. ). Then a sequence

in the underlying set XX (def. ) is said to converge to a point xXx_\infty \in X, denoted

if for every positive real number ϵ\epsilon, there exists a natural number nn, such that all elements in the sequence after the nnth one have distance less than ϵ\epsilon from xx_\infty.

Here the point xx_\infty is called the limit of the sequence. Often one writes limixi\underset{i \to \infty}{\lim}x_i for this point.

(Cauchy sequence)

Given a metric space (X,d)(X,d) (def. ), then a sequence of points in XX (def. )

is called a Cauchy sequence if for every positive real number ϵ\epsilon there exists a natural number nn \in \mathbb{N} such that the distance between any two elements of the sequence beyond the nnth one is less than ϵ\epsilon

(complete metric space)

A metric space (X,d)(X,d) (def. ), for which every Cauchy sequence (def. ) converges (def. ) is called a complete metric space.

A normed vector space, regarded as a metric space via prop. that is complete in this sense is called a Banach space.

(sequentially compact metric space)

A metric space (X,d)(X,d) (def. ) is called sequentially compact if every sequence in XX has a subsequence (def. ) which converges (def. ).

(sequentially compact metric spaces are equivalently compact metric spaces)

For a metric space (X,d)(X,d) (def. ) the following are equivalent:

1. XX is sequentially compact;

2. for every set {UiX}iI\{U_i \subset X\}_{i \in I} of open subsets UiU_i of XX (def. ) which cover XX in that X=iIUiX = \underset{i \in I}{\cup} U_i, then there exists a finite subset JIJ \subset I of these open subsets which still covers XX in that also X=iJIUiX = \underset{i \in J \subset I}{\cup} U_i.

(closure properties of open sets in a metric space)

The collection of open subsets of a metric space (X,d)(X,d) as in def. has the following properties:

1. The union of any set of open subsets is again an open subset.

2. The intersection of any finite number of open subsets is again an open subset.

(empty union and empty intersection)

Notice the degenerate case of unions iIUi\underset{i \in I}{\cup} U_i and intersections iIUi\underset{i \in I}{\cap} U_i of subsets UiXU_i \subset X for the case that they are indexed by the empty set I=I = \emptyset:

1. the empty union is the empty set itself;

2. the empty intersection is all of XX.

(The second of these may seem less obvious than the first. We discuss the general logic behind these kinds of phenomena below.)

This way prop. is indeed compatible with the degenerate cases of examples of open subsets in example .

(topological spaces)

Given a set XX, then a topology on XX is a collection τ\tau of subsets of XX called the open subsets, hence a subset of the power set P(X)P(X)

such that this is closed under forming

1. finite intersections;

2. arbitrary unions.

In particular (by remark ):

• the empty set X\emptyset \subset X is in τ\tau (being the union of no subsets)

and

• the whole set XXX \subset X itself is in τ\tau (being the intersection of no subsets).

A set XX equipped with such a topology is called a topological space.

In the field of topology it is common to eventually simply say space as shorthand for topological space. This is especially so as further qualifiers are added, such as Hausdorff space (def. below). But beware that there are other kinds of spaces in mathematics.

(neighbourhood)

Let (X,τ)(X,\tau) be a topological space and let xXx \in X be a point. A neighbourhood of xx is a subset UxXU_x \subset X which contains an open subset that still contains xx.

An open neighbourhood is a neighbourhood that is itself an open subset, hence an open neighbourhood of xx is the same as an open subset containing xx.

The simple definition of open subsets in def. and the simple implementation of the principle of continuity below in def. gives the field of topology its fundamental and universal flavor. The combinatorial nature of these definitions makes topology be closely related to formal logic. This becomes more manifest still for the sober topological space discussed below. For more on this perspective see the remark on locales below, remark . An introductory textbook amplifying this perspective is (Vickers 89).

(finer/coarser topologies)

Let XX be a set, and let τ1,τ2P(X)\tau_1, \tau_2 \in P(X) be two topologies on XX, hence two choices of open subsets for XX, making it a topological space. If

hence if every open subset of XX with respect to τ1\tau_1 is also regarded as open by τ2\tau_2, then one says that

• the topology τ2\tau_2 is finer than the topology τ2\tau_2

• the topology τ1\tau_1 is coarser than the topology τ1\tau_1.

(basis for the topology)

Let (X,τ)(X, \tau) be a topological space, def. , and let

be a subset of its set of open subsets. We say that

1. β\beta is a basis for the topology τ\tau if every open subset OτO \in \tau is a union of elements of β\beta;

2. β\beta is a sub-basis for the topology if every open subset OτO \in \tau is a union of finite intersections of elements of β\beta.

(recognition of topological bases)

Let XX be a set.

1. A collection βP(X)\beta \subset P(X) of subsets of XX is a basis for some topology τP(X)\tau \subset P(X) (def. ) precisely if

   1. every point of XX is contained in at least one element of β\beta;

   2. for every two subsets B1,B2βB_1, B_2 \in \beta and for every point xB1B2x \in B_1 \cap B_2 in their intersection, there exists a BβB \in \beta that contains xx and is contained in the intersection: xBB1B2x \in B \subset B_1 \cap B_2.

2. A subset BτB \subset \tau of open subsets is a sub-basis for a topology τ\tau on XX precisely if τ\tau is the coarsest topology (def. ) which contains BB.

(metric topology)

Let (X,d)(X,d) be a metric space (def. ). Then the collection of its open subsets in def. constitutes a topology on the set XX, making it a topological space in the sense of def. . This is called the metric topology.

The open balls in a metric space constitute a basis of a topology (def. ) for the metric topology.

(empty space and point space)

On the empty set there exists a unique topology τ\tau making it a topological space according to def. . We write also

for the resulting topological space, which we call the empty topological space.

On a singleton set {1}\{1\} there exists a unique topology τ\tau making it a topological space according to def. , namelyf

We write

for this topological space and call it the point topological space.

This is equivalently the metric topology (example ) on 0\mathbb{R}^0, regarded as the 0-dimensional Euclidean space (example ).

On the 2-element set {0,1}\{0,1\} there are (up to permutation of elements) three distinct topologies:

1. the codiscrete topology (def. ) τ={,{0,1}}\tau = \left\{ \emptyset, \{0,1\} \right\};

2. the discrete topology (def. ), τ={,{0},{1},{0,1}}\tau = \left\{ \emptyset, \{0\}, \{1\}, \{0,1\} \right\};

3. the Sierpinski space topology τ={,{1},{0,1}}\tau = \left\{\emptyset, \{1\}, \{0,1\} \right\}.

The following shows all the topologies on the 3-element set (up to permutation of elements)

graphics grabbed from Munkres 75

(discrete and co-discrete topology)

Let SS be any set. Then there are always the following two extreme possibilities of equipping XX with a topology τP(X)\tau \subset P(X) in the sense of def. , and hence making it a topological space:

1. τP(S)\tau \coloneqq P(S) the set of all open subsets;

   this is called the discrete topology on SS, it is the finest topology (def. ) on XX,

   we write Disc(S)Disc(S) for the resulting topological space;

2. τ{,S}\tau \coloneqq \{ \emptyset, S \} the set contaning only the empty subset of SS and all of SS itself;

   this is called the codiscrete topology on SS, it is the coarsest topology (def. ) on XX,

   we write CoDisc(S)CoDisc(S) for the resulting topological space.

The reason for this terminology is best seen when considering continuous functions into or out of these (co-)discrete topological spaces, we come to this in example below.

(cofinite topology)

Given a set XX, then the cofinite topology or finite complement topology on XX is the topology (def. ) whose open subsets are precisely

1. all cofinite subsets SXS \subset X (i.e. those such that the complement XSX \setminus S is a finite set);

2. the empty set.

If XX is itself a finite set (but not otherwise) then the cofinite topology on XX coincides with the discrete topology on XX (example ).

(disjoint union space)

For {(Xi,τi)}iI\{(X_i, \tau_i)\}_{i \in I} a set of topological spaces, then their disjoint union

is the topological space whose underlying set is the disjoint union of the underlying sets of the summand spaces, and whose open subsets are precisely the disjoint unions of the open subsets of the summand spaces.

In particular, for II any index set, then the disjoint union of II copies of the point space (example ) is equivalently the discrete topological space (example ) on that index set:

(subspace topology)

Let (X,τX)(X, \tau_X) be a topological space, and let SXS \subset X be a subset of the underlying set. Then the corresponding topological subspace has SS as its underlying set, and its open subsets are those subsets of SS which arise as restrictions of open subsets of XX.

(This is also called the initial topology of the inclusion map. We come back to this below in def. .)

The picture on the right shows two open subsets inside the square, regarded as a topological subspace of the plane 2\mathbb{R}^2:

graphics grabbed from Munkres 75

(quotient topological space)

Let (X,τX)(X,\tau_X) be a topological space (def. ) and let

be an equivalence relation on its underlying set. Then the quotient topological space has

• as underlying set the quotient set X/X/\sim, hence the set of equivalence classes,

and

• a subset OX/O \subset X/\sim is declared to be an open subset precisely if its preimage π1(O)\pi^{-1}(O) under the canonical projection map

  π:XX/\pi \;\colon\; X \to X/\sim

  is open in XX.

(This is also called the final topology of the projection π\pi. We come back to this below in def. . )

Often one considers this with input datum not the equivalence relation, but any surjection

of sets. Of course this identifies Y=X/Y = X/\sim with (x1x2)(π(x1)=π(x2))(x_1 \sim x_2) \Leftrightarrow (\pi(x_1) = \pi(x_2)). Hence the quotient topology on the codomain set of a function out of any topological space has as open subsets those whose pre-images are open.

To see that this indeed does define a topology on X/X/\sim it is sufficient to observe that taking pre-images commutes with taking unions and with taking intersections.

(binary product topological space)

For (X1,τX1)(X_1,\tau_{X_1}) and (X2,τX2)(X_2, \tau_{X_2}) two topological spaces, then their binary product topological space has as underlying set the Cartesian product X1×X2X_1 \times X_2 of the corresponding two underlying sets, and its topology is generated from the basis (def. ) given by the Cartesian products U1×U2U_1 \times U_2 of the opens UiτiU_i \in \tau_i.

graphics grabbed from Munkres 75

Beware for non-finite products, the descriptions of the product topology is not as simple. This we turn to below in example , after introducing the general concept of limits in the category of topological spaces.

Consider the real numbers \mathbb{R} as the 1-dimensional Euclidean space (example ) and hence as a topological space via the corresponding metric topology (example ). Moreover, consider the closed interval [0,1][0,1] \subset \mathbb{R} from example , regarded as a subspace (def. ) of \mathbb{R}.

The product space (example ) of this interval with itself

is a topological space modelling the closed square. The quotient space (example ) of that by the relation which identifies a pair of opposite sides is a model for the cylinder. The further quotient by the relation that identifies the remaining pair of sides yields a model for the torus.

graphics grabbed from Munkres 75

(spheres and disks)

For nn \in \mathbb{N} write

• DnD^n for the n-disk, the closed unit ball (def. ) in the nn-dimensional Euclidean space n\mathbb{R}^n (example ) and equipped with the induced subspace topology (example ) of the corresponding metric topology (example );

• Sn1S^{n-1} for the (n-1)-sphere (def. ) also equipped with the corresponding subspace topology;

• in:Sn1Dni_n \;\colon\; S^{n-1} \hookrightarrow D^n for the continuous function that exhibits this boundary inclusion.

Notice that

• S1=S^{-1} = \emptyset is the empty topological space (example );

• S0=**S^0 = \ast \sqcup \ast is the disjoint union space (example ) of the point topological space (example ) with itself, equivalently the discrete topological space on two elements (example ).

(Zariski topology on affine space)

Let kk be a field, let nn \in \mathbb{N}, and write k[X1,,Xn]k[X_1, \cdots, X_n] for the set of polynomials in nn variables over kk.

For k[X1,,Xn]\mathcal{F} \subset k[X_1, \cdots, X_n] a subset of polynomials, let the subset V()knV(\mathcal{F}) \subset k^n of the nn-fold Cartesian product of the underlying set of kk (the vanishing set of \mathcal{F}) be the subset of points on which all these polynomials jointly vanish:

These subsets are called the Zariski closed subsets.

Write

for the set of complements of the Zariski closed subsets. These are called the Zariski open subsets of knk^n.

The Zariski open subsets of knk^n form a topology (def. ), called the Zariski topology. The resulting topological space

is also called the nn-dimensional affine space over kk.

(Zariski topology on the prime spectrum of a commutative ring)

Let RR be a commutative ring. Write PrimeIdl(R)PrimeIdl(R) for its set of prime ideals. For R\mathcal{F} \subset R any subset of elements of the ring, consider the subsets of those prime ideals that contain \mathcal{F}:

These are called the Zariski closed subsets of PrimeIdl(R)PrimeIdl(R). Their complements are called the Zariski open subsets.

Then the collection of Zariski open subsets in its set of prime ideals

satisfies the axioms of a topology (def. ), the Zariski topology.

This topological space

is called (the space underlying) the prime spectrum of the commutative ring.

(closed subsets)

Let (X,τ)(X,\tau) be a topological space (def. ).

1. A subset SXS \subset X is called a closed subset if its complement XSX \setminus S is an open subset:

   (SXis closed)AAAA(XSXis open).\left( S \subset X\,\, \text{is closed} \right) \phantom{AA} \Leftrightarrow \phantom{AA} \left( X\setminus S \, \subset X \,\, \text{is open} \right) \,.

   graphics grabbed from Vickers 89

2. If a singleton subset {x}X\{x\} \subset X is closed, one says that xx is a closed point of XX.

3. Given any subset SXS \subset X, then its topological closure Cl(S)Cl(S) is the smallest closed subset containing SS:

   Cl(S)CXclosedSC(C).Cl(S) \;\coloneqq\; \underset{ {C \subset X\, \text{closed} } \atop {S \subset C } }{\cap} \left( C \right) \,.

4. A subset SXS \subset X such that Cl(S)=XCl(S) = X is called a dense subset of (X,τ)(X,\tau).

(alternative characterization of topological closure)

Let (X,τ)(X,\tau) be a topological space and let SXS \subset X be a subset of its underlying set. Then a point xXx \in X is contained in the topological closure Cl(S)Cl(S) (def. ) precisely if every open neighbourhood UxXU_x \subset X of xx (def. ) intersects SS:

Due to de Morgan duality (prop. ) we may rephrase the definition of the topological closure as follows:

(closure of a finite union is the union of the closures)

For II a finite set and {UiX}iI\{U_i \subset X\}_{i \in I} a finite set of subsets of a topological space, we have

By lemma we use that a point is in the closure of a set precisely if every open neighbourhood (def. ) of the point intersects the set.

Hence in one direction

because if every neighbourhood of a point intersects sopme UiU_i, then every neighbourhood intersects their union.

The other direction

is equivalent by de Morgan duality to

On left now we have the point for which there exists for each iIi \in I a neighbourhood Ux,iU_{x,i} which does not intersect UiU_i. Since II is finite, the intersection iIUx,i\underset{i \in I}{\cap} U_{x,i} is still an open neighbourhood of xx, and such that it intersects none of the UiU_i, hence such that it does not intersect their union. This implis that the given point is contained in the set on the right.

(topological interior and boundary)

Let (X,τ)(X,\tau) be a topological space (def. ) and let SXS \subset X be a subset. Then the topological interior of SS is the largest open subset Int(S)τInt(S) \in \tau still contained in SS, Int(S)SXInt(S) \subset S \subset X:

The boundary S\partial S of SS is the complement of its interior inside its topological closure (def. ):

(duality between closure and interior)

Let (X,τ)(X,\tau) be a topological space and let SXS \subset X be a subset. Then the topological interior of SS (def. ) is the same as the complement of the topological closure Cl(XS)Cl(X\setminus S) of the complement of SS:

and conversely

Using de Morgan duality (prop. ), we compute as follows:

Similarly for the other case.

(topological closure and interior of closed and open intervals)

Regard the real numbers as the 1-dimensional Euclidean space (example ) and equipped with the corresponding metric topology (example ) . Let a<ba \lt b \in \mathbb{R}. Then the topological interior (def. ) of the closed interval [a,b][a,b] \subset \mathbb{R} (example ) is the open interval (a,b)(a,b) \subset \mathbb{R}, moreover the closed interval is its own topological closure (def. ) and the converse holds (by lemma ):

Hence the boundary of the closed interval is its endpoints, while the boundary of the open interval is empty

(convergence in closed subspaces)

Let (X,d)(X,d) be a metric space (def. ), regarded as a topological space via example , and let VXV \subset X be a subset. Then the following are equivalent:

1. VXV \subset X is a closed subspace according to def. .

2. For every sequence xiVXx_i \in V \subset X (def. ) with elements in VV, which converges as a sequence in XX (def. ) to some xXx_\infty \in X, we have xVXx_\infty \in V \subset X.

First assume that VXV \subset X is closed and that xi⟶ixx_i \overset{i \to \infty}{\longrightarrow} x_{\infty} for some xXx_\infty \in X. We need to show that then xVx_\infty \in V. Suppose it were not, hence that xXVx_\infty \in X\setminus V. Since, by assumption on VV, this complement XVXX \setminus V \subset X is an open subset, it would follow that there exists a real number ϵ>0\epsilon \gt 0 such that the open ball around xx of radius ϵ\epsilon were still contained in the complement: Bx(ϵ)XVB^\circ_x(\epsilon) \subset X \setminus V. But since the sequence is assumed to converge in XX, this would mean that there exists NϵN_\epsilon such that all xi>Nϵx_{i \gt N_{\epsilon}} are in Bx(ϵ)B^\circ_x(\epsilon), hence in XVX\setminus V. This contradicts the assumption that all xix_i are in VV, and hence we have proved by contradiction that xVx_\infty \in V.

Conversely, assume that for all sequences in VV that converge to some xXx_\infty \in X then xVXx_\infty \in V \subset X. We need to show that then VV is closed, hence that XVXX \setminus V \subset X is an open subset, hence that for every xXVx \in X \setminus V we may find a real number ϵ>0\epsilon \gt 0 such that the open ball Bx(ϵ)B^\circ_x(\epsilon) around xx of radius ϵ\epsilon is still contained in XVX \setminus V. Suppose on the contrary that such ϵ\epsilon did not exist. This would mean that for each kk \in \mathbb{N} with k1k \geq 1 then the intersection Bx(1/k)VB^\circ_x(1/k) \cap V were non-empty. Hence then we could choose points xkBx(1/k)Vx_k \in B^\circ_x(1/k) \cap V in these intersections. These would form a sequence which clearly converges to the original xx, and so by assumption we would conclude that xVx \in V, which violates the assumption that xXVx \in X \setminus V. Hence we proved by contradiction XVX \setminus V is in fact open.

(subsets are closed in a closed subspace precisely if they are closed in the ambient space)

Let (X,τ)(X,\tau) be a topological space (def. ), and let CXC \subset X be a closed subset (def. ), regarded as a topological subspace (C,τsub)(C,\tau_{sub}) (example ). Then a subset SCS \subset C is a closed subset of (C,τsub)(C,\tau_{sub}) precisely if it is closed as a subset of (X,τ)(X,\tau).

If SCS \subset C is closed in (C,τsub)(C,\tau_{sub}) this means equivalently that there is an open subset VCV \subset C in (C,τsub)(C, \tau_{sub}) such that

But by the definition of the subspace topology, this means equivalently that there is a subset UXU \subset X which is open in (X,τ)(X,\tau) such that V=UCV = U \cap C. Hence the above is equivalent to the existence of an open subset UXU \subset X such that

But now the condition that CC itself is a closed subset of (X,τ)(X,\tau) means equivalently that there is an open subset WXW \subset X with C=XWC = X \setminus W. Hence the above is equivalent to the existence of two open subsets W,UXW,U \subset X such that

Since the union WUW \cup U is again open, this implies that SS is closed in (X,τ)(X,\tau).

Conversely, that SXS \subset X is closed in (X,τ)(X,\tau) means that there exists an open TXT \subset X with S=XTXS = X \setminus T \subset X. This means that S=SC=(XT)C=CT=C(TC)S = S \cap C = (X \setminus T) \cap C = C\setminus T = C \setminus (T \cap C), and since TCT \cap C is open in (C,τsub)(C,\tau_{sub}) by definition of the subspace topology, this means that SCS \subset C is closed in (C,τsub)(C, \tau_{sub}).

(irreducible closed subspace)

A closed subset SXS \subset X (def. ) of a topological space XX is called irreducible if it is non-empty and not the union of two closed proper (i.e. smaller) subsets. In other words, a non-empty closed subset SXS \subset X is irreducible if whenever S1,S2XS_1, S_2 \subset X are two closed subspace such that

then S1=SS_1 = S or S2=SS_2 = S.

(closures of points are irreducible)

For xXx \in X a point inside a topological space, then the closure Cl({x})Cl(\{x\}) of the singleton subset {x}X\{x\} \subset X is irreducible (def. ).

(no nontrivial closed irreducibles in metric spaces)

Let (X,d)(X,d) be a metric space, regarded as a topological space via its metric topology (example ). Then every point xXx \in X is closed (def ), hence every singleton subset {x}X\{x\} \subset X is irreducible according to def. .

Let \mathbb{R} be the 1-dimensional Euclidean space (example ) with its metric topology (example ). Then for a<ca \lt c \subset \mathbb{R} the closed interval [a,c][a,c] \subset \mathbb{R} (example ) is not irreducible, since for any bb \in \mathbb{R} with a<b<ca \lt b \lt c it is the union of two smaller closed subintervals:

In fact we will see below (prop. ) that in a metric space the singleton subsets are precisely the only irreducible closed subsets.

(irreducible closed subsets in terms of prime open subsets)

Let (X,τ)(X, \tau) be a topological space, and let PτP \in \tau be a proper open subset of XX, hence so that the complement FXPF \coloneqq X\setminus P is a non-empty closed subspace. Then FF is irreducible in the sense of def. precisely if whenever U1,U2τU_1,U_2 \in \tau are open subsets with U1U2PU_1 \cap U_2 \subset P then U1PU_1 \subset P or U2PU_2 \subset P:

The open subsets PXP \subset X with this property are also called the prime open subsets in τX\tau_X.

Observe that every closed subset FiFF_i \subset F may be exhibited as the complement

of some open subset UiτU_i \in \tau with respect to FF. Observe that under this identification the condition that U1U2PU_1 \cap U_2 \subset P is equivalent to the condition that F1F2=FF_1 \cup F_2 = F, because it is equivalent to the equation labeled ()(\star) in the following sequence of equations:

Similarly, the condition that UiPU_i \subset P is equivalent to the condition that Fi=FF_i = F, because it is equivalent to the equality ()(\star) in the following sequence of equalities:

Under these identifications, the two conditions are manifestly the same.

(frame homomorphisms)

Let (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y) be topological spaces (def. ). Then a function

between their sets of open subsets is called a frame homomorphism from τY\tau_Y to τX\tau_X if it preserves

1. arbitrary unions;

2. finiteintersections.

In other words, ϕ\phi is a frame homomorphism precisely if

1. for every set II and every II-indexed set {UiτY}iI\{U_i \in \tau_Y\}_{i \in I} of elements of τY\tau_Y, then

   ϕ(iIUi)=iIϕ(Ui)τX,\phi\left(\underset{i \in I}{\cup} U_i\right) \;=\; \underset{i \in I}{\cup} \phi(U_i)\;\;\;\;\in \tau_X \,,

2. for every finite set JJ and every JJ-indexed set {UjτY}jJ\{U_j \in \tau_Y\}_{j \in J} of elements in τY\tau_Y, then

   ϕ(jJUj)=jJϕ(Uj)τX.\phi\left(\underset{j \in J}{\cap} U_j\right) \;=\; \underset{j \in J}{\cap} \phi(U_j) \;\;\;\;\in \tau_X \,.

(frame homomorphisms preserve inclusions)

A frame homomorphism ϕ\phi as in def. necessarily also preserves inclusions in that

• for every inclusion U1U2U_1 \subset U_2 with U1,U2τYP(Y)U_1, U_2 \in \tau_Y \subset P(Y) then

  ϕ(U1)ϕ(U2)τX.\phi(U_1) \subset \phi(U_2) \;\;\;\;\;\;\; \in \tau_X \,.

This is because inclusions are witnessed by unions

or alternatively because inclusions are witnessed by finite intersections:

(pre-images of continuous functions are frame homomorphisms)

Let (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y) be two topological spaces. One way to obtain a function between their sets of open subsets

is to specify a function

of their underlying sets, and take ϕf1\phi \coloneqq f^{-1} to be the pre-image operation. A priori this is a function of the form

and hence in order for this to co-restrict to τXP(X)\tau_X \subset P(X) when restricted to τYP(Y)\tau_Y \subset P(Y) we need to demand that, under ff, pre-images of open subsets of YY are open subsets of ZZ. Below in def. we highlight these as the continuous functions between topological spaces.

In this case then

is a frame homomorphism from τY\tau_Y to τX\tau_X in the sense of def. , by prop. .

(irreducible closed subsets are equivalently frame homomorphisms to opens of the point)

For (X,τ)(X,\tau) a topological space, then there is a natural bijection between the irreducible closed subspaces of (X,τ)(X,\tau) (def. ) and the frame homomorphisms from τX\tau_X to τ*\tau_\ast, and this bijection is given by

where U(ϕ)U_\emptyset(\phi) is the union of all elements UτxU \in \tau_x such that ϕ(U)=\phi(U) = \emptyset:

First we need to show that the function is well defined in that given a frame homomorphism ϕ:τXτ*\phi \colon \tau_X \to \tau_\ast then XU(ϕ)X \setminus U_\emptyset(\phi) is indeed an irreducible closed subspace.

To that end observe that:

(*)(\ast) If there are two elements U1,U2τXU_1, U_2 \in \tau_X with U1U2U(ϕ)U_1 \cap U_2 \subset U_{\emptyset}(\phi) then U1U(ϕ)U_1 \subset U_{\emptyset}(\phi) or U2U(ϕ)U_2 \subset U_{\emptyset}(\phi).

This is because

where the first equality holds because ϕ\phi preserves finite intersections by def. , the inclusion holds because ϕ\phi respects inclusions by remark , and the second equality holds because ϕ\phi preserves arbitrary unions by def. . But in τ*={,{1}}\tau_\ast = \{\emptyset, \{1\}\} the intersection of two open subsets is empty precisely if at least one of them is empty, hence ϕ(U1)=\phi(U_1) = \emptyset or ϕ(U2)=\phi(U_2) = \emptyset. But this means that U1U(ϕ)U_1 \subset U_{\emptyset}(\phi) or U2U(ϕ)U_2 \subset U_{\emptyset}(\phi), as claimed.

Now according to prop. the condition (*)(\ast) identifies the complement XU(ϕ)X \setminus U_{\emptyset}(\phi) as an irreducible closed subspace of (X,τ)(X,\tau).

Conversely, given an irreducible closed subset XU0X \setminus U_0, define ϕ\phi by

This does preserve

1. arbitrary unions

   because ϕ(iUi)={}\phi(\underset{i}{\cup} U_i) = \{\emptyset\} precisely if iUiU0\underset{i}{\cup}U_i \subset U_0 which is the case precisely if all UiU0U_i \subset U_0, which means that all ϕ(Ui)=\phi(U_i) = \emptyset and because i=\underset{i}{\cup}\emptyset = \emptyset;

   while ϕ(iU1)={1}\phi(\underset{i}{\cup}U_1) = \{1\} as soon as one of the UiU_i is not contained in U0U_0, which means that one of the ϕ(Ui)={1}\phi(U_i) = \{1\} which means that iϕ(Ui)={1}\underset{i}{\cup} \phi(U_i) = \{1\};

2. finite intersections

   because if U1U2U0U_1 \cap U_2 \subset U_0, then by (*)(\ast) U1U0U_1 \subset U_0 or U2U0U_2 \subset U_0, whence ϕ(U1)=\phi(U_1) = \emptyset or ϕ(U2)=\phi(U_2) = \emptyset, whence with ϕ(U1U2)=\phi(U_1 \cap U_2) = \emptyset also ϕ(U1)ϕ(U2)=\phi(U_1) \cap \phi(U_2) = \emptyset;

   while if U1U2U_1 \cap U_2 is not contained in U0U_0 then neither U1U_1 nor U2U_2 is contained in U0U_0 and hence with ϕ(U1U2)={1}\phi(U_1 \cap U_2) = \{1\} also ϕ(U1)ϕ(U2)={1}{1}={1}\phi(U_1) \cap \phi(U_2) = \{1\} \cap \{1\} = \{1\}.

Hence this is indeed a frame homomorphism τXτ*\tau_X \to \tau_\ast.

Finally, it is clear that these two operations are inverse to each other.

principle of continuity

\, \, Continuous pre-Images of open subsets are open.

(continuous function)

A continuous function between topological spaces (def. )

is a function between the underlying sets,

such that pre-images under ff of open subsets of YY are open subsets of XX.

Let (X,τX)(X, \tau_X) and (Y,τY)(Y,\tau_Y) be two topological spaces (def. ). Then a function

between the underlying sets is continuous in the sense of def. precisely if pre-images under ff of closed subsets of YY (def. ) are closed subsets of XX.

This follows since taking pre-images commutes with taking complements.

(concrete category of topological spaces)

For X1,X2,X3X_1, X_2, X_3 three topological spaces and for

two continuous functions (def. ) then their composition

is clearly itself again a continuous function from X1X_1 to X3X_3.

Moreover, this composition operation is clearly associative, in that for

three continuous functions, then

Finally, the composition operation is also clearly unital, in that for each topological space XX there exists the identity function idX:XXid_X \colon X \to X and for f:X1X2f \colon X_1 \to X_2 any continuous function then

One summarizes this situation by saying that:

1. topological spaces constitute the objects,

2. continuous functions constitute the morphisms (homomorphisms)

of a category, called the category of topological spaces (Top for short).

It is useful to depict collections of objects with morphisms between them by diagrams, like this one:

\,

\,

\,

graphics grabbed from Lawvere-Schanuel 09.

There are other categories. For instance there is the category of sets (Set for short) whose

1. objects are sets,

2. morphisms are plain functions between these.

The two categories Top and Set are different, but related. After all,

1. an object of Top (hence a topological space) is an object of Set (hence a set) equipped with extra structure (namely with a topology);

2. a morphism in Top (hence a continuous function) is a morphism in Set (hence a plain function) with the extra property that it preserves this extra structure.

Hence we have the underlying set assigning function

from the class of topological spaces to the class of sets. But more is true: every continuous function between topological spaces is, by definition, in particular a function on underlying sets:

and this assignment (trivially) respects the composition of morphisms and the identity morphisms.

Such a function between classes of objects of categories, which is extended to a function on the sets of homomorphisms between these objects in a way that respects composition and identity morphisms is called a functor. If we write an arrow between categories

then it is understood that we mean not just a function between their classes of objects, but a functor.

The functor UU at hand has the special property that it does not do much except forgetting extra structure, namely the extra structure on a set XX given by a choice of topology τX\tau_X. One also speaks of a forgetful functor.

This is intuitively clear, and we may easily formalize it: The functor UU has the special property that as a function between sets of homomorphisms (hom sets, for short) it is injective. More in detail, given topological spaces (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y) then the component function of UU from the set of continuous function between these spaces to the set of plain functions between their underlying sets

is an injective function, including the continuous functions among all functions of underlying sets.

A functor with this property, that its component functions between all hom-sets are injective, is called a faithful functor.

A category equipped with a faithful functor to Set is called a concrete category.

Hence Top is canonically a concrete category.

(product topological space construction is functorial)

For 𝒞\mathcal{C} and 𝒟\mathcal{D} two categories as in remark (for instance Top or Set) then we obtain a new category denoted 𝒞×𝒟\mathcal{C} \times \mathcal{D} and called their product category whose

1. objects are pairs (c,d)(c,d) with cc an object of 𝒞\mathcal{C} and dd an object of 𝒟\mathcal{D};

• morphisms are pairs (f,g):(c,d)(c,d)(f,g) \;\colon\; (c,d) \to (c', d') with f:ccf \colon c \to c' a morphism of 𝒞\mathcal{C}

  and g:ddg \colon d \to d' a morphism of 𝒟\mathcal{D},

• composition of morphisms is defined pairwise (f,g)(f,g)(ff,gg)(f', g') \circ (f,g) \coloneqq ( f' \circ f, g' \circ g ).

This concept secretly underlies the construction of product topological spaces:

Let (X1,τX1)(X_1,\tau_{X_1}), (X2,τX2)(X_2, \tau_{X_2}), (Y1,τY1)(Y_1, \tau_{Y_1}) and (Y2,τY2)(Y_2, \tau_{Y_2}) be topological spaces. Then for all pairs of continuous functions

and

the canonically induced function on Cartesian products of sets

is clearly a continuous function with respect to the binary product space topologies (def. )

Moreover, this construction respects identity functions and composition of functions in both arguments.

In the language of category theory (remark ), this is summarized by saying that the product topological space construction ()×()(-) \times (-) extends to a functor from the product category of the category Top with itself to itself:

(point space is terminal)

For (X,τ)(X,\tau) any topological space, then there is a unique continuous function

1. from the empty topological space (def. ) XX

   ⟶AA!AAX\emptyset \overset{\phantom{AA} \exists ! \phantom{AA} }{\longrightarrow} X

2. from XX to the point topological space (def. ).

   X⟶AA!AA*X \overset{\phantom{AA} \exists ! \phantom{AA}}{\longrightarrow} \ast

In the language of category theory (remark ), this says that

1. the empty topological space is the initial object

2. the point space *\ast is the terminal object

in the category Top of topological spaces. We come back to this below in example .

(constant continuous functions)

For (X,τ)(X, \tau) a topological space then for xXx \in X any element of the underlying set, there is a unique continuous function (which we denote by the same symbol)

from the point topological space (def. ), whose image in XX is that element. Hence there is a natural bijection

between the continuous functions from the point to any topological space, and the underlying set of that topological space.

More generally, for (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y) two topological spaces, then a continuous function XYX \to Y between them is called a constant function with value some point yYy \in Y if it factors through the point spaces as

(locally constant function)

For (X,τX)(X,\tau_X), (Y,τY)(Y,\tau_Y) two topological spaces, then a continuous function f:(X,τX)(Y,τY)f \colon (X,\tau_X) \to (Y,\tau_Y) (def. ) is called locally constant if every point xXx \in X has a neighbourhood (def. ) on which the function is constant.

(continuous functions into and out of discrete and codiscrete spaces)

Let SS be a set and let (X,τ)(X,\tau) be a topological space. Recall from example

1. the discrete topological space Disc(S)Disc(S);

2. the co-discrete topological space CoDisc(S)CoDisc(S)

on the underlying set SS. Then continuous functions (def. ) into/out of these satisfy:

1. every function (of sets) Disc(S)⟶XDisc(S) \longrightarrow X out of a discrete space is continuous;

2. every function (of sets) X⟶CoDisc(S)X \longrightarrow CoDisc(S) into a codiscrete space is continuous.

Also:

• every continuous function (X,τ)⟶Disc(S)(X,\tau) \longrightarrow Disc(S) into a discrete space is locally constant (def. ).

(diagonal)

For XX a set, its diagonal ΔX\Delta_X is the function from XX to the Cartesian product of XX with itsef, given by

For (X,τ)(X,\tau) a topological space, then the diagonal is a continuous function to the product topological space (def. ) of XX with itself.

To see this, it is sufficient to see that the preimages of basic opens U1×U2U_1 \times U_2 in τX×X\tau_{X \times X} are in τX\tau_X. But these pre-images are the intersections U1U2XU_1 \cap U_2 \subset X, which are open by the axioms on the topology τX\tau_X.

(image factorization)

Let f:(X,τX)⟶(Y,τY)f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y) be a continuous function.

Write f(X)Yf(X) \subset Y for the image of ff on underlying sets, and consider the resulting factorization of ff through f(X)f(X) on underlying sets:

There are the following two ways to topologize the image f(X)f(X) such as to make this a sequence of two continuous functions:

1. By example f(X)f(X) inherits a subspace topology from (Y,τY)(Y,\tau_Y) which evidently makes the inclusion f(X)⟶Yf(X) \longrightarrow Y a continuous function.

   Observe that this also makes Xf(X)X \to f(X) a continuous function: An open subset of f(X)f(X) in this case is of the form UYf(X)U_Y \cap f(X) for UYτYU_Y \in \tau_Y, and f1(UYf(X))=f1(UY)f^{-1}( U_Y \cap f(X) ) = f^{-1}(U_Y), which is open in XX since ff is continuous.

2. By example f(X)f(X) inherits a quotient topology from (X,τX)(X,\tau_X) which evidently makes the surjection X⟶f(X)X \longrightarrow f(X) a continuous function.

   Observe that this also makes f(X)⟶Yf(X) \longrightarrow Y a continuous function: The preimage under this map of an open subset UYτYU_Y \in \tau_Y is the restriction UYf(X)U_Y \cap f(X), and the pre-image of that under Xf(X)X \to f(X) is f1(UY)f^{-1}(U_Y), as before, which is open since ff is continuous, and therefore UYf(X)U_Y \cap f(X) is open in the quotient topology.

Regard the real numbers \mathbb{R} as the 1-dimensional Euclidean space (example ) equipped with the metric topology (example ). For aa \in \mathbb{R} the constant function (example )

maps every open subset UU \subset \mathbb{R} to the singleton set {a}\{a\} \subset \mathbb{R}, which is not open.

Write Disc()Disc(\mathbb{R}) for the set of real numbers equipped with its discrete topology (def. ) and \mathbb{R} for the set of real numbers equipped with its Euclidean metric topology (example , example ). Then the identity function on the underlying sets

is a continuous function (a special case of example ). A singleton subset {a}Disc()\{a\} \in Disc(\mathbb{R}) is open, but regarded as a subset {a}\{a\} \in \mathbb{R} it is not open.

Consider the set of real numbers \mathbb{R} equipped with its Euclidean metric topology (example , example ). The exponential function

maps all of \mathbb{R} (which is a closed subset, since =\mathbb{R} = \mathbb{R} \setminus \emptyset) to the open interval (0,)(0,\infty) \subset \mathbb{R}, which is not closed.

(open maps and closed maps)

A continuous function f:(X,τX)(Y,τY)f \colon (X,\tau_X) \to (Y, \tau_Y) (def. ) is called

• an open map if the image under ff of an open subset of XX is an open subset of YY;

• a closed map if the image under ff of a closed subset of XX (def. ) is a closed subset of YY.

(image projections of open/closed maps are themselves open/closed)

If a continuous function f:(X,τX)(Y,τY)f \colon (X,\tau_X) \to (Y,\tau_Y) is an open map or closed map (def. ) then so its its image projection Xf(X)YX \to f(X) \subset Y, respectively, for f(X)Yf(X) \subset Y regarded with its subspace topology (example ).

If ff is an open map, and OXO \subset X is an open subset, so that f(O)Yf(O) \subset Y is also open in YY, then, since f(O)=f(O)f(X)f(O) = f(O) \cap f(X), it is also still open in the subspace topology, hence Xf(X)X \to f(X) is an open map.

If ff is a closed map, and CXC \subset X is a closed subset so that also f(C)Yf(C) \subset Y is a closed subset, then the complement Yf(C)Y \setminus f(C) is open in YY and hence (Yf(C))f(X)=f(X)f(C)(Y \setminus f(C)) \cap f(X) = f(X) \setminus f(C) is open in the subspace topology, which means that f(C)f(C) is closed in the subspace topology.

(projections are open continuous functions )

For (X1,τX1)(X_1,\tau_{X_1}) and (X2,τX2)(X_2,\tau_{X_2}) two topological spaces, then the projection maps

out of their product topological space (def. )

are open continuous functions (def. ).

This is because, by definition, every open subset OX1×X2O \subset X_1 \times X_2 in the product space topology is a union of products of open subsets UiX1U_i \in X_1 and ViX2V_i \in X_2 in the factor spaces

and because taking the image of a function preserves unions of subsets

(saturated subset)

Let f:X⟶Yf \;\colon\; X \longrightarrow Y be a function of sets. Then a subset SXS \subset X is called an ff-saturated subset (or just saturated subset, if ff is understood) if SS is the pre-image of its image:

Here f1(f(S))f^{-1}(f(S)) is also called the ff-saturation of SS.

(pre-images are saturated subsets)

For f:XYf \;\colon\; X \to Y any function of sets, and SYYS_Y \subset Y any subset of YY, then the pre-image f1(SY)Xf^{-1}(S_Y) \subset X is an ff-saturated subset of XX (def. ).

Let f:X⟶Yf \colon X \longrightarrow Y be a function. Then a subset SXS \subset X is ff-saturated (def. ) precisely if its complement XSX \setminus S is saturated.

(recognition of quotient topologies)

A continuous function (def. )

whose underlying function f:X⟶Yf \colon X \longrightarrow Y is surjective exhibits τY\tau_Y as the corresponding quotient topology (def. ) precisely if ff sends open and ff-saturated subsets in XX (def. ) to open subsets of YY. By lemma this is the case precisely if it sends closed and ff-saturated subsets to closed subsets.

(saturated open neighbourhoods of saturated closed subsets under closed maps)

Let

1. f:(X,τX)⟶(Y,τY)f \;\colon\; (X, \tau_X) \longrightarrow (Y, \tau_Y) be a closed map (def. );

2. CXC \subset X be a closed subset of XX (def. ) which is ff-saturated (def. );

3. UCU \supset C be an open subset containing CC;

then there exists a smaller open subset VV still containing CC

and such that VV is still ff-saturated.

We claim that the complement of XX by the ff-saturation (def. ) of the complement of XX by UU

has the desired properties. To see this, observe first that

1. the complement XUX \setminus U is closed, since UU is assumed to be open;

2. hence the image f(XU)f(X\setminus U) is closed, since ff is assumed to be a closed map;

3. hence the pre-image f1(f(XU))f^{-1}\left( f\left( X \setminus U \right)\right) is closed, since ff is continuous (using prop. ), therefore its complement VV is indeed open;

4. this pre-image f1(f(XU))f^{-1}\left( f\left( X \setminus U \right) \right) is saturated (by example ) and hence also its complement VV is saturated (by lemma ).

Therefore it now only remains to see that UVCU \supset V \supset C.

By de Morgan's law (prop. ) the inclusion UVU \supset V is equivalent to the inclusion f1(f(XU))XUf^{-1}\left( f\left( X \setminus U \right)\right) \supset X \setminus U, which is clearly the case.

The inclusion VCV \supset C is equivalent to f1(f(XU))C=f^{-1}\left( f\left( X \setminus U \right) \right) \,\cap \, C = \emptyset. Since CC is saturated by assumption, this is equivalent to f1(f(XU))f1(f(C))=f^{-1}\left( f\left( X \setminus U \right)\right) \,\cap \, f^{-1}(f(C)) = \emptyset. This in turn holds precisely if f(XU)f(C)=f\left( X \setminus U \right) \,\cap \, f(C) = \emptyset. Since CC is saturated, this holds precisely if XUC=X \setminus U \cap C = \emptyset, and this is true by the assumption that UCU \supset C.

(homeomorphisms)

An isomorphism in the category Top (remark ) of topological spaces (def. ) with continuous functions between them (def. ) is called a homeomorphism.

Hence a homeomorphism is a continuous function

between two topological spaces (X,τX)(X,\tau_X), (Y,τY)(Y,\tau_Y) such that there exists another continuous function the other way around

such that their composites are the identity functions on XX and YY, respectively:

graphics grabbed from Munkres 75

We notationally indicate that a continuous function is a homeomorphism by the symbol \simeq.

If there is some, possibly unspecified, homeomorphism between topological spaces (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y), then we also write

and say that the two topological spaces are homeomorphic.

A property/predicate PP of topological spaces which is invariant under homeomorphism in that

is called a topological property or topological invariant.

(notation for homeomorphisms)

Beware the following notation:

1. In topology the notation f1f^{-1} generally refers to the pre-image function of a given function ff, while if ff is a homeomorphism (def. ), it is also used for the inverse function of ff. This abuse of notation is convenient: If ff happens to be a homeomorphism, then the pre-image of a subsets under ff is its image under the inverse function f1f^{-1}.

2. Many authors strictly distinguish the symbols \cong and \simeq and use the former to denote homeomorphisms and the latter to refer to homotopy equivalences (which we consider in part 2). We use either symbol (but mostly \simeq) for isomorphism in whatever the ambient category may be and try to make that context always unambiguously explicit.

If f:(X,τX)(Y,τY)f \colon (X, \tau_X) \to (Y, \tau_Y) is a homeomorphism (def. ) with inverse coninuous function gg, then

1. also gg is a homeomophism, with inverse continuous function ff;

2. the underlying function of sets f:XYf \colon X \to Y of a homeomorphism ff is necessarily a bijection, with inverse bijection gg.

Consider the continuous function

from the half-open interval (def. ) to the unit circle S1S0(1)2S^1 \coloneqq S_0(1) \subset \mathbb{R}^2 (def. ), regarded as a topological subspace (example ) of the Euclidean plane (example ).

The underlying function of sets of ff is a bijection. The inverse function of sets however fails to be continuous at (1,0)S12(1,0) \in S^1 \subset \mathbb{R}^2. Hence this ff is not a homeomorphism.

Indeed, below we see that the two topological spaces [0,2π)[0,2\pi) and S1S^1 are distinguished by topological invariants, meaning that they cannot be homeomorphic via any (other) choice of homeomorphism. For example S1S^1 is a compact topological space (def. ) while [0,2π)[0,2\pi) is not, and S1S^1 has a non-trivial fundamental group, while that of [0,2π)[0,2\pi) is trivial (this prop.).

(homeomorphisms are the continuous and open bijections)

Let f:(X,τX)⟶(Y,τY)f \;\colon\; (X, \tau_X) \longrightarrow (Y,\tau_Y) be a continuous function between topological spaces (def. ). Then the following are equivalence:

1. ff is a homeomorphism;

2. ff is a bijection and an open map (def. );

3. ff is a bijection and a closed map (def. ).

It is clear from the definition that a homeomorphism in particular has to be a bijection. The condition that the inverse function YX:gY \leftarrow X \colon g be continuous means that the pre-image function of gg sends open subsets to open subsets. But by gg being the inverse to ff, that pre-image function is equal to ff, regarded as a function on subsets:

Hence g1g^{-1} sends opens to opens precisely if ff does, which is the case precisely if ff is an open map, by definition. This shows the equivalence of the first two items. The equivalence between the first and the third follows similarly via prop. .

(concrete point homeomorphic to abstract point space)

Let (X,τX)(X,\tau_X) be a non-empty topological space, and let xXx \in X be any point. Regard the corresponding singleton subset {x}X\{x\} \subset X as equipped with its subspace topology τ{x}\tau_{\{x\}} (example ). Then this is homeomorphic (def. ) to the abstract point space from example :

(open interval homeomorphic to the real line)

Regard the real line as the 1-dimensional Euclidean space (example ) with its metric topology (example ).

Then the open interval (1,1)(-1,1) \subset \mathbb{R} (def. ) regarded with its subspace topology (example ) is homeomorphic (def.) to all of the real line

An inverse pair of continuous functions is for instance given (via example ) by

and

But there are many other choices for ff and gg that yield a homeomorphism.

Similarly, for all a<ba \lt b \in \mathbb{R}

1. the open intervals (a,b)(a,b) \subset \mathbb{R} (example ) equipped with their subspace topology are all homeomorphic to each other,

2. the closed intervals [a,b][a,b] are all homeomorphic to each other,

3. the half-open intervals of the form [a,b)[a,b) are all homeomophic to each other;

4. the half-open intervals of the form (a,b](a,b] are all homeomophic to each other.

Generally, every open ball in n\mathbb{R}^n (def. ) is homeomorphic to all of n\mathbb{R}^n:

(homeomorphisms between iterated product spaces)

Let (X,τX)(X,\tau_X), (Y,τY)(Y,\tau_Y) and (Z,τZ)(Z, \tau_Z) be topological spaces.

Then:

1. There is an evident homeomorphism between the two ways of bracketing the three factors when forming their product topological space (def. ), called the associator:

   αX,Y,Z:((X,τX)×(Y,τY))×(Z,τZ)⟶AAAA(X,τX)×((Y,τY)×(Z,τZ)).\alpha_{X,Y,Z} \;\colon\; \left( (X, \tau_X) \times (Y, \tau_Y) \right) \times (Z, \tau_Z) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (X,\tau_X) \times \left( (Y,\tau_Y) \times (Z, \tau_Z) \right) \,.

2. There are evident homeomorphism between (X,τ)(X,\tau) and its product topological space (def. ) with the point space *\ast (example ), called the left and right unitors:

   λX:*×(X,τX)⟶AAAA(X,τX)\lambda_X \;\colon\; \ast \times (X, \tau_X) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (X,\tau_X)

   and

   ρX:(X,τX)×*⟶AAAA(X,τX).\rho_X \;\colon\; (X, \tau_X) \times \ast \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (X, \tau_X) \,.

3. There is an evident homeomorphism between the results of the two orders in which to form their product topological spaces (def. ), called the braiding:

   βX,Y:(X,τX)×(Y,τY)⟶AAAA(Y,τY)×(X,τX).\beta_{X,Y} \;\colon\; (X,\tau_X) \times (Y,\tau_Y) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} (Y,\tau_Y) \times (X, \tau_X) \,.

Moreover, all these homeomorphisms are compatible with each other, in that they make the following diagrams commute (recall remark ):

1. (triangle identity)

   (X×*)×Y⟶αX,*,YX×(*×Y)ρx×idYidX×λYX×Y\array{ & (X \times \ast ) \times Y &\stackrel{\alpha_{X,\ast,Y}}{\longrightarrow} & X \times (\ast \times Y) \\ & {}_{\rho_x \times id_Y}\searrow && \swarrow_{id_X \times \lambda_Y} & \\ && X \times Y }

2. (pentagon identity)

   (W×X)×(Y×Z)αW×X,Y,ZαW,X,Y×Z((W×X)×Y)×Z(W×(X×(Y×Z)))αW,X,Y×idZidW×αX,Y,Z(W×(X×Y))×Z⟶αW,X×Y,ZW×((X×Y)×Z)\array{ && (W \times X) \times (Y \times Z) \\ & {}^{\mathllap{\alpha_{W \times X, Y, Z}}}\nearrow && \searrow^{\mathrlap{\alpha_{W,X,Y \times Z}}} \\ ((W \times X ) \times Y) \times Z && && (W \times (X \times (Y \times Z))) \\ {}^{\mathllap{\alpha_{W,X,Y}} \times id_Z }\downarrow && && \uparrow^{\mathrlap{ id_W \times \alpha_{X,Y,Z} }} \\ (W \times (X \times Y)) \times Z && \underset{\alpha_{W,X \times Y, Z}}{\longrightarrow} && W \times ( (X \times Y) \times Z ) }

3. (hexagon identities)

   (X×Y)×Z⟶αX,Y,ZX×(Y×Z)⟶βX,Y×Z(Y×Z)×XβX,Y×idZαY,Z,X(Y×X)×Z⟶αY,X,ZY×(X×Z)⟶idY×βX,YY×(Z×X)\array{ (X \times Y) \times Z &\stackrel{\alpha_{X,Y,Z}}{\longrightarrow}& X \times (Y \times Z) &\stackrel{\beta_{X,Y \times Z}}{\longrightarrow}& (Y \times Z) \times X \\ \downarrow^{\beta_{X,Y} \times id_Z} &&&& \downarrow^{\alpha_{Y,Z,X}} \\ (Y \times X) \times Z &\stackrel{\alpha_{Y,X,Z}}{\longrightarrow}& Y \times (X \times Z) &\stackrel{id_Y \times \beta_{X,Y}}{\longrightarrow}& Y \times (Z \times X) }

   and

   X×(Y×Z)⟶αX,Y,Zinv(X×Y)×Z⟶βX×Y,ZZ×(X×Y)idX×βY,ZαZ,X,YinvX×(Z×Y)⟶αX,Z,Yinv(X×Z)×Y⟶βX,Z×id(Z×X)×Y,\array{ X \times (Y \times Z) &\stackrel{\alpha^{inv}_{X,Y,Z}}{\longrightarrow}& (X \times Y) \times Z &\stackrel{\beta_{X \times Y, Z}}{\longrightarrow}& Z \times (X \times Y) \\ \downarrow^{id_X \times \beta_{Y,Z}} &&&& \downarrow^{\alpha^{inv}_{Z,X,Y}} \\ X \times (Z \times Y) &\stackrel{\alpha^{inv}_{X,Z,Y}}{\longrightarrow}& (X \times Z) \times Y &\stackrel{\beta_{X,Z} \times id}{\longrightarrow}& (Z \times X) \times Y } \,,

4. (symmetry)

   βY,XβX,Y=id:(X1×X2τX1×X2)(X1×X2τX1×X2).\beta_{Y,X} \circ \beta_{X,Y} \;=\; id \;\colon\; (X_1 \times X_2 \tau_{X_1 \times X_2}) \to (X_1 \times X_2 \tau_{X_1 \times X_2}) \,.

In the language of category theory (remark ), all this is summarized by saying that the the functorial construction ()×()(-) \times (-) of product topological spaces (example ) gives the category Top of topological spaces the structure of a monoidal category which moreover is symmetrically braided.

From this, a basic result of category theory, the MacLane coherence theorem, guarantees that there is no essential ambiguity re-backeting arbitrary iterations of the binary product topological space construction, as long as the above homeomorphsims are understood.

Accordingly, we may write

for iterated product topological spaces without putting parenthesis.

(product of closed intervals homeomorphic to hypercubes)

Let nn \in \mathbb{N}, and let [ai,bi][a_i, b_i] \subset \mathbb{R} for i{1,,n}i \in \{1, \cdots, n\} be nn closed intervals in the real line (example ), regarded as topological subspaces of the 1-dimensional Euclidean space (example ) with its metric topology (example ). Then the product topological space (def. , example ) of all these intervals is homeomorphic (def. ) to the corresponding topological subspace of the nn-dimensional Euclidean space (example ):

Similarly for open intervals:

There is a canonical bijection between the underlying sets. It remains to see that this, as well and its inverse, are continuous functions. For this it is sufficient to see that under this bijection the defining basis (def. ) for the product topology is also a basis for the subspace topology. But this is immediate from lemma .

(closed interval glued at endpoints homeomorphic circle)

As topological spaces, the closed interval [0,1][0,1] (def. ) with its two endpoints identified is homeomorphic (def. ) to the standard circle:

More in detail: let

be the unit circle in the plane

equipped with the subspace topology (example ) of the plane 2\mathbb{R}^2, which is itself equipped with its standard metric topology (example ).

Moreover, let

be the quotient topological space (example ) obtained from the interval [0,1]1[0,1] \subset \mathbb{R}^1 with its subspace topology by applying the equivalence relation which identifies the two endpoints (and nothing else).

Consider then the function

given by

This has the property that f(0)=f(1)f(0) = f(1), so that it descends to the quotient topological space

We claim that f˜\tilde f is a homeomorphism (definition ).

First of all it is immediate that f˜\tilde f is a continuous function. This follows immediately from the fact that ff is a continuous function and by definition of the quotient topology (example ).

So we need to check that f˜\tilde f has a continuous inverse function. Clearly the restriction of ff itself to the open interval (0,1)(0,1) has a continuous inverse. It fails to have a continuous inverse on [0,1)[0,1) and on (0,1](0,1] and fails to have an inverse at all on [0,1], due to the fact that f(0)=f(1)f(0) = f(1). But the relation quotiented out in [0,1]/(01)[0,1]_{/(0 \sim 1)} is exactly such as to fix this failure.

(cylinder, Möbius strip and torus homeomorphic to quotients of the square)

The square [0,1]2[0,1]^2 with two of its sides identified is the cylinder, and with also the other two sides identified is the torus:

If the sides are identified with opposite orientation, the result is the Möbius strip:

graphics grabbed from Lawson 03

(stereographic projection)

For nn \in \mathbb{N} then there is a homeomorphism (def. ) between between the n-sphere SnS^n (example ) with one point pSnp \in S^n removed and the nn-dimensional Euclidean space n\mathbb{R}^n (example ) with its metric topology (example ):

This homeomorphism is given by stereographic projection: One thinks of both the nn-sphere as well as the Euclidean space n\mathbb{R}^n as topological subspaces (example ) of n+1\mathbb{R}^{n+1} in the standard way (example ), such that they intersect in the equator of the nn-sphere. For pSnp \in S^n one of the corresponding poles, then the homeomorphism is the function which sends a point xSn{p}x \in S^{n}\setminus \{p\} along the line connecting it with pp to the point yy where this line intersects tfhe equatorial plane.

In the canonical ambient coordinates this stereographic projection is given as follows:

First consider more generally the stereographic projection

of the entire ambient space minus the point pp onto the equatorial plane, still given by mapping a point xx to the unique point yy on the equatorial hyperplane such that the points pp, xx any yy sit on the same straight line.

This condition means that there exists dd \in \mathbb{R} such that

Since the only condition on yy is that y1=0y_1 = 0 this implies that

This equation has a unique solution for dd given by

and hence it follow that

Since rational functions are continuous (example ), this function σ\sigma is continuous and since the topology on Sn\pS^n\backslash p is the subspace topology under the canonical embedding Sn\pn+1\pS^n \backslash p \subset \mathbb{R}^{n+1} \backslash p it follows that the restriction

is itself a continuous function (because its pre-images are the restrictions of the pre-images of σ\sigma to Sn\pS^n\backslash p).

To see that σ|Sn\p\sigma \vert_{S^n \backslash p} is a bijection of the underlying sets we need to show that for every

there is a unique (x1,,xn+1)(x_1, \cdots , x_{n+1}) satisfying

1. (x1,,xn+1)Sn\{p}(x_1, \cdots, x_{n+1}) \in S^{n} \backslash \{p\}, hence

   1. x1<1x_1 \lt 1;

   2. i=1n+1(xi)2=1\underoverset{i = 1}{n+1}{\sum} (x_i)^2 = 1;

2. i{2,,n+1}(yi=xi1x1)\underset{i \in \{2, \cdots, n+1\}}{\forall} \left(y_i = \frac{x_i}{1-x_1} \right).

The last condition uniquely fixes the xi2x_{i \geq 2} in terms of the given yi2y_{i \geq 2} and the remaining x1x_1, as

With this, the second condition says that

hence equivalently that

By the quadratic formula the solutions of this equation are

The solution 2r2+22r2+2=1\frac{ 2 r^2 + 2 }{ 2 r^2 + 2 } = 1 violates the first condition above, while the solution 2r222r2+2<1\frac{ 2 r^2 - 2 }{ 2 r^2 + 2 } \lt 1 satisfies it.

Therefore we have a unique solution, given by

In particular therefore also an inverse function to the stereographic projection exists and is a rational function, hence continuous by example . So we have exhibited a homeomorphism as required.

(topological invariance of dimension)

For n1,n2n_1, n_2 \in \mathbb{N} but n1n2n_1 \neq n_2, then the Euclidean spaces n1\mathbb{R}^{n_1} and n2\mathbb{R}^{n_2} (example , example ) are not homeomorphic.

More generally, an open subset in n1\mathbb{R}^{n_1} is never homeomorphic to an open subset in n2\mathbb{R}^{n_2} if n1n2n_1 \neq n_2.

(real numbers quotiented by rational numbers)

Consider the real line \mathbb{R} regarded as the 1-dimensional Euclidean space (example ) with its metric topology (example ) and consider the equivalence relation \sim on \mathbb{R} which identifies two real numbers if they differ by a rational number:

Then the quotient topological space (def. )

is a codiscrete topological space (def. ), hence its topology does not distinguish any distinct proper subsets.

(open neighbourhoods in the Sierpinski space)

Consider the Sierpinski space from example , whose underlying set consists of two points {0,1}\{0,1\}, and whose open subsets form the set τ={,{1},{0,1}}\tau = \{ \emptyset, \{1\}, \{0,1\} \}. This means that the only (open) neighbourhood of the point {0}\{0\} is the entire space. Incidentally, also the topological closure of {0}\{0\} (def. ) is the entire space.

(line with two origins)

Consider the disjoint union space \mathbb{R} \sqcup \mathbb{R} (example ) of two copies of the real line \mathbb{R} regarded as the 1-dimensional Euclidean space (example ) with its metric topology (example ), which is equivalently the product topological space (example ) of \mathbb{R} with the discrete topological space on the 2-element set (example ):

Moreover, consider the equivalence relation on the underlying set which identifies every point xix_i in the iith copy of \mathbb{R} with the corresponding point in the other, the (1i)(1-i)th copy, except when x=0x = 0:

The quotient topological space by this equivalence relation (def. )

is called the line with two origins. These two origins are the points 000_0 and 010_1.

We claim that in this space every neighbourhood of 000_0 intersects every neighbouhood of 010_1.

Because, by definition of the quotient space topology, the open neighbourhoods of 0i()/0_i \in \left( \mathbb{R} \sqcup \mathbb{R} \right)/\sim are precisely those that contain subsets of the form

But this means that the two origins 000_0 and 010_1 may not be separated by neighbourhoods, since the intersection of (ϵ,ϵ)0(-\epsilon, \epsilon)_0 with (ϵ,ϵ)i(-\epsilon, \epsilon)_i is always non-empty:

(the first three separation axioms)

Let (X,τ)(X,\tau) be a topological space (def. ).

For xyXx \neq y \in X any two points in the underlying set of XX which are not equal as elements of this set, consider the following propositions:

• (T0) There exists a neighbourhood of one of the two points which does not contain the other point.

• (T1) There exist neighbourhoods of both points which do not contain the other point.

• (T2) There exists neighbourhoods_ of both points which do not intersect each other.

graphics grabbed from Vickers 89

The topological space XX is called a TnT_n-topological space or just TnT_n-space, for short, if it satisfies condition TnT_n above for all pairs of distinct points.

A T0T_0-topological space is also called a Kolmogorov space.

A T2T_2-topological space is also called a Hausdorff topological space.

For definiteness, we re-state these conditions formally. Write x,yXx,y \in X for points in XX, write Ux,UyτU_x, U_y \in \tau for open neighbourhoods of these points. Then:

• (T0) xy((Uy({x}Uy=))or(Ux(Ux{y}=)))\underset{x \neq y}{\forall} \left( \left( \underset{U_y}{\exists} \left( \{x\} \cap U_y = \emptyset \right) \right) \,\text{or}\, \left( \underset{U_x}{\exists} \left( U_x \cap \{y\} = \emptyset \right) \right) \right)

• ((T1) xy(Ux,Uy(({x}Uy=)and(Ux{y}=)))\underset{x \neq y}{\forall} \left(\underset{U_x,U_y}{\exists} \left(\left( \{x\} \cap U_y = \emptyset\right) \, \text{and} \, \left( U_x \cap \{y\} = \emptyset \right)\right) \right)

• (T2) xy(Ux,Uy(UxUy=))\underset{x \neq y}{\forall} \left( \underset{U_x, U_y}{\exists} \left( U_x \cap U_y = \emptyset\right) \right)

(TnT_n are topological properties of increasing strength)

The separation properties TnT_n from def. are topological properties in that if two topological spaces are homeomorphic (def. ) then one of them satisfies TnT_n precisely if the other does.

Moreover, these properties imply each other as

Examples of topological spaces that are not Hausdorff (def. ) include

1. the Sierpinski space (example ),

2. the line with two origins (example ),

3. the quotient topological space /\mathbb{R}/\mathbb{Q} (example ).

(finite T1T_1-spaces are discrete)

For a finite topological space (X,τ)(X,\tau), hence one for which the underlying set XX is a finite set, the following are equivalent:

1. (X,τ)(X,\tau) is T1T_1 (def. );

2. (X,τ)(X,\tau) is a discrete topological space (def. ).

(metric spaces are Hausdorff)

Every metric space (def ), regarded as a topological space via its metric topology (example ) is a Hausdorff topological space (def. ).

Because for xyXx \neq y \in X two distinct points, then the distance d(x,y)d(x,y) between them is positive number, by the non-degeneracy axiom in def. . Accordingly the open balls (def. )

are disjoint open neighbourhoods.

(subspace of TnT_n-space is TnT_n)

Let (X,τ)(X,\tau) be a topological space satisfying the TnT_n separation axiom for some n{0,1,2}n \in \{0,1,2\} according to def. . Then also every topological subspace SXS \subset X (example ) satisfies TnT_n.

(Beware that this fails for some higher nn discussed below in def. . Open subspaces of normal spaces need not be normal.)

(T0T_0 in terms of topological closures)

A topological space (X,τ)(X,\tau) is T0T_0 (def. ) precisely if the function Cl({})Cl(\{-\}) that forms topological closures (def. ) of singleton subsets from the underlying set of XX to the set of irreducible closed subsets of XX (def. , which is well defined according to example ), is injective:

Assume first that XX is T0T_0. Then we need to show that if x,yXx,y \in X are such that Cl({x})=Cl({y})Cl(\{x\}) = Cl(\{y\}) then x=yx = y. Hence assume that Cl({x})=Cl({y})Cl(\{x\}) = Cl(\{y\}). Since the closure of a point is the complement of the union of the open subsets not containing the point (lemma ), this means that the union of open subsets that do not contain xx is the same as the union of open subsets that do not contain yy:

But if the two points were distinct, xyx \neq y, then by T0T_0 one of the above unions would contain xx or yy, while the other would not, in contradiction to the above equality. Hence we have a proof by contradiction.

Conversely, assume that (Cl{x}=Cl{y})(x=y)\left( Cl\{x\} = Cl\{y\}\right) \Rightarrow \left( x = y\right), and assume that xyx \neq y. Hence by contraposition Cl({x})Cl({y})\mathrm{Cl}(\{x\}) \neq \mathrm{Cl}(\{y\}). We need to show that there exists an open set which contains one of the two points, but not the other.

Assume there were no such open subset, hence that every open subset containing one of the two points would also contain then other. Then by lemma this would mean that xCl({y})x \in \mathrm{Cl}(\{y\}) and that yCl({x})y \in \mathrm{Cl}(\{x\}). But this would imply that Cl({x})Cl({y})Cl(\{x\}) \subset \mathrm{Cl}(\{y\}) and that Cl({y})Cl({x})\mathrm{Cl}(\{y\}) \subset \mathrm{Cl}(\{x\}), hence that Cl({x})=Cl({y})\mathrm{Cl}(\{x\}) = \mathrm{Cl}(\{y\}). This is a proof by contradiction.

(T1T_1 in terms of topological closures)

A topological space (X,τ)(X,\tau) is T1T_1 (def. ) precisely if all its points are closed points (def. ).

We have

Here the first step is the reformulation of closure from lemma , the second is another application of the de Morgan law (prop. ), the third is the definition of union and complement, and the last one is manifestly by definition of T1T_1.

(T2T_2 in terms of topological closures)

A topological space (X,τX)(X,\tau_X) is T2T_2=Hausdorff precisely if the image of the diagonal

is a closed subset in the product topological space (X×X,τX×X)(X \times X, \tau_{X \times X}).

Observe that the Hausdorff condition is equivalently rephrased in terms of the product topology as: Every point (x,y)X(x,y) \in X which is not on the diagonal has an open neighbourhood U(x,y)×U(x,y)U_{(x,y)} \times U_{(x,y)} which still does not intersect the diagonal, hence:

Therefore if XX is Hausdorff, then the diagonal ΔX(X)X×X\Delta_X(X) \subset X \times X is the complement of a union of such open sets, and hence is closed:

Conversely, if the diagonal is closed, then (by lemma ) every point (x,y)X×X(x,y) \in X \times X not on the diagonal, hence with xyx \neq y, has an open neighbourhood U(x,y)×V(x,y)U_{(x,y)} \times V_{(x,y)} still not intersecting the diagonal, hence so that U(x,y)V(x,y)=U_{(x,y)} \cap V_{(x,y)} = \emptyset. Thus (X,τ)(X,\tau) is Hausdorff.

Let (X,τ)(X,\tau) be topological space (def. ).

Consider the following conditions

• (T3) The space (X,τ)(X,\tau) is T1T_1 (def. ) and for xXx \in X a point and CXC \subset X a closed subset (def. ) not containing xx, then there exist disjoint open neighbourhoods Ux{x}U_x \supset \{x\} and UCCU_C \supset C.

• (T4) The space (X,τ)(X,\tau) is T1T_1 (def. ) and for C1,C2XC_1, C_2 \subset X two disjoint closed subsets (def. ) then there exist disjoint open neighbourhoods UCiCiU_{C_i} \supset C_i.

If (X,τ)(X,\tau) satisfies T3T_3 it is said to be a T3T_3-space also called a regular Hausdorff topological space.

If (X,τ)(X,\tau) satisfies T4T_4 it is to be a T4T_4-space also called a normal Hausdorff topological space.

(metric spaces are normal Hausdorff)

Let (X,d)(X,d) be a metric space (def. ) regarded as a topological space via its metric topology (example ). Then this is a normal Hausdorff space (def. ).

By example metric spaces are T2T_2, hence in particular T1T_1. What we need to show is that given two disjoint closed subsets C1,C2XC_1, C_2 \subset X then their exists disjoint open neighbourhoods UC1C1U_{C_1} \subset C_1 and UC2C2U_{C_2} \supset C_2.

Recall the function

computing distances from a subset SXS \subset X (example ). Then the unions of open balls (def. )

and

have the required properties.

(TnT_n are topological properties of increasing strength)

The separation axioms from def. , def. are topological properties (def. ) which imply each other as

The implications

and

are immediate from the definitions. The remaining implication T3T2T_3 \Rightarrow T_2 follows with prop. : This says that by assumption of T1T_1 then all points in (X,τ)(X,\tau) are closed, and with this the condition T2T_2 is manifestly a special case of the condition for T3T_3.

(K-topology)

Write

for the subset of natural fractions inside the real numbers.

Define a topological basis βP()\beta \subset P(\mathbb{R}) on \mathbb{R} consisting of all the open intervals as well as the complements of KK inside them:

The topology τβP()\tau_{\beta} \subset P(\mathbb{R}) which is generated from this topological basis is called the K-topology.

We may denote the resulting topological space by

This is a Hausdorff topological space (def. ) which is not a regular Hausdorff space, hence (by prop. ) in particular not a normal Hausdorff space (def. ).

(T3T_3 in terms of topological closures)

A topological space (X,τ)(X,\tau) is a regular Hausdorff space (def. ), precisely if all points are closed and for all points xXx \in X with open neighbourhood U{x}U \supset \{x\} there exists a smaller open neighbourhood V{x}V \supset \{x\} whose topological closure Cl(V)Cl(V) is still contained in UU:

(T4T_4 in terms of topological closures)

A topological space (X,τ)(X,\tau) is normal Hausdorff space (def. ), precisely if all points are closed and for all closed subsets CXC \subset X with open neighbourhood UCU \supset C there exists a smaller open neighbourhood VCV \supset C whose topological closure Cl(V)Cl(V) is still contained in UU:

In one direction, assume that (X,τ)(X,\tau) is normal, and consider

It follows that the complement of the open subset UU is closed and disjoint from CC:

Therefore by assumption of normality of (X,τ)(X,\tau), there exist open neighbourhoods with

But this means that

and since the complement XWX \setminus W of the open set WW is closed, it still contains the closure of VV, so that we have

as required.

In the other direction, assume that for every open neighbourhood UCU \supset C of a closed subset CC there exists a smaller open neighbourhood VV with

Consider disjoint closed subsets

We need to produce disjoint open neighbourhoods for them.

From their disjointness it follows that

is an open neighbourhood. Hence by assumption there is an open neighbourhood VV with

Thus

are two disjoint open neighbourhoods, as required.

(Urysohn function)

Let (X,τ)(X,\tau) be a topological space, and let A,BXA,B \subset X be disjoint closed subsets. Then an Urysohn function separating AA from BBis

• a continuous function f:X[0,1]f \colon X \to [0,1]

to the closed interval equipped with its Euclidean metric topology (example , example ), such that

• it takes the value 0 on AA and the value 1 on BB:

  f(A)={0}AAAandAAAf(B)={1}.f(A) = \{0\} \phantom{AAA} \text{and} \phantom{AAA} f(B) = \{1\} \,.

(Urysohn's lemma)

Let XX be a normal Hausdorff topological space (def. ), and let A,BXA,B \subset X be two disjoint closed subsets of XX. Then there exists an Urysohn function separating AA from BB (def. ).

Beware, the Urysohn function in prop. may take the values 0 or 1 even outside of the two subsets. The condition that the function takes value 0 or 1, respectively, precisely on the two subsets corresponds to perfectly normal spaces.

of Urysohn's lemma, prop.

Set

Since by assumption

we have

That (X,τ)(X,\tau) is normal implies, by lemma , that every open neighbourhood UCU \supset C of a closed subset CC contains a smaller neighbourhood VV together with its topological closure Cl(V)Cl(V)

Apply this fact successively to the above situation to obtain the following infinite sequence of nested open subsets UrU_r and closed subsets CrC_r

and so on, labeled by the dyadic rational numbers dy\mathbb{Q}_{dy} \subset \mathbb{Q} within (0,1](0,1]

with the property

Define then the function

to assign to a point xXx \in X the infimum of the labels of those open subsets in this sequence that contain xx:

Here the limit is over the directed set of those UrU_r that contain xx, ordered by reverse inclusion.

This function clearly has the property that f(A)={0}f(A) = \{0\} and f(B)={1}f(B) = \{1\}. It only remains to see that it is continuous.

To this end, first observe that

Here it is immediate from the definition that (xUr)(f(x)r)(x \in U_r) \Rightarrow (f(x) \leq r) and that (f(x)<r)(xUrCl(Ur))(f(x) \lt r) \Rightarrow (x \in U_r \subset Cl(U_r)). For the remaining implication, it is sufficient to observe that

where UrCl(Ur)Ur\partial U_r \coloneqq Cl(U_r) \setminus U_r is the boundary of UrU_r.

This holds because the dyadic numbers are dense in \mathbb{R}. (And this would fail if we stopped the above decomposition into Ua/2nU_{a/2^n}-s at some finite nn.) Namely, in one direction, if xUrx \in \partial U_r then for every small positive real number ϵ\epsilon there exists a dyadic rational number rr' with r<r<r+ϵr \lt r' \lt r + \epsilon, and by construction UrCl(Ur)U_{r'} \supset Cl(U_r) hence xUrx \in U_{r'}. This implies that limUr{x}=r\underset{U_r \supset \{x\}}{\lim} = r.

Now we claim that for all α[0,1]\alpha \in [0,1] then

1. f1((α,1])=r>α(XCl(Ur))f^{-1}(\,(\alpha, 1]\,) = \underset{r \gt \alpha}{\cup} \left( X \setminus Cl(U_r) \right)

2. f1([0,α))=r<αUrf^{-1}(\,[0,\alpha)\,) = \underset{r \lt \alpha}{\cup} U_r

Thereby f1((α,1])f^{-1}(\,(\alpha, 1]\,) and f1([0,α))f^{-1}(\,[0,\alpha)\,) are exhibited as unions of open subsets, and hence they are open.

Regarding the first point:

and

Regarding the second point:

and

(In these derivations we repeatedly use that (0,1]dy(0,1] \cap \mathbb{Q}_{dy} is dense in [0,1][0,1] (def. ), and we use the contrapositions of ()(\star) and ()(\star \star).)

Now since the subsets {[0,α),(α,1]}α[0,1]\{ [0,\alpha), (\alpha,1]\}_{\alpha \in [0,1]} form a sub-base (def. ) for the Euclidean metric topology on [0,1][0,1], it follows that all pre-images of ff are open, hence that ff is continuous.

(normality equivalent to existence of Urysohn functions)

A T1T_1-space (def. ) is normal (def. ) precisely if it admits Urysohn functions (def ) separating every pair of disjoint closed subsets.

In one direction this is the statement of the Urysohn lemma, prop. .

In the other direction, assume the existence of Urysohn functions (def. ) separating all disjoint closed subsets. Let A,BXA, B \subset X be disjoint closed subsets, then we need to show that these have disjoint open neighbourhoods.

But let f:X[0,1]f \colon X \to [0,1] be an Urysohn function with f(A)={0}f(A) = \{0\} and f(B)={1}f(B) = \{1\} then the pre-images

are disjoint open neighbourhoods as required.

(TnT_n-reflection)

Let n{0,1,2}n \in \{0,1,2\}. Then for every topological space XX there exists

1. a TnT_n-topological space TnXT_n X

2. a continuous function

   tn(X):X⟶TnXt_n(X) \;\colon\; X \longrightarrow T_n X

   called the TnT_n-reflection of XX,

which is the closest approximation from the left to XX by a TnT_n-topological space, in that for YY any TnT_n-space, then continuous functions of the form

are in bijection with continuous function of the form

and such that the bijection is constituted by

• For n=0n = 0 this is known as the Kolmogorov quotient construction (see prop. below).

• For n=2n = 2 this is known as Hausdorff reflection or Hausdorffication or similar.

Moreover, the operation Tn()T_n(-) extends to continuous functions f:XYf \colon X \to Y

such as to preserve composition of functions as well as identity functions:

Finally, the comparison map is compatible with this in that

(reflective subcategories)

In the language of category theory (remark ) the TnT_n-reflection of prop. says that

1. Tn()T_n(-) is a functor Tn:Top⟶TopTnT_n \;\colon\; Top \longrightarrow Top_{T_n} from the category Top of topological spaces to the full subcategory TopTnιTopTop_{T_n} \overset{\iota}{\hookrightarrow} Top of Hausdorff topological spaces;

2. tn(X):XTnXt_n(X) \colon X \to T_n X is a natural transformation from the identity functor on Top to the functor ιTn\iota \circ T_n

3. TnT_n-topological spaces form a reflective subcategory of all topological spaces in that TnT_n is left adjoint to the inclusion functor ι\iota; this situation is denoted as follows:

   TopTnAιA⟵AHATop.Top_{T_n} \underoverset{\underset{\phantom{A}\iota\phantom{A}}{\hookrightarrow}}{\overset{\phantom{A}H\phantom{A}}{\longleftarrow}}{\bot} Top \,.

Generally, an adjunction between two functors

is for all pairs of objects c𝒞c \in \mathcal{C}, d𝒟d \in \mathcal{D} a bijection between sets of morphisms of the form

i.e.

and such that these bijections are natural in that they for all pairs of morphisms f:ccf \colon c' \to c and g:ddg \colon d \to d' then the folowing diagram commutes:

One calls the image under ϕc,L(c)\phi_{c,L(c)} of the identity morphism idL(x)id_{L(x)} the unit of the adjunction, written

One may show that it follows that the image f˜\tilde f under ϕc,d\phi_{c,d} of a general morphism f:cdf \colon c \to d (called the adjunct of ff) is given by this composite:

In the case of the reflective subcategory inclusion (Tnι)(T_n \dashv \iota) of the category of TnT_n-spaces into the category Top of all topological spaces this adjunction unit is precisely the TnT_n-reflection tn(X):Xι(Tn(X))t_n(X) \colon X \to \iota( T_n(X)) (only that we originally left the re-embedding ι\iota notationally implicit).

(TnT_n-reflection via explicit quotients)

Let n{0,1,2}n \in \{0,1,2\}. Let (X,τ)(X,\tau) be a topological space and consider the equivalence relation \sim on the underlying set XX for which x1x2x_1 \sim x_2 precisely if for every surjective continuous function f:XYf \colon X \to Y into any TnT_n-topological space YY (def. ) we have f(x1)=f(x2)f(x_1) = f(x_2):

Then

1. the set of equivalence classes

   TnXX/T_n X \coloneqq X /{\sim}

   equipped with the quotient topology (example ) is a TnT_n-topological space,

2. the quotient projection

   X⟶tn(X)X/xAAA[x]\array{ X & \overset{t_n(X)}{\longrightarrow} & X/{\sim} \\ x &\overset{\phantom{AAA}}{\mapsto}& [x] }

   exhibits the TnT_n-reflection of XX, according to prop. .

First we observe that every continuous function f:X⟶Yf \colon X \longrightarrow Y into a TnT_n-topological space YY factors uniquely, via tn(X)t_n(X) through a continuous function f˜\tilde f (this makes use of the universal property of the quotient topology, which we dwell on a bit more below in example ):

Clearly this continuous function f˜\tilde f is unique if it exists, because its underlying function of sets must be given by

First observe that this is indeed well defined as a function of underlying sets. To that end, factor ff through its image f(X)f(X)

equipped with its subspace topology as a subspace of YY (example ). By prop. also the image f(X)f(X) is a TnT_n-topological space, since YY is. This means that if two elements x1,x2Xx_1, x_2 \in X have the same equivalence class, then, by definition of the equivalence relation, they have the same image under all comntinuous surjective functions into a TnT_n-space, hence in particular they have the same image under f:X⟶surjectivef(X)Yf \colon X \overset{\text{surjective}}{\longrightarrow} f(X) \hookrightarrow Y:

This shows that f˜\tilde f is well defined as a function between sets.

To see that f˜\tilde f is also continuous, consider UYU \in Y an open subset. We need to show that the pre-image f˜1(U)\tilde f^{-1}(U) is open in X/X/\sim. But by definition of the quotient topology (example ), this is open precisely if its pre-image under the quotient projection tn(X)t_n(X) is open, hence precisely if

is open in XX. But this is the case by the assumption that ff is continuous. Hence f˜\tilde f is indeed the unique continuous function as required.

What remains to be seen is that TnXT_n X as constructed is indeed a TnT_n-topological space. Hence assume that [x][y]TnX[x] \neq [y] \in T_n X are two distinct points. Depending on the value of nn, need to produce open neighbourhoods around one or both of these points not containing the other point and possibly disjoint to each other.

Now by definition of TnXT_n X the assumption [x][y][x] \neq [y] means that there exists a TnT_n-topological space YY and a surjective continuous function f:X⟶surjectiveYf \colon X \overset{surjective}{\longrightarrow} Y such that f(x)f(y)Yf(x) \neq f(y) \in Y:

Accordingly, since YY is TnT_n, there exist the respective kinds of neighbourhoods around f(x1)f(x_1) and f(x2)f(x_2) in YY. Moreover, by the previous statement there exists the continuous function f˜:TnXY\tilde f \colon T_n X \to Y with f˜([x1])=f(x1)\tilde f([x_1]) = f(x_1) and f˜([x2])=f(x2)\tilde f([x_2]) = f(x_2). By the nature of continuous functions, the pre-images of these open neighbourhoods in YY are still open in XX and still satisfy the required disjunction properties. Therefore TnXT_n X is a TnT_n-space.

(Kolmogorov quotient)

Let (X,τ)(X,\tau) be a topological space. Consider the relation on the underlying set by which x1x2x_1 \sim x_2 precisely if neither xix_i has an open neighbourhood not containing the other. This is an equivalence relation. The quotient topological space XX/X \to X/\sim by this equivalence relation (def. ) exhibits the T0T_0-reflection of XX according to prop. .

(more explicit Hausdorff reflection)

For (Y,τY)(Y,\tau_Y) a topological space, write rYY×Yr_Y \subset Y \times Y for the transitive closure of the relation given by the topological closure Cl(ΔY)Cl(\Delta_Y) of the image of the diagonal ΔY:YY×Y\Delta_Y \colon Y \hookrightarrow Y \times Y.

Now for (X,τX)(X,\tau_X) a topological space, define by induction for each ordinal number α\alpha an equivalence relation rαr^\alpha on XX as follows, where we write qα:XHα(X)q^\alpha \colon X \to H^\alpha(X) for the corresponding quotient topological space projection:

We start the induction with the trivial equivalence relation:

• rX0ΔXr^0_X \coloneqq \Delta_X;

For a successor ordinal we set

• rXα+1{(a,b)X×X|(qα(a),qα(b))rHα(X)}r_X^{\alpha+1} \coloneqq \left\{ (a,b) \in X \times X \,\vert\, (q^\alpha(a), q^\alpha(b)) \in r_{H^\alpha(X)} \right\}

and for a limit ordinal α\alpha we set

• rXαβ<αrXβr_X^\alpha \coloneqq \underset{\beta \lt \alpha}{\cup} r_X^\beta.

Then:

1. there exists an ordinal α\alpha such that rXα=rXα+1r_X^\alpha = r_X^{\alpha+1}

2. for this α\alpha then Hα(X)=H(X)H^\alpha(X) = H(X) is the Hausdorff reflection from prop. .

(Hausdorff reflection of the line with two origins)

The Hausdorff reflection (T2T_2-reflection, prop. )

of the line with two origins from example is the real line itself:

(sober topological space)

A topological space (X,τ)(X,\tau) is called a sober topological space precisely if every irreducible closed subspace (def. ) is the topological closure (def. ) of a unique point, hence precisely if the function

from the underlying set of XX to the set of irreducible closed subsets of XX (def. , well defined according to example ) is bijective.

(sober implies T0T_0)

Every sober topological space (def. ) is T0T_0 (def. ).

By prop. .

(Hausdorff spaces are sober)

Every Hausdorff topological space (def. ) is a sober topological space (def. ).

More specifically, in a Hausdorff topological space the irreducible closed subspaces (def. ) are precisely the singleton subspaces (def. ).

Hence, by example , in particular every metric space with its metric topology (example ) is sober.

The second statement clearly implies the first. To see the second statement, suppose that FF is an irreducible closed subspace which contained two distinct points xyx \neq y. Then by the Hausdorff property there would be disjoint neighbourhoods Ux,UyU_x, U_y, and hence it would follow that the relative complements FUxF \setminus U_x and FUyF \setminus U_y were distinct closed proper subsets of FF with

in contradiction to the assumption that FF is irreducible.

This proves by contradiction that every irreducible closed subset is a singleton. Conversely, generally the topological closure of every singleton is irreducible closed, by example .

The intersection of the classes of sober topological spaces (def. ) and T1T_1-topological spaces (def. ) is not empty, but neither class is contained within the other.

(T1T_1 neither implies nor is implied by sobriety)

• The Sierpinski space (def. ) is sober, but not T1T_1.

• The cofinite topology (example ) on a non-finite set is T1T_1 but not sober.

(schemes are sober but in general not Hausdorff)

The Zariski topology on an affine space (example ) or more generally on the prime spectrum of a commutative ring (example ) is

1. sober (def );

2. in general not Hausdorff (def. ).

For details see at Zariski topology this prop and this example.

If (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y) are sober topological spaces (def. ), then for every frame homomorphism (def. )

there is a unique continuous function f:XYf \colon X \to Y such that ϕ\phi is the function of forming pre-images under ff:

We first consider the special case of frame homomorphisms of the form

and show that these are in bijection to the underlying set XX, identified with the continuous functions *(X,τ)\ast \to (X,\tau) via example .

By prop. , the frame homomorphisms ϕ:τXτ*\phi \colon \tau_X \to \tau_\ast are identified with the irreducible closed subspaces XU(ϕ)X \setminus U_\emptyset(\phi) of (X,τX)(X,\tau_X). Therefore by assumption of sobriety of (X,τ)(X,\tau) there is a unique point xXx \in X with XU=Cl({x})X \setminus U_{\emptyset} = Cl(\{x\}). In particular this means that for UxU_x an open neighbourhood of xx, then UxU_x is not a subset of U(ϕ)U_\emptyset(\phi), and so it follows that ϕ(Ux)={1}\phi(U_x) = \{1\}. In conclusion we have found a unique xXx \in X such that

This is precisely the inverse image function of the continuous function *X\ast \to X which sends 1x1 \mapsto x.

Hence this establishes the bijection between frame homomorphisms of the form τ*⟵τX\tau_\ast \longleftarrow \tau_X and continuous functions of the form *(X,τ)\ast \to (X,\tau).

With this it follows that a general frame homomorphism of the form τX⟵ϕτY\tau_X \overset{\phi}{\longleftarrow} \tau_Y defines a function of sets X⟶fYX \overset{f}{\longrightarrow} Y by composition:

By the previous analysis, an element UYτYU_Y \in \tau_Y is sent to {1}\{1\} under this composite precisely if the corresponding point *X⟶fY\ast \to X \overset{f}{\longrightarrow} Y is in UYU_Y, and similarly for an element UXτXU_X \in \tau_X. It follows that ϕ(UY)τX\phi(U_Y) \in \tau_X is precisely that subset of points in XX which are sent by ff to elements of UYU_Y, hence that ϕ=f1\phi = f^{-1} is the pre-image function of ff. Since ϕ\phi by definition sends open subsets of YY to open subsets of XX, it follows that ff is indeed a continuous function. This proves the claim in generality.

(locales)

Proposition is often stated as saying that sober topological spaces are equivalently the locales with enough points (Johnstone 82, II 1.). Here locale refers to a concept akin to topological spaces where one considers just a frame of open subsets τX\tau_X, without requiring that its elements be actual subsets of some ambient set. The natural notion of homomorphism between such generalized topological spaces are clearly the frame homomorphisms τXτY\tau_X \leftarrow \tau_Y from def. .

From this perspective, prop. says that sober topological spaces (X,τX)(X, \tau_X) are entirely characterized by their frames of opens τX\tau_X and just so happen to have enough points such that these are actual open subsets of some ambient set, namely of XX.

(sober reflection)

Let (X,τ)(X,\tau) be a topological space.

Define SXS X to be the set

of frame homomorphisms (def. ) from the frame of opens of XX to that of the point. Define a topology τSXP(SX)\tau_{S X} \subset P(S X) on this set by declaring it to have one element U˜\tilde U for each element UτXU \in \tau_X and given by

Consider the function

which sends an element xXx \in X to the function which assigns inverse images of the constant function constx:{1}Xconst_x \;\colon\; \{1\} \to X on that element.

We are going to call this function the sober reflection of XX.

(sober reflection is well defined)

The construction (SX,τSX)(S X, \tau_{S X}) in def. is a topological space, and the function sX:XSXs_X \colon X \to S X is a continuous function

To see that τSXP(SX)\tau_{S X} \subset P(S X) is closed under arbitrary unions and finite intersections, observe that the function

in fact preserves arbitrary unions and finite intersections. Whith this the statement follows by the fact that τX\tau_X is closed under these operations.

To see that ()˜\widetilde{(-)} indeed preserves unions, observe that (e.g. Johnstone 82, II 1.3 Lemma)

where we used that the frame homomorphism p:τXτ*p \colon \tau_X \to \tau_\ast preserves unions. Similarly for intersections, now with II a finite set:

where we used that the frame homomorphism pp preserves finite intersections.

To see that sXs_X is continuous, observe that sX1(U˜)=Us_X^{-1}(\tilde U) = U, by construction.

(sober reflection detects T0T_0 and sobriety)

For (X,τX)(X, \tau_X) a topological space, the function sX:XSXs_X \colon X \to S X from def. is

1. an injection precisely if (X,τX)(X,\tau_X) is T0T_0 (def. );

2. a bijection precisely if (X,τY)(X,\tau_Y) is sober (def. ), in which case sXs_X is in fact a homeomorphism (def. ).

By lemma there is an identification SXIrrClSub(X)S X \simeq IrrClSub(X) and via this sXs_X is identified with the map xCl({x})x \mapsto Cl(\{x\}).

Hence the second statement follows by definition, and the first statement by prop. .

That in the second case sXs_X is in fact a homeomorphism follows from the definition of the opens U˜\tilde U: they are identified with the opens UU in this case (expand).

(soberification lands in sober spaces, e.g. Johnstone 82, lemma II 1.7)

For (X,τ)(X,\tau) a topological space, then the topological space (SX,τSX)(S X, \tau_{S X}) from def. , lemma is sober.

Let SXU˜S X \setminus \tilde U be an irreducible closed subspace of (SX,τSX)(S X, \tau_{S X}). We need to show that it is the topological closure of a unique element ϕSX\phi \in S X.

Observe first that also XUX \setminus U is irreducible.

To see this use prop. , saying that irreducibility of XUX \setminus U is equivalent to U1U2U(U1U)or(U2U)U_1 \cap U_2 \subset U \Rightarrow (U_1 \subset U) or (U_2 \subset U). But if U1U2UU_1 \cap U_2 \subset U then also U˜1U˜2U˜\tilde U_1 \cap \tilde U_2 \subset \tilde U (as in the proof of lemma ) and hence by assumption on U˜\tilde U it follows that U˜1U˜\tilde U_1 \subset \tilde U or U˜2U˜\tilde U_2 \subset \tilde U. By lemma this in turn implies U1UU_1 \subset U or U2UU_2 \subset U. In conclusion, this shows that also XUX \setminus U is irreducible .

By lemma this irreducible closed subspace corresponds to a point pSXp \in S X. By that same lemma, this frame homomorphism p:τXτ*p \colon \tau_X \to \tau_\ast takes the value \emptyset on all those opens which are inside UU. This means that the topological closure of this point is just SXU˜S X \setminus \tilde U.

This shows that there exists at least one point of which XU˜X \setminus \tilde U is the topological closure. It remains to see that there is no other such point.

So let p1p2SXp_1 \neq p_2 \in S X be two distinct points. This means that there exists UτXU \in \tau_X with p1(U)p2(U)p_1(U) \neq p_2(U). Equivalently this says that U˜\tilde U contains one of the two points, but not the other. This means that (SX,τSX)(S X, \tau_{S X}) is T0. By prop. this is equivalent to there being no two points with the same topological closure.

(unique factorization through soberification)

For (X,τX)(X, \tau_X) any topological space, for (Y,τYsob)(Y,\tau_Y^{sob}) a sober topological space, and for f:(X,τX)⟶(Y,τY)f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y) a continuous function, then it factors uniquely through the soberification sX:(X,τX)⟶(SX,τSX)s_X \colon (X, \tau_X) \longrightarrow (S X, \tau_{S X}) from def. , lemma

By the construction in def. , we find that the outer part of the following square commutes:

By lemma and lemma , the right vertical morphism sSXs_{S X} is an isomorphism (a homeomorphism), hence has an inverse morphism. This defines the diagonal morphism, which is the desired factorization.

To see that this factorization is unique, consider two factorizations f˜,f¯::(SX,τSX)(Y,τYsob)\tilde f, \overline{f} \colon \colon (S X, \tau_{S X}) \to (Y, \tau_Y^{sob}) and apply the soberification construction once more to the triangles

Here on the right we used again lemma to find that the vertical morphism is an isomorphism, and that f˜\tilde f and f¯\overline{f} do not change under soberification, as they already map between sober spaces. But now that the left vertical morphism is an isomorphism, the commutativity of this triangle for both f˜\tilde f and f¯\overline{f} implies that f˜=f¯\tilde f = \overline{f}.

(sober reflection)

For every topological space XX there exists

1. a sober topological spaces SXS X;

2. a continuous function sn:X⟶SXs_n \colon X \longrightarrow S X

such that

(sober topological spaces are a reflective subcategory)

In the language of category theory (remark ) and in terms of the concept of adjoint functors (remark ), proposition simply says that sober topological spaces form a reflective subcategory TopsobTop_{sob} of the category Top of all topological spaces

(universal property of binary product topological space)

Let X1,X2X_1, X_2 be topological spaces. Consider their product topological space X1×X2X_1 \times X_2 from example . By example the two projections out of the product space are continuous functions

Now let YY be any other topological space. Then, by composition, every continuous function YX1×X2Y \to X_1 \times X_2 into the product space yields two continuous component functions f1f_1 and f2f_2:

But in fact these two components completely characterize the function into the product: There is a (natural) bijection between continuous functions into the product space and pairs of continuous functions into the two factor spaces:

(universal property of disjoint union spaces)

Let X1,X2X_1, X_2 be topological spaces. Consider their disjoint union space X1X2X_1 \sqcup X_2 from example . By definition, the two inclusions into the disjoint union space are clearly continuous functions

Now let YY be any other topological space. Then by composition a continuous function X1X2⟶YX_1 \sqcup X_2 \longrightarrow Y out of the disjoint union space yields two continuous component functions f1f_1 and f2f_2:

But in fact these two components completely characterize the function out of the disjoint union: There is a (natural) bijection between continuous functions out of disjoint union spaces and pairs of continuous functions out of the two summand spaces:

(universal property of quotient topological spaces)

Let XX be a topological space, and let \sim be an equivalence relation on its underlying set. Then the corresponding quotient topological space X/X/\sim together with the corresponding qutient continuous function p:XX/p \colon X \to X/\sim has the following universal property:

Given f:X⟶Yf \colon X \longrightarrow Y any continuous function out of XX with the property that it respects the given equivalence relation, in that

then there is a unique continuous function f˜:X/⟶Y\tilde f \colon X/\sim \longrightarrow Y such that

(We already made use of this universal property in the construction of the TnT_n-reflection in the proof of prop. .)

First observe that there is a unique function f˜\tilde f as claimed on the level of functions of the underlying sets: In order for f=f˜pf = \tilde f \circ p to hold, f˜\tilde f must send an equivalence class in X/X/\sim to one of its members

and that this is well defined and independent of the choice of representative xx is guaranteed by the condition on ff above.

Hence it only remains to see that f˜\tilde f defined this way is continuous, hence that for UYU \subset Y an open subset, then its pre-image f˜1(U)X/\tilde f^{-1}(U) \subset X/\sim is open in the quotient topology. By definition of the quotient topology (example ), this is the case precisely if its further pre-image under pp is open in XX. But by the fact that f=f˜pf = \tilde f \circ p, this is the case by the continuity of ff:

(free diagram of sets/topological spaces)

A free diagram XX_\bullet of sets or of topological spaces is

1. an indexed set {Xi}iI\{ X_i \}_{i \in I} of sets or of topological spaces, respectively;

2. for every pair (i,j)I×I(i,j) \in I \times I of labels, a set {Xi⟶fαXj}αIi,j\{ X_i \overset{ f_\alpha }{\longrightarrow} X_j\}_{\alpha \in I_{i,j}} of functions of continuous functions, respectively, between these.

(discrete diagram and empty diagram)

Let II be any set, and for each (i,j)I×I(i,j) \in I \times I let Ii,j=I_{i,j} = \emptyset be the empty set.

The corresponding free diagrams (def. ) are simply a set of sets/topological spaces with no specified (continuous) functions between them. This is called a discrete diagram.

For example for I={1,2,3}I = \{1,2,3\} the set with 3-elements, then such a diagram looks like this:

Notice that here the index set may be empty set, I=I = \emptyset, in which case the corresponding diagram consists of no data. This is also called the empty diagram.

(parallel morphisms diagram)

Let I={a,b}I = \{a, b\} be the set with two elements, and consider the sets

The corresponding free diagrams (def. ) are called pairs of parallel morphisms. They may be depicted like so:

(span and cospan diagram)

Let I={a,b,c}I = \{a,b,c\} the set with three elements, and set

The corresponding free diagrams (def. ) look like so:

These are called span diagrams.

Similary, there is the cospan diagram of the form

(tower diagram)

Let I=I = \mathbb{N} be the set of natural numbers and consider

The corresponding free diagrams (def. ) are called tower diagrams. They look as follows:

Similarly there are co-tower diagram

(cone over a free diagram)

Consider a free diagram of sets or of topological spaces (def. )

Then

1. a cone over this diagram is

   1. a set or topological space X˜\tilde X (called the tip of the cone);

   2. for each iIi \in I a function or continuous function X˜⟶piXi\tilde X \overset{p_i}{\longrightarrow} X_i

   such that

   • for all (i,j)I×I(i,j) \in I \times I and all αIi,j\alpha \in I_{i,j} then the condition

     fαpi=pjf_{\alpha} \circ p_i = p_j

     holds, which we depict as follows:

     X˜pipjXi⟶fαXj\array{ && \tilde X \\ & {}^{\mathllap{p_i}}\swarrow && \searrow^{\mathrlap{p_j}} \\ X_i && \underset{f_\alpha}{\longrightarrow} && X_j }

2. a co-cone over this diagram is

   1. a set or topological space X˜\tilde X (called the tip of the co-cone);

   2. for each iIi \in I a function or continuous function qi:Xi⟶X˜q_i \colon X_i \longrightarrow \tilde X;

   such that

   • for all (i,j)I×I(i,j) \in I \times I and all αIi,j\alpha \in I_{i,j} then the condition

     qjfα=qiq_j \circ f_{\alpha} = q_i

     holds, which we depict as follows:

     Xi⟶fαXjqiqjX˜.\array{ X_i && \overset{f_\alpha}{\longrightarrow} && X_j \\ & {}_{\mathllap{q_i}}\searrow && \swarrow_{\mathrlap{q_j}} \\ && \tilde X } \,.

(solutions to equations are cones)

Let f,g:f,g \colon \mathbb{R} \to \mathbb{R} be two functions from the real numbers to themselves, and consider the corresponding parallel morphism diagram of sets (example ):

Then a cone (def. ) over this free diagram with tip the singleton set *\ast is a solution to the equation f(x)=g(x)f(x) = g(x)

Namely the components of the cone are two functions of the form

hence equivalently two real numbers, and the conditions on these are

(limiting cone over a diagram)

Consider a free diagram of sets or of topological spaces (def. ):

Then

1. its limiting cone (or just limit for short, also inverse limit, for historical reasons) is the cone

   {lim⟵kXkpipjXi⟶fαXj}\left\{ \array{ && \underset{\longleftarrow}{\lim}_k X_k \\ & {}^{\mathllap{p_i}}\swarrow && \searrow^{\mathrlap{p_j}} \\ X_i && \underset{f_\alpha}{\longrightarrow} && X_j } \right\}

   over this diagram (def. ) which is universal among all possible cones, in that for

   {X˜pipjXi⟶fαXj}\left\{ \array{ && \tilde X \\ & {}^{\mathllap{p'_i}}\swarrow && \searrow^{\mathrlap{p'_j}} \\ X_i && \underset{f_\alpha}{\longrightarrow} && X_j } \right\}

   any other cone, then there is a unique function or continuous function, respectively

   ϕ:X˜⟶lim⟶iXi\phi \;\colon\; \tilde X \overset{}{\longrightarrow} \underset{\longrightarrow}{\lim}_i X_i

   that factors the given cone through the limiting cone, in that for all iIi \in I then

   pi=piϕp'_i = p_i \circ \phi

   which we depict as follows:

   X˜!ϕpilim⟶iXi⟶piXi\array{ \tilde X \\ {}^{\mathllap{ \exists !\, \phi}}\downarrow & \searrow^{\mathrlap{p'_i}} \\ \underset{\longrightarrow}{\lim}_i X_i &\underset{p_i}{\longrightarrow}& X_i }

2. its colimiting cocone (or just colimit for short, also direct limit, for historical reasons) is the cocone

   {Xi⟶fαXjqiqjlim⟶iXi}\left\{ \array{ X_i && \overset{f_\alpha}{\longrightarrow} && X_j \\ & {}^{\mathllap{q_i}}\searrow && \swarrow^{\mathrlap{q_j}} \\ \\ && \underset{\longrightarrow}{\lim}_i X_i } \right\}

   under this diagram (def. ) which is universal among all possible co-cones, in that it has the property that for

   {Xi⟶fαXjqiqjX˜}\left\{ \array{ X_i && \overset{f_\alpha}{\longrightarrow} && X_j \\ & {}^{\mathllap{q'_i}}\searrow && \swarrow_{\mathrlap{q'_j}} \\ && \tilde X } \right\}

   any other cocone, then there is a unique function or continuous function, respectively

   ϕ:lim⟶iXi⟶X˜\phi \;\colon\; \underset{\longrightarrow}{\lim}_i X_i \overset{}{\longrightarrow} \tilde X

   that factors the given co-cone through the co-limiting cocone, in that for all iIi \in I then

   qi=ϕqiq'_i = \phi \circ q_i

   which we depict as follows:

   Xi⟶qilim⟶iXiqi!ϕX˜\array{ X_i &\overset{q_i}{\longrightarrow}& \underset{\longrightarrow}{\lim}_i X_i \\ & {}_{q'_i}\searrow & \downarrow^{\mathrlap{\exists ! \phi}} \\ && \tilde X }

(initial object and terminal object)

Consider the empty diagram (def. ).

1. A cone over the empty diagram is just an object XX, with no further structure or condition. The universal property of the limit \top over the empty diagram is hence that for every object XX, there is a unique map of the form XX \to \top, with no further condition. Such an object \top is called a terminal object.

2. A co-cone over the empty diagram is just an object XX, with no further structure or condition. The universal property of the colimit \bot over the empty diagram is hence that for every object XX, there is a unique map of the form X\bot \to X. Such an object \bot is called an initial object.

(Cartesian product and coproduct)

Let {Xi}iI\{X_i\}_{i \in I} be a discrete diagram (example ), i.e. just a set of objects.

1. The limit over this diagram is called the Cartesian product, denoted iIXi\underset{i \in I}{\prod} X_i;

2. The colimit over this diagram is called the coproduct, denoted iIXi\underset{i \in I}{\coprod} X_i.

(equalizer)

Let

be a free diagram of the shape pair of parallel morphisms (example ).

A limit over this diagram according to def. is also called the equalizer of the maps f1f_1 and f2f_2. This is a set or topological space eq(f1,f2)eq(f_1,f_2) equipped with a map eq(f1,f2)⟶p1X1eq(f_1,f_2) \overset{p_1}{\longrightarrow} X_1, so that f1p1=f2p1f_1 \circ p_1 = f_2 \circ p_1 and such that if YX1Y \to X_1 is any other map with this property

then there is a unique factorization through the equalizer:

In example we have seen that a cone over such a pair of parallel morphisms is a solution to the equation f1(x)=f2(x)f_1(x) = f_2(x).

The equalizer above is the space of all solutions of this equation.

(pullback/fiber product and coproduct)

Consider a cospan diagram (example )

The limit over this diagram is also called the fiber product of XX with YY over ZZ, and denoted X×ZYX \underset{Z}{\times}Y. Thought of as equipped with the projection map to XX, this is also called the pullback of ff along gg

Dually, consider a span diagram (example )

The colimit over this diagram is also called the pushout of ff along gg, denoted XZYX \underset{Z}{\sqcup}Y:

(limits and colimits of sets)

Let

be a free diagram of sets (def. ). Then

1. its limit cone (def. ) is given by the following subset of the Cartesian product iIXi\underset{i \in I}{\prod} X_i of all the sets XiX_i appearing in the diagram

   lim⟵iXiAAAiIXi\underset{\longleftarrow}{\lim}_i X_i \,\overset{\phantom{AAA}}{\hookrightarrow}\, \underset{i \in I}{\prod} X_i

   on those tuples of elements which match the graphs of the functions appearing in the diagram:

   lim⟵iXi{(xi)iI|i,jIαIi,j(fα(xi)=xj)}\underset{\longleftarrow}{\lim}_{i} X_i \;\simeq\; \left\{ (x_i)_{i \in I} \,\vert\, \underset{ {i,j \in I} \atop { \alpha \in I_{i,j} } }{\forall} \left( f_{\alpha}(x_i) = x_j \right) \right\}

   and the projection functions are pi:(xj)jIxip_i \colon (x_j)_{j \in I} \mapsto x_i.

2. its colimiting co-cone (def. ) is given by the quotient set of the disjoint union iIXi\underset{i \in I}{\sqcup} X_i of all the sets XiX_i appearing in the diagram

   iIXi⟶AAAlim⟶iIXi\underset{i \in I}{\sqcup} X_i \,\overset{\phantom{AAA}}{\longrightarrow}\, \underset{\longrightarrow}{\lim}_{i \in I} X_i

   with respect to the equivalence relation which is generated from the graphs of the functions in the diagram:

   lim⟶iXi(iIXi)/((xx)(i,jIαIi,j(fα(x)=x)))\underset{\longrightarrow}{\lim}_i X_i \;\simeq\; \left( \underset{i \in I}{\sqcup} X_i \right)/ \left( (x \sim x') \Leftrightarrow \left( \underset{ {i,j \in I} \atop { \alpha \in I_{i,j} } }{\exists} \left( f_\alpha(x) = x' \right) \right) \right)

   and the injection functions are the evident maps to equivalence classes:

   qi:xi[xi].q_i \;\colon\; x_i \mapsto [x_i] \,.

We dicuss the proof of the first case. The second is directly analogous.

First observe that indeed, by construction, the projection maps pip_i as given do make a cone over the free diagram, by the very nature of the relation that is imposed on the tuples:

We need to show that this is universal, in that every other cone over the free diagram factors universally through this one. First consider the case that the tip of a given cone is a singleton:

As shown on the right, the data in such a cone is equivantly: for each iIi \in I an element xiXix'_i \in X_i, such that for all i,jIi, j \in I and αIi,j\alpha \in I_{i,j} then fα(xi)=xjf_\alpha(x'_i) = x'_j. But this is precisely the relation used in the construction of the limit above and hence there is a unique map

such that for all iIi \in I we have

namely that map is the one that picks the element (xi)iI(x'_i)_{i \in I}.

This shows that every cone with tip a singleton factors uniquely through the claimed limiting cone. But then for a cone with tip an arbitrary set YY, this same argument applies to all the single elements of YY.

(initial topology and final topology)

Let {(Xi,τi)}iI\{(X_i, \tau_i)\}_{i \in I} be a set of topological spaces, and let SS be a bare set. Then

• For

  {S⟶AApiAAXi}iI\{S \overset{\phantom{AA}p_i\phantom{AA}}{\longrightarrow} X_i \}_{i \in I}

  a set of functions out of SS, the initial topology τinitial({pi}iI)\tau_{initial}(\{p_i\}_{i \in I}) is the coarsest topology on SS (def. ) such that all fi:(S,τinitial({pi}iI))⟶Xif_i \colon (S,\tau_{initial}(\{p_i\}_{i \in I})) \longrightarrow X_i are continuous.

  By lemma this is equivalently the topology whose open subsets are the unions of finite intersections of the preimages of the open subsets of the component spaces under the projection maps, hence the topology generated from the sub-base

  βini({pi})={pi1(Ui)|iI,UiXiopen}.\beta_{ini}(\{p_i\} ) = \left\{ p_i^{-1}(U_i) \;\vert\; i \in I,\, U_i \subset X_i \, \text{open} \right\} \,.

• For

  {Xi⟶AAfiAAS}iI\{X_i \overset{\phantom{AA}f_i\phantom{AA}}{\longrightarrow} S\}_{i \in I}

  a set of functions into SS, the final topology τfinal({fi}iI)\tau_{final}(\{f_i\}_{i \in I}) is the finest topology on SS (def. ) such that all qi:Xi⟶(S,τfinal({fi}iI))q_i \colon X_i \longrightarrow (S,\tau_{final}(\{f_i\}_{i \in I})) are continuous.

  Hence a subset USU \subset S is open in the final topology precisely if for all iIi \in I then the pre-image qi1(U)Xiq_i^{-1}(U) \subset X_i is open.

Beware a variation of synonyms that is in use:

(subspace topology as an initial topology)

For (X,τ)(X,\tau) a single topological space, and q:SXq \colon S \hookrightarrow X a subset of its underlying set, then the initial topology τintial(p)\tau_{intial}(p), def. , is the subspace topology from example , making

a topological subspace inclusion.

(quotient topology as a final topology)

Conversely, for (X,τ)(X,\tau) a topological space and for q:X⟶Sq \colon X \longrightarrow S a surjective function out of its underlying set, then the final topology τfinal(q)\tau_{final}(q) on SS, from def. , is the quotient topology from example , making qq a continuous function:

(limits and colimits of topological spaces)

Let

be a free diagram of topological spaces (def. ).

1. The limit over this free diagram (def. ) is given by the topological space

   1. whose underlying set is the limit of the underlying sets according to prop. ;

   2. whose topology is the initial topology, def. , for the functions pip_i which are the limiting cone components:

   lim⟵kIXkpipjXi⟶Xj.\array{ && \underset{\longleftarrow}{\lim}_{k \in I} X_k \\ & {}^{\mathllap{p_i}}\swarrow && \searrow^{\mathrlap{p_j}} \\ X_i && \underset{}{\longrightarrow} && X_j } \,.

   Hence

   lim⟵iI(Xi,τi)(lim⟵iIXi,τinitial({pi}iI))\underset{\longleftarrow}{\lim}_{i \in I} (X_i, \tau_i) \;\simeq\; \left(\underset{\longleftarrow}{\lim}_{i \in I} X_i,\; \tau_{initial}\left(\{p_i\}_{i \in I} \right) \right)

2. The colimit over the free diagram (def. ) is the topological space

   1. whose underlying set is the colimit of sets of the underlying diagram of sets according to prop. ,

   2. whose topology is the final topology, def. for the component maps ιi\iota_i of the colimiting cocone

   Xi⟶Xjqiqjlim⟶kIXk.\array{ X_i && \longrightarrow && X_j \\ & {}_{\mathllap{q_i}}\searrow && \swarrow_{\mathrlap{q_j}} \\ && \underset{\longrightarrow}{\lim}_{k \in I} X_k } \,.

   Hence

   lim⟶iI(Xi,τi)(lim⟶iIXi,τfinal({qi}iI))\underset{\longrightarrow}{\lim}_{i \in I} (X_i, \tau_i) \;\simeq\; \left(\underset{\longrightarrow}{\lim}_{i \in I} X_i,\; \tau_{final}(\{q_i\}_{i \in I})\right)

We discuss the first case, the second is directly analogous:

Consider any cone over the given free diagram:

By the nature of the limiting cone of the underlying diagram of underlying sets, which always exists by prop. , there is a unique function of underlying sets of the form

satisfying the required conditions piϕ=pip_i \circ \phi = p'_i. Since this is already unique on the underlying sets, it is sufficient to show that this function is always continuous with respect to the initial topology.

Hence let Ulim⟵iXiU \subset \underset{\longleftarrow}{\lim}_i X_i be in τinitial({pi})\tau_{initial}( \{p_i\} ). By def. , this means that UU is a union of finite intersections of subsets of the form pi1(Ui)p_i^{-1}(U_i) with UiXiU_i \subset X_i open. But since taking pre-images preserves unions and intersections (prop. ), and since unions and intersections of opens in (X˜,τX˜)(\tilde X, \tau_{\tilde X}) are again open, it is sufficient to consider UU of the form U=pi1(Ui)U = p_i^{-1}(U_i). But then by the condition that piϕ=pip_i \circ \phi = p'_i we find

and this is open by the assumption that pip'_i is continuous.

(functions into a limit cone are the limit of the functions into the diagram)

Let {Xi⟶fαXj}i,jI,αIi,j\{X_i \overset{f_\alpha}{\longrightarrow} X_j\}_{i,j \in I, \alpha \in I_{i,j}} be a free diagram (def. ) of sets or of topological spaces.

1. If the limit lim⟵iXi𝒞\underset{\longleftarrow}{\lim}_i X_i \in \mathcal{C} exists (def. ), then the set of (continuous) function into this limiting object is the limit over the sets Hom(,)Hom(-,-) of (continuous) functions (homomorphisms) into the components XiX_i:

   Hom(Y,lim⟵iXi)lim⟵i(Hom(Y,Xi)).Hom\left(\,Y\,, \,\underset{\longleftarrow}{\lim}_i X_i \, \right) \;\simeq\; \underset{\longleftarrow}{\lim}_i \left( Hom\left(\,Y\,, \,X_i\, \right) \right) \,.

   Here on the right we have the limit over the free diagram of sets given by the operations fα()f_\alpha \circ (-) of post-composition with the maps in the original diagram:

   {Hom(Y,Xi)⟶Afα()AHom(Y,Xj)}i,jI,αIi,j.\left\{ Hom(Y,X_i) \overset{\phantom{A} f_\alpha \circ (-) \phantom{A} }{\longrightarrow} Hom(Y, X_j) \right\}_{i,j \in I, \alpha \in I_{i,j}} \,.

2. If the colimit lim⟶iXi𝒞\underset{\longrightarrow}{\lim}_i X_i \in \mathcal{C} exists, then the set of (continuous) functions out of this colimiting object is the limit over the sets of morphisms out of the components of XiX_i:

   Hom(lim⟶iXi,Y)lim⟵(Hom(Xi,Y)).Hom\left(\,\underset{\longrightarrow}{\lim}_i X_i\,,\, Y\,\right) \;\simeq\; \underset{\longleftarrow}{\lim} \left( Hom\left(\, X_i\,, \,Y\, \right) \right) \,.

   Here on the right we have the colimit over the free diagram of sets given by the operations ()fα(-) \circ f_\alpha of pre-composition with the original maps:

   {Hom(Xi,Y)⟶A()fαAHom(Xj,Y)}i,jI,αIi,j.\left\{ Hom(X_i, Y) \overset{\phantom{A} (-)\circ f_\alpha \phantom{A}}{\longrightarrow} Hom(X_j, Y) \right\}_{i,j \in I, \alpha \in I_{i,j}} \,.

We give the proof of the first statement. The proof of the second statement is directly analogous (just reverse the direction of all maps).

First observe that, by the very definition of limiting cones, maps out of some YY into them are in natural bijection with the set Cones(Y,{XifαXj})Cones\left(Y, \{X_i \overset{f_\alpha}{\to} X_j\} \right) of cones over the corresponding diagram with tip YY:

Hence it remains to show that there is also a natural bijection like so:

Now, again by the very definition of limiting cones, a single element in the limit on the right is equivalently a cone of the form

This is equivalently for each iIi \in I a choice of map pi:YXip_i \colon Y \to X_i , such that for each i,jIi,j \in I and αIi,j\alpha \in I_{i,j} we have fαpi=pjf_\alpha \circ p_i = p_j. And indeed, this is precisely the characterization of an element in the set Cones(Y,{XifαXj})Cones\left( Y, \{X_i \overset{f_\alpha}{\to} X_j\} \right).

(limits and colimits in categories of nice topological spaces)

Recall from remark the concept of adjoint functors

witnessed by natural isomorphisms

Then these adjoints preserve (co-)limits in that

1. the left adjoint functor LL preserve colimits (def. )

   in that for every diagram {XifαXj}\{X_i \overset{f_\alpha}{\to} X_j\} in 𝒟\mathcal{D} there is a natural isomorphism of the form

   L(lim⟶iXi)lim⟶iL(Xi)L \left( \underset{\longrightarrow}{\lim}_i X_i \right) \;\simeq\; \underset{\longrightarrow}{\lim}_i L(X_i)

2. the right adjoint functor RR preserve limits (def. )

   in that for every diagram {XifαXj}\{X_i \overset{f_\alpha}{\to} X_j\} in 𝒞\mathcal{C} there is a natural isomorphism of the form

   R(lim⟵iXi)lim⟵iR(Xi).R \left( \underset{\longleftarrow}{\lim}_i X_i \right) \;\simeq\; \underset{\longleftarrow}{\lim}_i R(X_i) \,.

This implies that if we have a reflective subcategory of topological spaces

(such as with Tn2T_{n \leq 2}-spaces according to remark or with sober spaces according to remark )

then

1. limits in TopniceTop_{nice} are computed as limits in TopTop;

2. colimits in TopniceTop_{nice} are computed as the reflection LL of the colimit in TopTop.

For example let {(Xi,τi)fα(Xj,τj)}\{(X_i, \tau_i) \overset{f_\alpha}{\to} (X_j, \tau_j) \} be a diagram of Hausdorff spaces, regarded as a diagram of general topological spaces. Then

1. not only is the limit of topological spaces lim⟵i(Xi,τi)\underset{\longleftarrow}{\lim}_i (X_i, \tau_i) according to prop. again a Hausdorff space, but it also satisfies its universal property with respect to the category of Hausdorff spaces;

2. not only is the reflection T2(lim⟶iXi)T_2 \left( \underset{\longrightarrow}{\lim}_i X_i \right) of the colimit as topological spaces a Hausdorff space (while the colimit as topological spaces in general is not), but this reflection does satisfy the universal property of a colimit with respect to the category of Hausdorff spaces.

First to see that right/left adjoint functors preserve limits/colimits: We discuss the case of the right adjoint functor preserving limits. The other case is directly anlogous (just reverse the direction of all arrows).

So let lim⟵iXi\underset{\longleftarrow}{\lim}_i X_i be the limit over some diagram {XifαXj}i,jI,αIi,j\left\{ X_i \overset{f_\alpha}{\to} X_j\right\}_{i,j \in I, \alpha \in I_{i,j}}. To test what a right adjoint functor does to this, we may map any object YY into it. Using prop. this yields

Since this is true for all YY, it follows that

Now to see that limits/colimits in the reflective subcategory are computed as claimed;

()

(empty space and point space as empty colimit and limit)

Consider the empty diagram (example ) as a diagram of topological spaces. By example the limit and colimit (def. ) over this type of diagram are the terminal object and initial object, respectively. Applied to topological spaces we find:

1. The limit of topological spaces over the empty diagram is the point space *\ast (example ).

2. The colimit of topological spaces over the empty diagram is the empty topological space \emptyset (example ).

This is because for an empty diagram, the a (co-)cone is just a topological space, without any further data or properties, and it is universal precisely if there is a unique continuous function to (respectively from) this space to any other space XX. This is the case for the point space (respectively empty space) by example :

(binary product topological space and disjoint union space as limit and colimit)

Consider a discrete diagram consisting of two topological spaces (X,τX),(Y,τY)(X,\tau_X), (Y,\tau_Y) (example ). Generally, it limit and colimit is the product X×YX \times Y and coproduct XYX \sqcup Y, respectively (example ).

1. In topological space this product is the binary product topological space from example , by the universal property observed in example :

   (X,τX)×(Y,τY)(X×Y,τX×Y).(X,\tau_X) \times (Y, \tau_Y) \simeq (X \times Y, \tau_{X \times Y}) \,.

2. In topological spaces, this coproduct is the disjoint union space from example , by the universal property observed in example :

   (X,τX)(Y,τY)(XY,τXY).(X, \tau_X) \sqcup (Y, \tau_Y) \simeq ( X \sqcup Y, \tau_{X \sqcup Y} ) \,.

(general product topological spaces with Tychonoff topology)

Consider an arbitrary discrete diagram of topological spaces (def. ), hence a set {(Xi,τi)}iI\{(X_i, \tau_i)\}_{i \in I} of topological spaces, indexed by any set II, not necessarily a finite set.

The limit over this diagram (a Cartesian product, example ) is called the product topological space of the spaces in the diagram, and denoted

By prop. and prop. , the underlying set of this product space is just the Cartesian product of the underlying sets, hence the set of tuples (xiXi)iI(x_i \in X_i)_{i \in I}. This comes for each iIi \in I with the projection map

By prop. and def. , the topology on this set is the coarsest topology such that the pre-images pri(U)pr_i(U) of open subsets UXiU \subset X_i under these projection maps are open. Now one such pre-image is a Cartesian product of open subsets of the form

The coarsest topology that contains these open subsets ist that generated by these subsets regarded as a sub-basis for the topology (def. ), hence the arbitrary unions of finite intersections of subsets of the above form.

Observe that a binary intersection of these generating open is (for iji \neq j):

and generally for a finite subset JIJ \subset I then

Therefore the open subsets of the product topology are unions of those of this form. Hence the product topology is equivalently that generated by these subsets when regarded as a basis for the topology (def. ).

This is also known as the Tychonoff topology.

Notice the subtlety: Naively we could have considered as open subsets the unions of products iIUi\underset{i \in I}{\prod}U_i of open subsets of the factors, without the constraint that only finitely many of them differ from the corresponding total space. This also defines a topology, called the box topology. For a finite index set II the box topology coincides with the product space (Tychinoff) topology, but for non-finite II it is strictly finer (def. ).

(Cantor space)

Write Disc({0,1})Disc(\{0,1\}) for the the discrete topological space with two points (example ). Write nDisc({0,2})\underset{n \in \mathbb{N}}{\prod} Disc(\{0,2\}) for the product topological space (example ) of a countable set of copies of this discrete space with itself (i.e. the corresponding Cartesian product of sets n{0,1}\underset{n \in \mathbb{N}}{\prod} \{0,1\} equipped with the Tychonoff topology induced from the discrete topology of {0,1}\{0,1\}).

Notice that due to the nature of the Tychonoff topology, this product space is not itself discrete.

Consider the function

which sends an element in the product space, hence a sequence of binary digits, to the value of the power series as shown on the right.

One checks that this is a continuous function (from the product topology to the Euclidean metric topology on the closed interval). Moreover with its image κ(n{0,1})[0,1]\kappa\left( \underset{n \in \mathbb{N}}{\prod} \{0,1\}\right) \subset [0,1] equipped with its subspace topology, then this is a homeomorphism onto its image:

This image is called the Cantor space.

(equalizer of continuous functions)

The equalizer (example ) of two continuous functions f,g:(X,τX)⟶⟶(Y,τY)f, g \colon (X,\tau_X) \stackrel{\longrightarrow}{\longrightarrow} (Y,\tau_Y) is the equalizer of the underlying functions of sets

(hence the largest subset of YY on which both functions coincide) and equipped with the subspace topology from example .

(coequalizer of continuous functions)

The coequalizer of two continuous functions f,g:(X,τX)⟶⟶(Y,τY)f, g \colon (X,\tau_X) \stackrel{\longrightarrow}{\longrightarrow} (Y,\tau_Y) is the coequalizer of the underlying functions of sets

(hence the quotient set by the equivalence relation generated by the relation f(x)g(x)f(x) \sim g(x) for all xXx \in X) and equipped with the quotient topology, example .

(union of two open or two closed subspaces is pushout)

Let XX be a topological space and let A,BXA,B \subset X be subspaces such that

1. A,BXA,B \subset X are both open subsets or are both closed subsets;

2. they constitute a cover: X=ABX = A \cup B

Write iA:AXi_A \colon A \to X and iB:BXi_B \colon B \to X for the corresponding inclusion continuous functions.

Then the commuting square

is a pushout square in Top (example ).

By the universal property of the pushout this means in particular that for YY any topological space then a function of underlying sets

is a continuous function as soon as its two restrictions

are continuous.

More generally if {UiX}iI\{U_i \subset X\}_{i \in I} is a cover of XX by an arbitrary set of open subsets or by a finite set of closed subsets, then a function f:XYf \colon X \to Y is continuous precisely if all its restrictions f|Uif\vert_{U_i} for iIi \in I are continuous.

By prop. the underlying diagram of underlying sets is clearly a pushout in Set. Therefore, by prop. , we need to show that the topology on XX is the final topology (def. ) induced by the set of functions {iA,iB}\{i_A, i_B\}, hence that a subset SXS \subset X is an open subset precisely if the pre-images (restrictions)

are open subsets of AA and BB, respectively.

Now by definition of the subspace topology, if SXS \subset X is open, then the intersections ASAA \cap S \subset A and BSBB \cap S \subset B are open in these subspaces.

Conversely, assume that ASAA \cap S \subset A and BSBB \cap S \subset B are open. We need to show that then SXS \subset X is open.

Consider now first the case that A;BXA;B \subset X are both open open. Then by the nature of the subspace topology, that ASA \cap S is open in AA means that there is an open subset SAXS_A \subset X such that AS=ASAA \cap S = A \cap S_A. Since the intersection of two open subsets is open, this implies that ASAA \cap S_A and hence ASA \cap S is open. Similarly BSB \cap S. Therefore

is the union of two open subsets and therefore open.

Now consider the case that A,BXA,B \subset X are both closed subsets.

Again by the nature of the subspace topology, that ASAA \cap S \subset A and BSBB \cap S \subset B are open means that there exist open subsets SA,SBXS_A, S_B \subset X such that AS=ASAA \cap S = A \cap S_A and BS=BSBB \cap S = B \cap S_B. Since A,BXA,B \subset X are closed by assumption, this means that AS,BSXA \setminus S, B \setminus S \subset X are still closed, hence that X(AS),X(BS)XX \setminus (A \setminus S), X \setminus (B \setminus S) \subset X are open.

Now observe that (by de Morgan duality)

This exhibits SS as the intersection of two open subsets, hence as open.

(attachment spaces)

Consider a cospan diagram (example ) of continuous functions

The colimit under this diagram called the pushout (example )

Consider on the disjoint union set XYX \sqcup Y the equivalence relation generated by the relation

Then prop. implies that the pushout is equivalently the quotient topological space (example ) by this equivalence relation of the disjoint union space (example ) of XX and YY:

If gg is an topological subspace inclusion AXA \subset X, then in topology its pushout along ff is traditionally written as

and called the attachment space (sometimes: attaching space or adjunction space) of AXA \subset X along ff.

(graphics from Aguilar-Gitler-Prieto 02)

(n-sphere as pushout of the equator inclusions into its hemispheres)

As an important special case of example , let

be the canonical inclusion of the standard (n-1)-sphere as the boundary of the standard n-disk (example ).

Then the colimit of topological spaces under the span diagram,

is the topological n-sphere SnS^n (example ):

(graphics from Ueno-Shiga-Morita 95)

(single cell attachment)

For XX any topological space and for nn \in \mathbb{N}, then an nn-cell attachment to XX is the result of gluing an n-disk to XX, along a prescribed image of its bounding (n-1)-sphere (def. ):

Let

be a continuous function, then the space attachment (example )

is the topological space which is the pushout of the boundary inclusion of the nn-sphere along ϕ\phi, hence the universal space that makes the following diagram commute:

(discrete topological spaces from 0-cell attachment to the empty space)

A single cell attachment of a 0-cell, according to example is the same as forming the disjoint union space X*X \sqcup \ast with the point space *\ast:

In particular if we start with the empty topological space X=X = \emptyset itself (example ), then by attaching 0-cells we obtain a discrete topological space. To this then we may attach higher dimensional cells.

(attaching many cells at once)

If we have a set of attaching maps {Sni1⟶ϕiX}iI\{S^{n_i-1} \overset{\phi_i}{\longrightarrow} X\}_{i \in I} (as in def. ), all to the same space XX, we may think of these as one single continuous function out of the disjoint union space of their domain spheres

Then the result of attaching all the corresponding nn-cells to XX is the pushout of the corresponding disjoint union of boundary inclusions:

(relative cell complexes and CW-complexes)

Let XX be a topological space, then A topological relative cell complex of countable height based on XX is a continuous function

and a sequential diagram of topological space of the form

such that

1. each XkXk+1X_k \hookrightarrow X_{k+1} is exhibited as a cell attachment according to def. , hence presented by a pushout diagram of the form

   iISni1⟶(ϕi)iIXk(po)iIDni⟶Xk+1.\array{ \underset{i \in I}{\sqcup} S^{n_i - 1} &\overset{(\phi_i)_{i \in I}}{\longrightarrow}& X_k \\ \downarrow &(po)& \downarrow \\ \underset{i \in I}{\sqcup} D^{n_i} &\longrightarrow& X_{k+1} } \,.

2. Y=kXkY = \underset{k\in \mathbb{N}}{\cup} X_k is the union of all these cell attachments, and f:XYf \colon X \to Y is the canonical inclusion; or stated more abstractly: the map f:XYf \colon X \to Y is the inclusion of the first component of the diagram into its colimiting cocone lim⟶kXk\underset{\longrightarrow}{\lim}_k X_k:

   X=X0⟶X1⟶X2⟶fY=lim⟶X\array{ X = X_0 &\longrightarrow& X_1 &\longrightarrow& X_2 &\longrightarrow& \cdots \\ & {}_{\mathllap{f}}\searrow & \downarrow & \swarrow && \cdots \\ && Y = \underset{\longrightarrow}{\lim} X_\bullet }

If here X=X = \emptyset is the empty space then the result is a map Y\emptyset \hookrightarrow Y, which is equivalently just a space YY built form attaching cells to nothing. This is then called just a topological cell complex of countable hight.

Finally, a topological (relative) cell complex of countable hight is called a CW-complex is the (k+1)(k+1)-st cell attachment XkXk+1X_k \to X_{k+1} is entirely by (k+1)(k+1)-cells, hence exhibited specifically by a pushout of the following form:

Given a CW-complex, then XnX_n is also called its nn-skeleton.

A finite CW-complex is one which admits a presentation in which there are only finitely many attaching maps, and similarly a countable CW-complex is one which admits a presentation with countably many attaching maps.

(connected topological space)

A topological space (X,τ)(X,\tau) (def. ) is called connected if the following equivalent conditions hold:

1. For all pairs of topological spaces (X1,τ1),(X2,τ2)(X_1, \tau_1), (X_2, \tau_2) such that (X,τ)(X, \tau) is homeomorphic (def. ) to their disjoint union space (def. )

   (X,τ)(X1,τ1)(X2,τ2)(X,\tau) \simeq (X_1,\tau_1) \sqcup (X_2,\tau_2)

   then exactly one of the two spaces is the empty space (example ).

2. For all pairs of open subsets U1,U2XU_1, U_2 \subset X if

   U1U2=XAandAU1U2=U_1 \cup U_2 = X \phantom{A}\text{and} \phantom{A} U_1 \cap U_2 = \emptyset

   then exactly one of the two subsets is the empty set;

3. if a subset COXCO \subseteq X is both an open subset and a closed subset (def. ) then CO=XCO = X if and only if COCO is non-empty.

The conditions in def. are indeed equivalent.

First consider the equivalence of the first two statements:

Suppose that in every disjoint union decomposition of (X,τ)(X,\tau) exactly one summand is empty. Now consider two disjoint open subsets U1,U2XU_1, U_2 \subset X whose union is XX and whose intersection is empty. We need to show that exactly one of the two subsets is empty.

Write (U1,τ1)(U_1, \tau_{1}) and (U2,τ2)(U_2, \tau_2) for the corresponding topological subspaces. Then observe that from the definition of subspace topology (example ) and of the disjoint union space (example ) we have a homeomorphism

because by assumption every open subset UXU \subset X is the disjoint union of open subsets of U1U_1 and U2U_2, respectively:

which is the definition of the disjoint union topology.

Hence by assumption exactly one of the two summand spaces is the empty space and hence the underlying set is the empty set.

Conversely, suppose that for every pair of open subsets U1,U2UU_1, U_2 \subset U with U1U2=XU_1 \cup U_2 = X and U1U2=U_1 \cap U_2 = \emptyset then exactly one of the two is empty. Now consider a homeomorphism of the form (X,τ)(X1,τ1)(X2,τ2)(X,\tau) \simeq (X_1, \tau_1) \sqcup (X_2,\tau_2). By the nature of the disjoint union space this means that X1,X2XX_1, X_2 \subset X are disjoint open subsets of XX which cover XX. So by assumption precisely one of the two subsets is the empty set and hence precisely one of the two topological spaces is the empty space.

Now regarding the equivalence to the third statement:

If a subset COXCO \subset X is both closed and open, this means equivalently that it is open and that its complement XCOX \setminus CO is also open, hence equivalently that there are two open subsets CO,X\COXCO, X \backslash CO \subset X whose union is XX and whose intersection is empty. This way the third condition is equivalent to the second, hence also to the first.

(empty space is not connected)

According to def. the empty topological space \emptyset (def. ) is not connected, since \emptyset \simeq \emptyset \sqcup \emptyset, where both instead of exactly one of the summands are empty.

Of course it is immediate to change def. so that it would regard the empty space as connected. This is a matter of convention.

(connected subspaces of the real line are the intervals)

Regard the real line with its Euclidean metric topology (example , ). Then a subspace SS \subset \mathbb{R} (example ) is connected (def. ) precisely if it is an interval, hence precisely if

Suppose on the contrary that we had x<r<yx \lt r \lt y but rSr \notin S. Then by the nature of the subspace topology there would be a decomposition of SS as a disjoint union of disjoint open subsets:

But since x<rx \lt r and r<yr \lt y both these open subsets were non-empty, thus contradicting the assumption that SS is connected. This yields a proof by contradiction.

(continuous images of connected spaces are connected)

Let XX be a connected topological space (def. ), let YY be any topological space, and let

be a continuous function (def. ). This factors via continuous functions through the image

for f(X)f(X) equipped either with he subspace topology relative to YY or the quotient topology relative to XX (example ). In either case:

If XX is a connected topological space (def. ), then so is f(X)f(X).

In particular connectedness is a topological property (def. ).

Let U1,U2f(X)U_1,U_2 \subset f(X) be two open subsets such that U1U2=f(X)U_1 \cup U_2 = f(X) and U1U2=U_1 \cap U_2 = \emptyset. We need to show that precisely one of them is the empty set.

Since pp is a continuous function, also the pre-images p1(U1),p1(U2)Xp^{-1}(U_1), p^{-1}(U_2) \subset X are open subsets and are still disjoint. Since pp is surjective it also follows that p1(U1)p1(U2)=Xp^{-1}(U_1) \cup p^{-1}(U_2) = X. Since XX is connected, it follows that one of these two pre-images is the empty set. But again sicne pp is surjective, this implies that precisely one of U1,U2U_1, U_2 is empty, which means that f(X)f(X) is connected.

(intermediate value theorem)

Regard the real numbers \mathbb{R} with their Euclidean metric topology (example , example ), and consider a closed interval [a,b][a,b] \subset \mathbb{R} (example ) equipped with its subspace topology (example ).

Then a continuous function (def. )

takes every value in between f(a)f(a) and f(b).

By example the interval [a,b][a,b] is connected. By prop. also its image f([a,b])f([a,b]) \subset \mathbb{R} is connected. By example that image is hence itself an interval. This implies the claim.

(product space of connected spaces is connected)

Let {Xi}iI\{X_i\}_{i \in I} be a set of connected topological spaces (def. ). Then also their product topological space iIXi\underset{i \in I}{\prod}X_i (example ) is connected.

Let U1,U2iIXiU_1, U_2 \subset \underset{i \in I}{\prod}X_i be an open cover of the product space by two disjoint open subsets. We need to show that precisely one of the two is empty. Since each XiX_i is connected and hence non-empty, the product space is not empty, and hence it is sufficient to show that at least one of the two is empty.

Assume on the contrary that both U1U_1 and U2U_2 were non-empty.

Observe first that if so, then we could find x1U1x_1 \in U_1 and x2U2x_2 \in U_2 whose coordinates differed only in a finite subset of II. This is since by the nature of the Tychonoff topology πi(U1)=Xi\pi_i(U_1) = X_i and πi(U2)=Xi\pi_i(U_2) = X_i for all but a finite number of iii \in i.

Next observe that we then could even find x1U1x'_1 \in U_1 that differed only in a single coordinate from x2x_2: Because pick one coordinate in which x1x_1 differs from x2x_2 and change it to the corresponding coordinate of x2x_2. Since U1U_1 and U2U_2 are a cover, the resulting point is either in U1U_1 or in U2U_2. If it is in U2U_2, then x1x_1 already differed in only one coordinate from x2x_2 and we may take x1x1x'_1 \coloneqq x_1. If instead the new point is in U1U_1, then rename it to x1x_1 and repeat the argument. By induction this finally yields an x1x'_1 as claimed.

Therefore it is now sufficient to see that it leads to a contradiction to assume that there are points x1U1x_1 \in U_1 and x2U2x_2 \in U_2 that differ in only the i0i_0th coordinate, for some i0Ii_0 \in I, in that this would imply that x1=x2x_1 = x_2.

Observe that the inclusion

which is the identity on the i0i_0th component and is otherwise constant on the iith component of x1x_1 or equivalently of x2x_2 is a continuous function, by the nature of the Tychonoff topology (example ).

Therefore also the restrictions ι1(U1)\iota^{-1}(U_1) and ι1(U2)\iota^{-1}(U_2) are open subsets. Moreover they are still disjoint and cover Xi0X_{i_0}. Hence by the connectedness of Xi0X_{i_0}, precisely one of them is empty. This means that the i0i_0-component of both x1x_1 and x2x_2 must be in the other subset of XiX_i, and hence that x1x_1 and x2x_2 must both be in U1U_1 or both in U2U_2, contrary to the assumption.

(connected components)

For (X,τ)(X,\tau) a topological space, then its connected components are the equivalence classes under the equivalence relation on XX which regards two points as equivalent if they both sit in some subset which, as a topological subspace (example ), is connected (def. ):

Equivalently, the connected components are the maximal elements in the pre-ordered set of connected subspaces, pre-ordred by inclusion.

(connected components of disjoint union spaces)

For {Xi}iI\{X_i\}_{i \in I} an II-indexed set of connected topological spaces, then the set of connected components (def. ) of their disjoint union space iIXi\underset{i \in I}{\sqcup}X_i (example ) is the index set II.

(topological closure of connected subspace is connected)

Let (X,τ)(X,\tau) be a topological space and let SXS \subset X be a subset which, as a subspace, is connected (def. ). Then also the topological closure Cl(S)XCl(S) \subset X is connected

Suppose that Cl(S)=ABCl(S) = A \sqcup B with A,BXA,B \subset X disjoint open subsets. We need to show that one of the two is empty.

But also the intersections AS,BSSA \cap S\,,B \cap S \subset S are disjoint subsets, open as subsets of the subspace SS with S=(AS)(BS)S = (A \cap S) \sqcup (B \cap S). Hence by the connectedness of SS, one of ASA \cap S or BSB \cap S is empty. Assume BSB \cap S is empty, otherwise rename. Hence AS=SA \cap S = S, or equivalently: SAS \subset A. By disjointness of AA and BB this means that SCl(S)BS \subset Cl(S) \setminus B. But since BB is open, Cl(S)BCl(S) \setminus B is still closed, so that

This means that B=B = \emptyset, and hence that Cl(S)Cl(S) is connected.

(connected components are closed)

Let (X,τ)(X,\tau) be a topological space. Then its connected components (def. ) are closed subsets.

By definition, the connected components are maximal elements in the set of connected subspaces pre-ordered by inclusion. Assume a connected component UU were not closed. By prop. its closure Cl(U)Cl(U) is still closed, and would be strictly larger, contradicting the maximality of UU. This yields a proof by contradiction.

Prop. implies that when a space has a finite set of connected components, then they are not just closed but also open, hence clopen subsets (because then each is the complement of a finite union of closed subspaces). This in turn means that the space is the disjoint union space of its connected components.

For a non-finite set of connected components this remains true if the space is locally connected see def. , lemma below.

(totally disconnected topological space)

A topological space is called totally disconnected if all its connected components (def. ) are singletons, hence point spaces (example ).

(discrete topological spaces are totally disconnected)

Every discrete topological space (example ) is a totally disconnected topological space (def. ).

(the rational numbers are totally disconnected but non-discrete topological space)

The rational numbers \mathbb{Q} \subset \mathbb{R} equipped with their subspace topology (def. ) inherited from the Euclidean metric topology (example , example ) on the real numbers, form a totally disconnected space (def. ), but not a discrete topological space (example ).

It is clear that the subspace topology is not discrete, since the singletons {q}\{q\} \subset \mathbb{Q} are not open subsets (because their pre-image in \mathbb{R} are still singletons, and the points in a metric space are closed, by example and prop. ).

What we need to see is that \mathbb{Q} \subset \mathbb{R} is nevertheless totally disconnected:

By construction, a base for the topology (def. ) is given by the open subsets which are restrictions of open intervals of real numbers to the rational numbers

for a<ba \lt b \in \mathbb{R}.

Now for any such a<ba \lt b there exists an irrational number r\r \in \mathbb{R}\backslash \mathbb{Q} with a<r<ba \lt r \lt b. This being irrational implies that (a,r)(a,r)_{\mathbb{Q}} \subset \mathbb{Q} and (r,b)(r,b)_{\mathbb{Q}} \subset \mathbb{Q} are disjoint subsets. Therefore every basic open subset is the disjoint union of (at least) two open subsets:

Hence no non-empty open subspace of the rational numbers is connected.

(Cantor space is totally disconnected but non-discrete)

The Cantor space nDisc({0,1})\underset{n \in \mathbb{N}}{\prod} Disc(\{0,1\}) (example ) is a totally disconnected topological space (def. ) but is not a discrete topological space.

The base opens (def. ) of the product topological space nDisc({1,2})\underset{n \in \mathbb{N}}{\prod} Disc(\{1,2\}) (example ) are of the form

for II \subset \mathbb{N} a finite subset.

First of all this is not the discrete topology, for that also contains infinite products of proper subsets of {1,2}\{1,2\} as open subsets, hence is strictly finer.

On the other hand, since II \subset \mathbb{N} is finite, I\mathbb{N}\setminus I is non-empty and hence there exists some k0k_0 \in \mathbb{N} such that the corresponding factor in the above product is the full set {1,2}\{1,2\}. But then the above subset is the disjoint union of the open subsets

In particular if xyx \neq y are two distinct points in the original open subset, them being distinct means that there is a smallest k0k_0 \in \mathbb{N} such that they have different coordinates in {1,2}\{1,2\} in that position. By the above this implies that they belong to different connected components.

(locally connected topological spaces)

A topological space (X,τ)(X,\tau) is called locally connected if the following equivalent conditions hold:

1. For every point xx and every neighbourhood Ux{x}U_x \supset \{x\} there is a connected (def. ) open neighbourhood CnxUxCn_x \subset U_x.

2. Every connected component of every open subspace of XX is open.

3. Every open subspace (example ) is the disjoint union space (def. ) of its connected components (def. ).

The conditions in def. are indeed all equivalent.

\,

1) \Rightarrow 2)

Assume every neighbourhood of XX contains a connected neighbourhood and let UXU \subset X be an open subset with U0UU_0 \subset U a connected component. We need to show that U0U_0 is open.

Consider any point xU0x \in U_0. Since then also xUx \in U, there is a connected open neighbourhood Ux,0U_{x,0} of xx in XX. Observe that this must be contained in U0U_0, for if it were not then U0Ux,0U_0 \cup U_{x,0} were a larger open connected open neighbourhood, contradicting the maximality of the connected component U0U_0.

Hence U0=xU0Ux,0U_0 = \underset{x \in U_0}{\cup} U_{x,0} is a union of open subsets, and hence itself open.

2) \Rightarrow 3)

Now assume that every connected component of every open subset UU is open. Since the connected components generally constitute a cover of XX by disjoint subsets, this means that now they form an open cover by disjoint subsets. But by forming intersections with the cover this implies that every open subset of UU is the disjoint union of open subsets of the connected components (and of course every union of open subsets of the connected components is still open in UU), which is the definition of the topology on the disjoint union space of the connected components.

3) \Rightarrow 1)

Finally assume that every open subspace is the disjoint union of its connected components. Let xx be a point and Ux{x}U_x \supset \{x\} a neighbourhood. We need to show that UxU_x contains a connected neighbourhood of xx.

But, by definition, UxU_x contains an open neighbourhood of xx and by assumption this decomposes as the disjoint union of its connected components. One of these contains xx. Since in a disjoint union space all summands are open, this is the required connected open neighbourhod.

(Euclidean space is locally connected)

For nn \in \mathbb{N} the Euclidean space n\mathbb{R}^n (example ) (with its metric topology, example ) is locally connected (def. ).

By nature of the Euclidean metric topology, every neighbourhood UxU_x of a point xx contains an open ball containing xx (def. ). Moreover, every open ball clearly contains an open cube, hence a product space i{1,,n}(xiϵ,xi+ϵ)\underset{i \in \{1, \cdots, n\}}{\prod} (x_i-\epsilon, x_i + \epsilon) of open intervals which is still a neighbourhood of xx (example ).

Now intervals are connected by example and product spaces of connected spaces are connected by example . This shows that ever open neighbourhood contains a connected neighbourhood, which is the characterization of local connectedness in the first item of def. .

(open subspace of locally connected space is locally connected)

Every open subspace (example ) of a locally connected topological space (example ) is itself locally connected

This is immediate from the first item of def. .

(path)

Let XX be a topological space. Then a path or continuous curve in XX is a continuous function

from the closed interval (example ) equipped with its Euclidean metric topology (example , example ).

We say that this path connects its endpoints γ(0),γ(1)X\gamma(0), \gamma(1) \in X.

(being connected by a path is equivalence relation)

Let (X,τ)(X,\tau) be a topological space. Being connected by a path (def. ) is an equivalence relation pcon\sim_{pcon} on the underlying set of points XX.

We need to show that the relation is reflexive, symmetric and transitive.

For xXx \in X a point, then the constant function with value xx

is continuous (example ). Therefore xpconxx \sim_{pcon} x for all points (reflexivity).

For x,yXx,y \in X two points and

a path connecting them, then the reverse path

is continuous (the function [0,1]1()[0,1][0,1] \overset{1-(-)}{\to} [0,1] is continuous because polynomials are continuous ). Hence with xpconyx \sim_{pcon} y also ypconxy \sim_{pcon} x (symmetry).

For x,y,zXx,y,z \in X three points and for

two paths with γ1(0)=x\gamma_1(0) = x, γ1(1)=γ2(0)=y\gamma_1(1) = \gamma_2(0) = y and γ2(1)=z\gamma_2(1) = z

consider the function

This is a continuous function by example , hence this constitutes a path connecting xx with zz (the concatenated path). Therefore xpconyx \sim_{pcon} y and ypconzy \sim_{pcon} z implies xpconzx \sim_{pcon} z (transitivity).

(path-connected components)

Let XX be a topological space. The equivalence classes of the equivalence relation connected by a path (def. , lemma ) are called the path-connected components of XX. The set of the path-connected components is usually denoted

(The notation reflects the fact that this is the degree-zero case of a more general concept of homotopy groups πn\pi_n for all nn \in \mathbb{N}. We discuss the fundamental group π1\pi_1 in part 2. The higher homotopy groups are discussed in Introduction to Homotopy Theory).

If there is a single path-connected component (π0(*)*\pi_0(\ast) \simeq \ast), then XX is called a path-connected topological space.

(Euclidean space is path-connected)

For nn \in \mathbb{N} then Euclidean space n\mathbb{R}^n is a path-connected topological space (def. ).

Because for x,yn\vec x, \vec y \in \mathbb{R}^n, consider the function

This clearly has the property that γ(0)=x\gamma(0) = \vec x and γ(1)=y\gamma(1) = \vec y. Moreover, it is a polynomial function and polynomials are continuous functions (example ).

(continuous image of path-connected space is path-connected)

Let XX be a path-connected topological space and let

be a continuous function. Then also the image f(X)f(X) of XX

with either of its two possible topologies (example ) is path-connected.

In particular path-connectedness is a topological property (def. ).

Let x,yXx,y \in X be two points. Since p:Xf(X)p \;\colon\; X \to f(X) is surjective, there are pre-images p1(x),p1(y)Xp^{-1}(x), p^{-1}(y) \in X. Since XX is path-connected, there is a continuous function

with γ(0)=p1(x)\gamma(0) = p^{-1}(x) and γ(1)=p1(y)\gamma(1) = p^{-1}(y). Since the composition of continuous functions is continuous, it follows that pγ:[0,1]f(X)p \circ \gamma \;\colon\; [0,1] \to f(X) is a path connecting xx with yy.

(path space)

Let XX be a topological space. Since the interval [0,1][0,1] is a locally compact topological space (example ) there is the mapping space

hence the set of paths in XX (def. ) equipped with the compact-open topology (def. ).

This is often called the path space of XX.

By functoriality of the mapping space (remark ) the two endpoint inclusions

induce two continuous functions of the form

The coequalizer (example ) of these two functions is the set π0(X)\pi_0(X) of path-connected components (def. ) topologized with the corresponding final topology (def. ).

(path-connected spaces are connected)

A path connected topological space XX (def. ) is connected (def. ).

Assume it were not, then it would be covered by two disjoint non-empty open subsets U1,U2XU_1, U_2 \subset X. But by path connectedness there were a continuous path γ:[0,1]X\gamma \colon [0,1] \to X from a point in one of the open subsets to a point in the other. The continuity would imply that γ1(U1),γ1(U2)[0,1]\gamma^{-1}(U_1), \gamma^{-1}(U_2) \subset [0,1] were a disjoint open cover of the interval. This would be in contradiction to the fact that intervals are connected. Hence we have a proof by contradiction.

(locally path-connected topological space)

A topological space XX is called locally path-connected if for every point xXx \in X and every neighbourhood Ux{x}U_x \supset \{x\} there exists a neighbourhood CxUxC_x \subset U_x which, as a subspace, is path-connected (def. ).

(Euclidean space is locally path-connected)

For nn \in \mathbb{N} then Euclidean space n\mathbb{R}^n (with its metric topology) is locally path-connected, since each open ball is a path-connected topological space (example ).

(open subspace of locally path-connected space is locally path-connected)

Every open subspace of a locally path-connected topological space is itself locally path-connected.

Let XX be a locally path-connected topological space (def. ). Then each of its path-connected components is an open set and a closed set.

To see that every path connected component PxP_x is open, it is sufficient to show that every point yPxy \in P_x has an neighbourhood UyU_y which is still contained in PxP_x. But by local path connectedness, yy has a neighbourhood VyV_y which is path connected. It follows by concatenation of paths (as in the proof of lemma ) that VyPxV_y \subset P_x.

Now each path-connected component PxP_x is the complement of the union of all the other path-connected components. Since these are all open, their union is open, and hence the complement PxP_x is closed.

(in a locally path-connected space connected components coincide with path-connected components)

Let XX be a locally path-connected topological space (def. ). Then the connected components of XX according to def. agree with the path-connected components according to def. .

In particular, locally path connected spaces are locally connected topological spaces (def. ).

A path connected component is always connected by lemma , and in a locally path-connected space it is also open, by prop. . This implies that the path-connected components are maximal connected subspaces, and hence must be the connected components.

Conversely let UU be a connected component. It is now sufficient to see that this is path-connected. Suppose it were not, then it would be covered by more than one disjoint non-empty path-connected components. But by prop. these would all be open. This would be in contradiction with the assumption that UU is connected. Hence we have a proof by contradiction.

(embedding of topological spaces)

Let (X,τX)(X,\tau_X) and (Y,τY)(Y,\tau_Y) be topological spaces. A continuous function f:X⟶Yf \;\colon\; X \longrightarrow Y is called an embedding of topological spaces if in its image factorization (example )

with the image f(X)Yf(X) \hookrightarrow Y equipped with the subspace topology (example ), we have that Xf(X)X \to f(X) is a homeomorphism (def. ).

(open/closed continuous injections are embeddings)

A continuous function f:(X,τX)(Y,τY)f \colon (X, \tau_X) \to (Y,\tau_Y) which is

1. an injective function

2. an open map or a closed map (def. )

is an embedding of topological spaces (def. ).

This is called a closed embedding if the image f(X)Yf(X) \subset Y is a closed subset.

If ff is injective, then the map onto its image Xf(X)YX \to f(X) \subset Y is a bijection. Moreover, it is still continuous with respect to the subspace topology on f(X)f(X) (example ). Now a bijective continuous function is a homeomorphism precisely if it is an open map or a closed map, by prop. . But the image projection of ff has this property, respectively, if ff does, by prop .

(injective proper maps to locally compact spaces are equivalently the closed embeddings)

Let

1. XX be a topological space,

2. YY a locally compact topological space (def. ),

3. f:XYf \colon X \to Y a continuous function.

Then the following are equivalent:

1. ff is an injective proper map (prop. ),

2. ff is a closed embedding of topological spaces (def ).

In one direction, if ff is an injective proper map, then since proper maps to locally compact spaces are closed (prop. ), we have that ff is also closed map. The claim then follows since closed injections are embeddings (prop. ), and since the image of a closed map is closed, by definition.

Conversely, if ff is a closed embedding, we only need to show that the embedding map is proper. So for CYC \subset Y a compact subspace, we need to show that the pre-image f1(C)Xf^{-1}(C) \subset X is also compact. But since ff is an injection (being an embedding), that pre-image is equivalently just the intersection f1(C)Cf(X)Yf^{-1}(C) \simeq C \cap f(X) \subset Y, regarded as a subspace of YY.

To see that this is compact, let {ViX}iI\{V_i \subset X\}_{i \in I} be an open cover of the subspace Cf(X)C \cap f(X), hence, by the nature of the subspace topology, let {UiY}iI\{U_i \subset Y\}_{i \in I} be a set of open subsets of YY, which cover Cf(X)YC \cap f(X) \subset Y and with ViV_i the restriction of UiU_i to Cf(X)C \cap f(X). Now since f(X)Yf(X) \subset Y is closed by assumption, it follows that the complement Yf(X)Y \setminus f(X) is open and hence that

is an open cover of CYC \subset Y. By compactness of CC this has a finite subcover. Since restricting that finite subcover back to Cf(X)C \cap f(X) makes the potential element Yf(X)Y \setminus f(X) disappear, this restriction is a finite subcover of {ViCf(X)}\{V_i \subset C \cap f(X)\}. This shows that Cf(X)C \cap f(X) is compact.

(open cover)

An open cover of a topological space (X,τ)(X,\tau) (def. ) is a set {UiX}iI\{U_i \subset X\}_{i \in I} of open subsets UiU_i of XX, indexed by some set II, such that their union is all of XX: iIUi=X\underset{i \in I}{\cup} U_i \;=\; X.

A subcover of a cover is a subset JIJ \subset I such that {UiX}iJI\{U_i \subset X\}_{i \in J \subset I} is still a cover.

(compact topological space)

A topological space XX (def. ) is called a compact topological space if every open cover {UiX}iI\{U_i \subset X\}_{i \in I} (def. ) has a finite subcover in that there is a finite subset JIJ \subset I such that {UiX}iJ\{U_i \subset X\}_{i \in J} is still a cover of XX in that also iJUi=X\underset{i \in J}{\cup} U_i = X.

(varying terminology regarding compact)

Beware the following terminology issue which persists in the literature:

Some authors use compact to mean Hausdorff and compact. To disambiguate this, some authors (mostly in algebraic geometry, but also for instance Waldhausen) say quasi-compact for what we call compact in def. .

(compactness in terms of closed subsets)

Let (X,τ)(X,\tau) be a topological space. Then the following are equivalent:

1. (X,τ)(X,\tau) is compact in the sense of def. .

2. Let {CiX}iI\{C_i \subset X\}_{i \in I} be a set of closed subsets (def. ) such that their intersection is empty iICi=\underset{i \in I}{\cap} C_i = \emptyset, then there is a finite subset JIJ \subset I such that the corresponding finite intersection is still empty iJiCi=\underset{i \in J \subset i}{\cap} C_i = \emptyset.

3. Let {CiX}iI\{C_i \subset X\}_{i \in I} be a set of closed subsets (def. ) such that it enjoys the finite intersection property, meaning that for every finite subset JIJ \subset I then the corresponding finite intersection is non-empty iJICi\underset{i \in J \subset I}{\cap} C_i \neq \emptyset. Then also the total intersection is non-empty, iICi\underset{i \in I}{\cap} C_i \neq \emptyset.

The equivalence between the first and the second statement is immediate from the definitions after expressing open subsets as complements of closed subsets Ui=XCiU_i = X \setminus C_i and applying de Morgan's law (prop. ).

We discuss the equivalence between the first and the third statement:

In one direction, assume that (X,τ)(X,\tau) is compact in the sense of def. , and that {CiX}iI\{C_i \subset X\}_{i \in I} satisfies the finite intersection property. We need to show that then iICi\underset{i \in I}{\cap} C_i \neq \emptyset.

Assume that this were not the case, hence assume that iICi=\underset{i \in I}{\cap} C_i = \emptyset. This would imply that the open complements were an open cover of XX (def. )

because (using de Morgan's law, prop. )

But then by compactness of (X,τ)(X,\tau) there were a finite subset JIJ \subset I such that {UiX}iJI\{ U_i \subset X\}_{i \in J \subset I} were still an open cover, hence that iJIUi=X\underset{i \in J \subset I}{\cup} U_i = X. Translating this back through the de Morgan's law again this would mean that

This would be in contradiction with the finite intersection property of {CiX}iI\{C_i \subset X\}_{i \in I}, and hence we have proof by contradiction.

Conversely, assume that every set of closed subsets in XX with the finite intersection property has non-empty total intersection. We need to show that the every open cover {UiX}iI\{U_i \subset X\}_{i \in I} of XX has a finite subcover.

Write CiXUiC_i \coloneqq X \setminus U_i for the closed complements of these open subsets.

Assume on the contrary that there were no finite subset JIJ \subset I such that iJIUi=X\underset{i \in J \subset I}{\cup} U_i = X, hence no finite subset such that iJICi=\underset{i \in J \subset I}{\cap} C_i = \emptyset. This would mean that {CiX}iI\{C_i \subset X\}_{i \in I } satisfied the finite intersection property.

But by assumption this would imply that iICi\underset{i \in I}{\cap} C_i \neq \emptyset, which, again by de Morgan, would mean that iIUiX\underset{i \in I}{\cup} U_i \neq X. But this contradicts the assumption that the {UiX}iI\{U_i \subset X\}_{i \in I} are a cover. Hence we have a proof by contradiction.

(finite discrete spaces are compact)

A discrete topological space (def. ) is compact (def. ) precisely if its underlying set is a finite set.

(closed intervals are compact topological spaces)

For any a<ba \lt b \in \mathbb{R} the closed interval (example )

regarded with its subspace topology of Euclidean space (example ) with its metric topology (example ) is a compact topological space (def. ).

Since all the closed intervals are homeomorphic (by example ) it is sufficient to show the statement for [0,1][0,1]. Hence let {Ui[0,1]}iI\{U_i \subset [0,1]\}_{i \in I} be an open cover (def. ). We need to show that it has an open subcover.

Say that an element x[0,1]x \in [0,1] is admissible if the closed sub-interval [0,x][0,x] is covered by finitely many of the UiU_i. In this terminology, what we need to show is that 11 is admissible.

Observe from the definition that

1. 0 is admissible,

2. if y<x[0,1]y \lt x \in [0,1] and xx is admissible, then also yy is admissible.

This means that the set of admissible xx forms either

1. an open interval [0,g)[0,g)

2. or a closed interval [0,g][0,g],

for some g[0,1]g \in [0,1]. We need to show that the latter is true, and for g=1g = 1. We do so by observing that the alternatives lead to contradictions:

1. Assume that the set of admissible values were an open interval [0,g)[0,g). Pick an i0Ii_0 \in I such that gUi0g \in U_{i_0} (this exists because of the covering property). Since such Ui0U_{i_0} is an open neighbourhood of gg, there is a positive real number ϵ\epsilon such that the open ball Bg(ϵ)Ui0B^\circ_g(\epsilon) \subset U_{i_0} is still contained in the patch. It follows that there is an element xBg(ϵ)[0,g)Ui0[0,g)x \in B^\circ_g(\epsilon) \cap [0,g) \subset U_{i_0} \cap [0,g) and such that there is a finite subset JIJ \subset I with {Ui[0,1]}iJI\{U_i \subset [0,1]\}_{i \in J \subset I} a finite open cover of [0,x)[0,x). It follows that {Ui[0,1]}iJI{Ui0}\{U_i \subset [0,1]\}_{i \in J \subset I} \sqcup \{U_{i_0}\} were a finite open cover of [0,g][0,g], hence that gg itself were still admissible, in contradiction to the assumption.

2. Assume that the set of admissible values were a closed interval [0,g][0,g] for g<1g \lt 1. By assumption there would then be a finite set JIJ \subset I such that {Ui[0,1]}iJI\{U_i \subset [0,1]\}_{i \in J \subset I} were a finite cover of [0,g][0,g]. Hence there would be an index igJi_g \in J such that gUigg \in U_{i_g}. But then by the nature of open subsets in the Euclidean space \mathbb{R}, this UigU_{i_g} would also contain an open ball Bg(ϵ)=(gϵ,g+ϵ)B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon). This would mean that the set of admissible values includes the open interval [0,g+ϵ)[0,g+ \epsilon), contradicting the assumption.

This gives a proof by contradiction.

(Euclidean space is non-compact)

For all nn \in \mathbb{N}, n>0n \gt 0, the Euclidean space n\mathbb{R}^n (example ), regarded with its metric topology (example ), is not a compact topological space (def. ).

Pick any ϵ(0,1/2)\epsilon \in (0,1/2). Consider the open cover of n\mathbb{R}^n given by

This is not a finite cover, and removing any one of its patches UnU_n, it ceases to be a cover, since the points of the form (n+ϵ,x2,x3,,xn)(n + \epsilon, x_2, x_3, \cdots, x_n) are contained only in UnU_n and in no other patch.

(unions and intersection9] of [[compact spaces?)

Let (X,τ)(X,\tau) be a topological space and let

be a set of compact subspaces.

1. If II is a finite set, then the union iIKiX\underset{i \in I}{\cup} K_i \subset X is itself a compact subspace;

2. If all KiXK_i \subset X are also closed subsets then their intersection iIKiX\underset{i \in I}{\cap} K_i \subset X is itself a compact subspace.

(complement of compact by open subspaces is compact)

Let XX be a topological space. Let

1. KXK\subset X be a compact subspace;

2. UXU \subset X be an open subset.

Then the complement

is itself a compact subspace.

(continuous surjections out of compact spaces have compact codomain)

Let f:(X,τX)⟶(Y,τY)f \colon (X,\tau_X) \longrightarrow (Y,\tau_Y) be a continuous function between topological spaces such that

1. (X,τX)(X,\tau_X) is a compact topological space (def. );

2. f:XYf \colon X \to Y is a surjective function.

Then also (Y,τY)(Y,\tau_Y) is compact.

Let {UiY}iI\{U_i \subset Y\}_{i \in I} be an open cover of YY (def. ). We need show that this has a finite sub-cover.

By the continuity of ff the pre-images f1(Ui)f^{-1}(U_i) form an open cover {f1(Ui)X}iI\{f^{-1}(U_i) \subset X\}_{i \in I} of XX. Hence by compactness of XX, there exists a finite subset JIJ \subset I such that {f1(Ui)X}iJI\{f^{-1}(U_i) \subset X\}_{i \in J \subset I} is still an open cover of XX. Finally, by surjectivity of ff it it follows that

where we used that images of unions are unions of images.

This means that also {UiY}iJI\{U_i \subset Y\}_{i \in J \subset I} is still an open cover of YY, and in particular a finite subcover of the original cover.

(continuous images of compact spaces are compact)

If f:X⟶Yf \colon X \longrightarrow Y is a continuous function out of a compact topological space XX (def. ) which is not necessarily surjective, then we may consider its image factorization

as in example . Now by construction Xf(X)X \to f(X) is surjective, and so lemma implies that f(X)f(X) is compact.

(proper maps)

A continuous function f:(X,τX)(Y,τY)f \colon (X, \tau_X) \to (Y, \tau_Y) is called proper if for CYC \in Y a compact topological subspace of YY, then also its pre-image f1(C)f^{-1}(C) is compact in XX.

(extreme value theorem)

Let CC be a compact topological space (def. ), and let

be a continuous function to the real numbers equipped with their Euclidean metric topology.

Then ff attains its maximum and its minimum in that there exist xmin,xmaxCx_{min}, x_{max} \in C such that

Since continuous images of compact spaces are compact (prop. ) the image f([a,b])f([a,b]) \subset \mathbb{R} is a compact subspace.

Suppose this image did not contain its maximum. Then {(,x)}xf([a,b])\{(-\infty,x)\}_{x \in f([a,b])} were an open cover of the image, and hence, by its compactness, there would be a finite subcover, hence a finite set (x1<x2<<xn)(x_1 \lt x_2 \lt \cdots \lt x_n) of points xif([a,b])x_i \in f([a,b]), such that the union of the (,xi)(-\infty,x_i) and hence the single set (,xn)(-\infty, x_n) alone would cover the image. This were in contradiction to the assumption that xnf([a,b])x_n \in f([a,b]) and hence we have a proof by contradiction.

Similarly for the minimum.

(traditional extreme value theorem)

Let

be a continuous function from a bounded closed interval (a<ba \lt b \in \mathbb{R}) (def. ) regarded as a topological subspace (example ) of real numbers to the real numbers, with the latter regarded with their Euclidean metric topology (example , example ).

Then ff attains its maximum and minimum: there exists xmax,xmin[a,b]x_{max}, x_{min} \in [a,b] such that for all x[a,b]x \in [a,b] we have

Since continuous images of compact spaces are compact (prop. ) the image f([a,b])f([a,b]) \subset \mathbb{R} is a compact subspace (def. , example ). By the Heine-Borel theorem (prop. ) this is a bounded closed subset (def. , def. ). By the nature of the Euclidean metric topology, the image is hence a union of closed intervals. Finally by continuity of ff it needs to be a single closed interval, hence (being bounded) of the form

(closed-projection characterization of compactness)

Let (X,τX)(X,\tau_X) be a topological space. The following are equivalent:

1. (X,τX)(X,\tau_X) is a compact topological space according to def. ;

2. For every topological space (Y,τY)(Y,\tau_Y) then the projection map out of the product topological space (example , example )

   πY:(Y,τY)×(X,τX)⟶(Y,τY)\pi_Y \;\colon\; (Y, \tau_Y) \times (X, \tau_X) \longrightarrow (Y, \tau_Y)

   is a closed map.

(due to Todd Trimble)

In one direction, assume that (X,τX)(X,\tau_X) is compact and let CY×XC \subset Y \times X be a closed subset. We need to show that πY(C)Y\pi_Y(C) \subset Y is closed.

By lemma this is equivalent to showing that every point yYπY(C)y \in Y \setminus \pi_Y(C) in the complement of πY(C)\pi_Y(C) has an open neighbourhood Vy{y}V_y \supset \{y\} which does not intersect πY(C)\pi_Y(C):

This is clearly equivalent to

and this is what we will show.

To this end, consider the set

Observe that this is an open cover of XX: For every xXx \in X then (y,x)C(y,x) \notin C by assumption of YY, and by closure of CC this means that there exists an open neighbourhood of (y,x)(y,x) in Y×XY \times X not intersecting CC, and by nature of the product topology this contains an open neighbourhood of the form V×UV \times U.

Hence by compactness of XX, there exists a finite subcover {UjX}jJ\{ U_j \subset X \}_{j \in J} of XX and a corresponding set {VjY}jJ\{V_j \subset Y\}_{j \in J} with Vj×UjC=V_j \times U_j \cap C = \emptyset.

The resulting open neighbourhood

of yy has the required property:

Now for the converse:

Assume that πY:Y×XX\pi_Y \colon Y \times X \to X is a closed map for all YY. We need to show that XX is compact. By prop. this means equivalently that for every set {CiX}iI\{C_i \subset X\}_{i \in I} of closed subsets and satisfying the finite intersection property, we need to show that iICi\underset{i \in I}{\cap} C_i \neq \emptyset.

So consider such a set {CiX}iI\{C_i \subset X\}_{i \in I} of closed subsets satisfying the finite intersection property. Construct a new topological space (Y,τY)(Y, \tau_Y) by setting

1. YX{}Y \coloneqq X \sqcup \{\infty\};

2. βYP(X){(Ci{})Y}iI\beta_Y \coloneqq P(X) \sqcup \left\{ (C_i \cup \{\infty\}) \subset Y \right\}_{i \in I} a sub-base for τY\tau_Y (def. ).

Then consider the topological closure Cl(Δ)Cl(\Delta) of the diagonal Δ\Delta in Y×XY \times X

We claim that there exists xXx \in X such that

This is because

by the assumption that πY\pi_Y is a closed map, and

by construction. So if \infty were not in πY(Cl(Δ))\pi_Y(Cl(\Delta)), then, by lemma , it would have an open neighbourhood not intersecting XX. But by definition of τY\tau_Y, the open neighbourhoods of \infty are the unions of finite intersections of Ci{}C_i \cup \{\infty\}, and by the assumed finite intersection property all their finite intersections do still intersect XX.

Since thus (,x)Cl(Δ)(\infty,x) \in Cl(\Delta), lemma gives again that all of its open neighbourhoods intersect the diagonal. By the nature of the product topology (example ) this means that for all iIi \in I and all open neighbourhoods Ux{x}U_x \supset \{x\} we have that

By definition of Δ\Delta this means equivalently that

for all open neighbourhoods Ux{x}U_x \supset \{x\}.

But by closure of CiC_i and using lemma , this means that

for all ii, hence that

as required.

(tube lemma)

Let

1. (X,τX)(X,\tau_X) be a topological space,

2. (Y,τY)(Y,\tau_Y) a compact topological space (def. ),

3. xXx \in X a point,

4. WopenX×YW \underset{\text{open}}{\subset} X \times Y an open subset in the product topology (example , example ),

such that the YY-fiber over xx is contained in WW:

Then there exists an open neighborhood UxU_x of xx such that the tube Ux×YU_x \times Y around the fiber {x}×Y\{x \} \times Y is still contained:

Let

be the complement of WW. Since this is closed, by prop. also its projection pX(C)Xp_X(C) \subset X is closed.

Now

and hence by the closure of pX(C)p_X(C) there is (by lemma ) an open neighbourhood Ux{x}U_x \supset \{x\} with

This means equivalently that Ux×YC=U_x \times Y \cap C = \emptyset, hence that Ux×YWU_x \times Y \subset W.

(Tychonoff theorem the product space of compact spaces is compact)

Let {(Xi,τi)}iI\{(X_i, \tau_i)\}_{i \in I} be a set of compact topological spaces (def. ). Then also their product space iI(Xi,τi)\underset{i \in I}{\prod}(X_i, \tau_i) (example ) is compact.

By prop. it is sufficient to show that for every topological space (Y,τY)(Y,\tau_Y) then the projection

is a closed map. We proceed by induction. For n=0n = 0 the statement is obvious. Suppose it has been proven for some nn \in \mathbb{N}. Then the projection for n+1n+1 factors is the composite of two consecutive projections

By prop. , the first map here is closed since (Xn+1,τn+1)(X_{n+1}, \tau_{n+1}) is compact by the assumption of the proposition, and similarly the second is closed by induction assumtion. Hence the composite is a closed map.

(converging sequence in a topological space)

Let (X,τ)(X,\tau) be a topological space (def. ) and let (xn)n(x_n)_{n \in \mathbb{N}} be a sequence of points (xn)(x_n) in XX (def. ). We say that this sequence converges in (X,τ)(X,\tau) to a point xXx_\infty \in X, denoted

if for each open neighbourhood UxU_{x_\infty} of xx_\infty there exists a kk \in \mathbb{N} such that for all nkn \geq k then xnUxx_n \in U_{x_\infty}:

(sequentially compact topological space)

Let (X,τ)(X,\tau) be a topological space (def. ). It is called sequentially compact if for every sequence of points (xn)(x_n) in XX (def. ) there exists a sub-sequence (xnk)k(x_{n_k})_{k \in \mathbb{N}} which converges acording to def. .

(sequentially compact metric spaces are equivalently compact metric spaces)

If (X,d)(X,d) is a metric space (def. ), regarded as a topological space via its metric topology (example ), then the following are equivalent:

1. (X,d)(X,d) is a compact topological space (def. ).

2. (X,d)(X,d) is a sequentially compact metric space (def. ) hence a sequentially compact topological space (def. ).

of prop. and prop.

Assume first that (X,d)(X,d) is a compact topological space. Let (xk)k(x_k)_{k \in \mathbb{N}} be a sequence in XX. We need to show that it has a sub-sequence which converges.

Consider the topological closures of the sub-sequences that omit the first nn elements of the sequence

and write

for their open complements.

Assume now that the intersection of all the FnF_n were empty

or equivalently that the union of all the UnU_n were all of XX

hence that {UnX}n\{U_n \subset X\}_{n \in \mathbb{N}} were an open cover. By the assumption that XX is compact, this would imply that there were a finite subset {i1<i2<<ik}\{i_1 \lt i_2 \lt \cdots \lt i_k\} \subset \mathbb{N} with

This in turn would mean that Fik=F_{i_k} = \empty, which contradicts the construction of FikF_{i_k}. Hence we have a proof by contradiction that assumption (*)(\ast) is wrong, and hence that there must exist an element

By definition of topological closure this means that for all nn the open ball Bx(1/(n+1))B^\circ_x(1/(n+1)) around xx of radius 1/(n+1)1/(n+1) must intersect the nnth of the above subsequences:

If we choose one point (xn)(x'_n) in the nnth such intersection for all nn this defines a sub-sequence, which converges to xx.

In summary this proves that compact implies sequentially compact for metric spaces.

For the converse, assume now that (X,d)(X,d) is sequentially compact. Let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover of XX. We need to show that there exists a finite sub-cover.

Now by the Lebesgue number lemma, there exists a positive real number δ>0\delta \gt 0 such that for each xXx \in X there is ixIi_x \in I such that Bx(δ)UixB^\circ_x(\delta) \subset U_{i_x}. Moreover, since sequentially compact metric spaces are totally bounded, there exists then a finite set SXS \subset X such that

Therefore {UisX}sS\{U_{i_s} \to X\}_{s \in S} is a finite sub-cover as required.

(neither compactness nor sequential compactness implies the other)

Beware, in contrast to prop. , general topological spaces being sequentially compact neither implies nor is implied by being compact.

1. The product topological space (example ) r[0,1)Disc({0,1})\underset{r \in [0,1)}{\prod} Disc(\{0,1\}) of copies of the discrete topological space (example ) indexed by the elements of the half-open interval is compact by the Tychonoff theorem (prop. ), but the sequence xnx_n with

   πr(xn)=nth digit of the binary expansion ofr\pi_r(x_n) = n\text{th digit of the binary expansion of}\, r

   has no convergent subsequence.

2. conversely, there are spaces that are sequentially compact, but not compact, see for instance Vermeeren 10, prop. 18.

(nets fix the shortcomings of sequences)

That compactness of topological spaces is not detected by convergence of sequences (remark ) may be regarded as a shortcoming of the concept of sequence. While a sequence is indexed over the natural numbers, the concept of convergence of sequnces only invokes that the natural numbers form a directed set. Hence the concept of convergence immediately generalizes to sets of points in a space which are indexed over an arbitrary directed set. This is called a net.

And with these the expected statement does become true (for a proof see here):

\;\;\;\;A topological space (X,τ)(X,\tau) is compact precisely if every net in XX has a converging subnet.

In fact convergence of nets also detects closed subsets in topological spaces (hence their topology as such), and it detects the continuity of functions between topological spaces. It also detects for instance the Hausdorff property. (For detailed statements and proofs see here.) Hence when analysis is cast in terms of nets instead of just sequences, then it raises to the same level of generality as topology.

(closed subspaces of compact Hausdorff spaces are equivalently compact subspaces)

Let

1. (X,τ)(X,\tau) be a compact Hausdorff topological space (def. , def. )

2. YXY \subset X be a topological subspace (example ).

Then the following are equivalent:

1. YXY \subset X is a closed subspace (def. );

2. YY is a compact topological space (def. ).

By lemma and lemma below.

(closed subspaces of compact spaces are compact)

Let

1. (X,τ)(X,\tau) be a compact topological space (def. ),

2. YXY \subset X be a closed topological subspace (def. , example ).

Then also YY is compact.

Let {ViY}iI\{V_i \subset Y\}_{i \in I} be an open cover of YY (def. ). We need to show that this has a finite sub-cover.

By definition of the subspace topology, there exist open subsets UiXU_i \subset X with

By the assumption that YY is closed, the complement XYXX \setminus Y \subset X is an open subset of XX, and therefore

is an open cover of XX (def. ). Now by the assumption that XX is compact, this latter cover has a finite subcover, hence there exists a finite subset JIJ \subset I such that

is still an open cover of XX, hence in particular restricts to a finite open cover of YY. But since Y(XY)=Y \cap ( X \setminus Y ) = \empty, it follows that

is a cover of YY, and in indeed a finite subcover of the original one.

(compact subspaces in Hausdorff spaces are separated by neighbourhoods from points)

Let

1. (X,τ)(X,\tau) be a Hausdorff topological space (def. );

2. YXY \subset X a compact subspace (def. , example ).

Then for every xXYx \in X \setminus Y there exists

1. an open neighbourhood Ux{x}U_x \supset \{x\};

2. an open neighbourhood UYYU_Y \supset Y

such that

• they are still disjoint: UxUY=U_x \cap U_Y = \emptyset.

By the assumption that (X,τ)(X,\tau) is Hausdorff, we find for every point yYy \in Y disjoint open neighbourhoods Ux,y{x}U_{x,y} \supset \{x\} and Uy{y}U_y \supset \{y\}. By the nature of the subspace topology of YY, the restriction of all the UyU_y to YY is an open cover of YY:

Now by the assumption that YY is compact, there exists a finite subcover, hence a finite set SYS \subset Y such that

is still a cover.

But the finite intersection

of the corresponding open neighbourhoods of xx is still open, and by construction it is disjoint from all the UsU_{s}, hence also from their union

Therefore UxU_x and UYU_Y are two open subsets as required.

(compact subspaces of Hausdorff spaces are closed)

Let

1. (X,τ)(X,\tau) be a Hausdorff topological space (def. )

2. CXC \subset X be a compact (def. ) topological subspace (example ).

Then CXC \subset X is also a closed subspace (def. ).

Let xXCx \in X \setminus C be any point of XX not contained in CC. By lemma we need to show that there exists an open neighbourhood of xx in XX which does not intersect CC. This is implied by lemma .

(Heine-Borel theorem)

For nn \in \mathbb{N}, consider n\mathbb{R}^n as the nn-dimensional Euclidean space via example , regarded as a topological space via its metric topology (example ).

Then for a topological subspace SnS \subset \mathbb{R}^n the following are equivalent:

1. SS is compact (def. );

2. SS is closed (def. ) and bounded (def. ).

First consider a subset SnS \subset \mathbb{R}^n which is closed and bounded. We need to show that regarded as a topological subspace it is compact.

The assumption that SS is bounded by (hence contained in) some open ball Bx(ϵ)B^\circ_x(\epsilon) in n\mathbb{R}^n implies that it is contained in {(xi)i=1nn|ϵxiϵ}\{ (x_i)_{i = 1}^n \in \mathbb{R}^n \,\vert\, -\epsilon \leq x_i \leq \epsilon \}. By example , this topological subspace is homeomorphic to the nn-cube

hence to the product topological space (example ) of nn copies of the closed interval with itself.

Since the closed interval [ϵ,ϵ][-\epsilon, \epsilon] is compact by example , the Tychonoff theorem (prop. ) implies that this nn-cube is compact.

Since subsets are closed in a closed subspace precisely if they are closed in the ambient space (lemma ) the closed subset SnS \subset \mathbb{R}^n is also closed as a subset S[ϵ,ϵ]nS \subset [-\epsilon, \epsilon]^n. Since closed subspaces of compact spaces are compact (lemma ) this implies that SS is compact.

Conversely, assume that SnS \subset \mathbb{R}^n is a compact subspace. We need to show that it is closed and bounded.

The first statement follows since the Euclidean space n\mathbb{R}^n is Hausdorff (example ) and since compact subspaces of Hausdorff spaces are closed (prop. ).

Hence what remains is to show that SS is bounded.

To that end, choose any positive real number ϵ>0\epsilon \in \mathbb{R}_{\gt 0} and consider the open cover of all of n\mathbb{R}^n by the open n-cubes

for n-tuples of integers (k1,k2,,kn)n(k_1, k_2 , \cdots, k_n ) \in \mathbb{Z}^n. The restrictions of these to SS hence form an open cover of the subspace SS. By the assumption that SS is compact, there is then a finite subset of nn-tuples of integers such that the corresponding nn-cubes still cover SS. But the union of any finite number of bounded closed nn-cubes in n\mathbb{R}^n is clearly a bounded subset, and hence so is SS.

(examples of compact Hausdorff spaces)

We list some basic examples of compact Hausdorff spaces (def. , def. )

1. For nn \in \mathbb{N}, the n-sphere SnS^n may canonically be regarded as a topological subspace of Euclidean space n+1\mathbb{R}^{n+1} (example ).

These are clearly closed and bounded subspaces of Euclidean space, hence they are compact topological space, by the Heine-Borel theorem, prop. .

(maps from compact spaces to Hausdorff spaces are closed and proper)

Let f:(X,τX)⟶(Y,τY)f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y) be a continuous function between topological spaces such that

1. (X,τX)(X,\tau_X) is a compact topological space (def. );

2. (Y,τY)(Y,\tau_Y) is a Hausdorff topological space (def. ).

Then ff is

1. a closed map (def. );

2. a proper map (def. ).

For the first statement, we need to show that if CXC \subset X is a closed subset of XX, then also f(C)Yf(C) \subset Y is a closed subset of YY.

Now

1. since closed subspaces of compact spaces are compact (lemma ) it follows that CXC \subset X is also compact;

2. since continuous images of compact spaces are compact (cor. ) it then follows that f(C)Yf(C) \subset Y is compact;

3. since compact subspaces of Hausdorff spaces are closed (prop. ) it finally follow that f(C)f(C) is also closed in YY.

For the second statement we need to show that if CYC \subset Y is a compact subset, then also its pre-image f1(C)f^{-1}(C) is compact.

Now

1. since compact subspaces of Hausdorff spaces are closed (prop. ) it follows that CYC \subset Y is closed;

2. since pre-images under continuous functions of closed subsets are closed (prop. ), also f1(C)Xf^{-1}(C) \subset X is closed;

3. since closed subspaces of compact spaces are compact (lemma ), it follows that f1(C)f^{-1}(C) is compact.

(continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms)

Let f:(X,τX)⟶(Y,τY)f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y) be a continuous function between topological spaces such that

1. (X,τX)(X,\tau_X) is a compact topological space (def. );

2. (Y,τY)(Y,\tau_Y) is a Hausdorff topological space (def. ).

3. f:X⟶Yf \;\colon\; X \longrightarrow Y is a bijection of sets.

Then ff is a homeomorphism (def. )

In particular then both (X,τX)(X,\tau_X) and (Y,τY)(Y, \tau_Y) are compact Hausdorff spaces.

By prop. it is sufficient to show that ff is a closed map. This is the case by prop. .

(compact Hausdorff spaces are normal)

Every compact Hausdorff topological space (def. , def. ) is a normal topological space (def. ).

First we claim that (X,τ)(X,\tau) is regular. To show this, we need to find for each point xXx \in X and each closed subset YXY \in X not containing xx disjoint open neighbourhoods Ux{x}U_x \supset \{x\} and UYYU_Y \supset Y. But since closed subspaces of compact spaces are compact (lemma ), the subset YY is in fact compact, and hence this is the statement of lemma .

Next to show that (X,τ)(X,\tau) is indeed normal, we apply the idea of the proof of lemma once more:

Let Y1,Y2XY_1, Y_2 \subset X be two disjoint closed subspaces. By the previous statement then for every point y1Yy_1 \in Y we find disjoint open neighbourhoods Uy1{y1}U_{y_1} \supset \{y_1\} and UY2,y1Y2U_{Y_2,y_1} \supset Y_2. The union of the Uy1U_{y_1} is a cover of Y1Y_1, and by compactness of Y1Y_1 there is a finite subset SYS \subset Y such that

is an open neighbourhood of Y1Y_1 and

is an open neighbourhood of Y2Y_2, and both are disjoint.

(continuous surjections from compact spaces to Hausdorff spaces are quotient projections)

Let

be a continuous function between topological spaces such that

1. (X,τX)(X,\tau_X) is a compact topological space (def. );

2. (Y,τY)(Y, \tau_Y) is a Hausdorff topological space (def. );

3. π:X⟶Y\pi \;\colon\; X \longrightarrow Y is a surjective function.

Then τY\tau_Y is the quotient topology inherited from τX\tau_X via the surjection ff (def. ).

We need to show that a subset UYU \subset Y is an open subset of (Y,τY)(Y , \tau_Y) precisely if its pre-image π1(U)X\pi^{-1}(U) \subset X is an open subset in (X,τX)(X,\tau_X). Equivalenty, as in prop. , we need to show that UU is a closed subset precisely if π1(U)\pi^{-1}(U) is a closed subset. The implication

follows via prop. from the continuity of π\pi. The implication

follows since π\pi is a closed map by prop. .

(quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff)

Let

be a continuous function between topological spaces such that

1. (X,τ)(X, \tau) is a compact Hausdorff topological space (def. , def. );

2. π\pi is a surjection and τY\tau_Y is the corresponding quotient topology (def. ).

Then the following are equivalent

1. (Y,τY)(Y, \tau_Y) is itself a Hausdorff topological space (def. );

2. π\pi is a closed map (def. ).

The implicaton ((Y,τY)Hausdorff)(πclosed)\left( (Y, \tau_Y)\, \text{Hausdorff} \right) \Rightarrow \left( \pi \, \text{closed} \right) is given by prop. . We need to show the converse.

Hence assume that π\pi is a closed map. We need to show that for every pair of distinct points y1y2Yy_1 \neq y_2 \in Y there exist open neighbourhoods Uy1,Uy2τYU_{y_1}, U_{y_2} \in \tau_Y which are disjoint, Uy1Uy2=U_{y_1} \cap U_{y_2} = \emptyset.

First notice that the singleton subsets {x},{y}Y\{x\}, \{y\} \in Y are closed. This is because they are images of singleton subsets in XX, by surjectivity of ff, and because singletons in a Hausdorff space are closed by prop, and prop. , and because images under ff of closed subsets are closed, by the assumption that ff is a closed map.

It follows that the pre-images

are closed subsets of XX.

Now again since compact Hausdorff spaces are normal (prop. ) it follows (by def. ) that we may find disjoint open subset U1,U2τXU_1, U_2 \in \tau_X such that

Moreover, by lemma we may find these UiU_i such that they are both saturated subsets (def. ). Therefore finally lemma says that the images π(Ui)\pi(U_i) are open in (Y,τY)(Y,\tau_Y). These are now clearly disjoint open neighbourhoods of y1y_1 and y2y_2.

Consider the function

• from the quotient topological space (def. ) of the closed interval (def. ) by the equivalence relation which identifies the two endpoints

  (xy)((x=y)or((x{0,2π})and(y{0,2π})))(x \sim y) \;\Leftrightarrow\; \left( \left( x = y\right) \,\text{or}\, \left( \left( x \in \{0,2\pi\}\right) \,\text{and}\, \left( y\in \{0, 2\pi\} \right) \right) \right)

• to the unit circle S1=S0(1)2S^1 = S_0(1) \subset \mathbb{R}^2 (def. ) regarded as a topological subspace of the 2-dimensional Euclidean space (example ) equipped with its

metric topology (example ).

This is clearly a continuous function and a bijection on the underlying sets. Moreover, since continuous images of compact spaces are compact (cor. ) and since the closed interval [0,1][0,1] is compact (example ) we also obtain another proof that the circle is compact.

Hence by prop. the above map is in fact a homeomorphism

Compare this to the counter-example , which observed that the analogous function

is not a homeomorphism, even though this, too, is a bijection on the the underlying sets. But the half-open interval [0,2π)[0,2\pi) is not compact (for instance by the Heine-Borel theorem, prop. ), and hence prop. does not apply.

(locally compact topological space)

A topological space XX is called locally compact if for every point xXx \in X and every open neighbourhood Ux{x}U_x \supset \{x\} there exists a smaller open neighbourhood VxUxV_x \subset U_x whose topological closure is compact (def. ) and still contained in UU:

(varying terminology regarding locally compact)

On top of the terminology issue inherited from that of compact, remark (regarding whether or not to require Hausdorff with compact; we do not), the definition of locally compact is subject to further ambiguity in the literature. There are various definitions of locally compact spaces alternative to def. , we consider one such alternative definition below in def. .

For Hausdorff topological spaces all these definitions happen to be equivalent (prop. below), but in general they are not.

(discrete spaces are locally compact)

Every discrete topological space (example ) is locally compact (def. ).

(Euclidean space is locally compact)

For nn \in \mathbb{N} then Euclidean space n\mathbb{R}^n (example ) regarded as a topological space via its metric topology (def. ), is locally compact (def. ).

Let xXx \in X be a point and Ux{x}U_x \supset \{x\} an open neighbourhood. By definition of the metric topology (example ) this means that UxU_x contains an open ball Bx(ϵ)B^\circ_x(\epsilon) (def. ) around xx of some radius ϵ\epsilon. This ball also contains the open ball VxBx(ϵ/2)V_x \coloneqq B^\circ_x(\epsilon/2) and its topological closure, which is the closed ball Bx(ϵ/2)B_x(\epsilon/2). This closed ball is compact, for instance by the Heine-Borel theorem (prop. ).

(open subspaces of compact Hausdorff spaces are locally compact)

Every open topological subspace XopenKX \underset{\text{open}}{\subset} K of a compact (def. ) Hausdorff space (def. ) is a locally compact topological space (def. ).

In particular compact Hausdorff spaces themselves are locally compact.

Let XX be a topological space such that it arises as a topological subspace XKX \subset K of a compact Hausdorff space. We need to show that XX is a locally compact topological space (def. ).

Let xXx \in X be a point and let UxXU_x \subset X an open neighbourhood. We need to produce a smaller open neighbourhood whose closure is compact and still contained in UxU_x.

By the nature of the subspace topology there exists an open subset VxKV_x\subset K such that Ux=XVxU_x = X \cap V_x. Since XKX \subset K is assumed to be open, it follows that UxU_x is also open as a subset of KK. Since compact Hausdorff spaces are normal (prop. ) it follows by prop. that there exists a smaller open neighbourhood WxKW_x \subset K whose topological closure is still contained in UxU_x, and since closed subspaces of compact spaces are compact (prop. ), this topological closure is compact:

The intersection of this situation with XX is the required smaller compact neighbourhood Cl(Wx)XCl(W_x) \cap X:

(finite product space of locally compact spaces is locally compact)

The product topological space (example ) iJ(Xi,τi)\underset{i \in J}{\prod} (X_i, \tau_i) of a a finite set {(Xi,τi)}iI\{ (X_i, \tau_i) \}_{i \in I} of locally compact topological spaces (Xi,τi)(X_i, \tau_i) (def. ) it itself locally compact.

(countably infinite products of non-compact spaces are not locally compact)

Let XX be a topological space which is not compact (def. ). Then the product topological space (example ) of a countably infinite set of copies of XX

is not a locally compact space (def. ).

Since the continuous image of a compact space is compact (prop. ), and since the projection maps pi:X⟶Xp_i \;\colon\; \underset{\mathbb{N}}{\prod} X \longrightarrow X are continuous (by nature of the initial topology/Tychonoff topology), it follows that every compact subspace of the product space is contained in one of the form

for KiXK_i \subset X compact.

But by the nature of the Tychonoff topology, a base for the topology on X\underset{\mathbb{N}}{\prod} X is given by subsets of the form

with UiXU_i \subset X open. Hence every compact neighbourhood in X\underset{\mathbb{N}}{\prod} X contains a subset of this kind, but if XX itself is non-compact, then none of these is contained in a product of compact subsets.

(local compactness via compact neighbourhood base)

A topological space is locally compact if for for every point xXx \in X every open neighbourhood Ux{x}U_x \supset \{x\} contains a compact neighbourhood KxUxK_x \subset U_x.

(equivalence of definitions of local compactness for Hausdorff spaces)

If XX is a Hausdorff topological space, then the two definitions of local compactness of XX

1. definition (every open neighbourhood contains a compact neighbourhood),

2. definition (every open neighbourhood contains a compact neighbourhood that is the topological closure of an open neighbourhood)

are equivalent.

Generally, definition directly implies definition . We need to show that Hausdorffness implies the converse.

Hence assume that for every point xXx \in X then every open neighbourhood Ux{x}U_x \supset \{x\} contains a compact neighbourhood. We need to show that it then also contains the closure Cl(Vx)Cl(V_x) of a smaller open neighbourhood and such that this closure is compact.

So let KxUxK_x \subset U_x be a compact neighbourhood. Being a neighbourhood, it has a non-trivial interior which is an open neighbouhood

Since compact subspaces of Hausdorff spaces are closed (lemma ), it follows that KxXK_x \subset X is a closed subset. This implies that the topological closure of its interior as a subset of XX is still contained in KxK_x (since the topological closure is the smallest closed subset containing the given subset, by def. ): Cl(Int(Kx))KxCl(Int(K_x)) \subset K_x. Since subsets are closed in a closed subspace precisely if they are closed in the ambient space (lemma ), Cl(Int(Kx))Cl(Int(K_x)) is also closed as a subset of the compact subspace KxK_x. Now since closed subspaces of compact spaces are compact (lemma ), it follows that this closure is also compact as a subspace of KxK_x, and since continuous images of compact spaces are compact (prop. ), it finally follows that it is also compact as a subspace of XX:

(topological mapping space with compact-open topology)

For

1. (X,τX)(X, \tau_X) a locally compact topological space (in the sense of def. or just in the sense of def. )

2. (Y,τY)(Y,\tau_Y) any topological space

then the mapping space

is the topological space

• whose underlying set HomTop(X,Y)Hom_Top(X,Y) is the set of continuous functions XYX \to Y;

• whose topology τcpt-op\tau_{\text{cpt-op}} is generated from the sub-basis for the topology (def. ) which is given by subsets to denoted

  UKHomTop(X,Y)U^K \subset Hom_{Top}(X,Y) for labels

  • KXK \subset X a compact subset,

  • UYU \subset Y an open subset

  and defined to be those subsets of all those continuous functions ff that take KK to UU:

  UK{f:X⟶continuousY|f(K)U}.U^K \;\coloneqq\; \left\{ f \colon X \overset{\text{continuous}}{\longrightarrow} Y \;\;\vert\;\; f(K) \subset U \right\} \,.

Accordingly this topology τcpt-op\tau_{\text{cpt-op}} is called the compact-open topology on the set of functions.

(universal property of the mapping space)

Let (X,τX)(X,\tau_X), (Y,τY)(Y, \tau_Y), (Z,τZ)(Z, \tau_Z) be topological spaces, with XX locally compact (def. or just def. ). Then

1. The evaluation function

   (X,τX)×Maps((X,τX),(Y,τY))⟶AAevAA(Y,τY)(x,f)AAAf(x)\array{ (X, \tau_X) \times Maps((X,\tau_X), (Y, \tau_Y)) & \overset{\phantom{AA} ev \phantom{AA}}{\longrightarrow}& (Y, \tau_Y) \\ (x, f) &\overset{\phantom{AAA}}{\mapsto}& f(x) }

   is a continuous function.

2. The natural bijection of function sets

   {X×Z⟶Y}HomSet(X×Z,Y)⟶AAAA{Z⟶HomSet(X,Y)}HomSet(Z,HomSet(X,Y))(f:(x,z)f(x,z))AAAf˜:z(xf(x,z))\array{ \underset {Hom_{Set}(X \times Z, Y) } {\underbrace{ \left\{ X \times Z \longrightarrow Y \right\} }} &\underoverset{}{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow}& \underset {Hom_{Set}\left( Z, Hom_{Set}(X,Y) \right)} {\underbrace{ \left\{ Z \longrightarrow Hom_{Set}(X,Y) \right\} }} \\ \left(f \colon (x,z) \mapsto f(x,z) \right) &\overset{\phantom{AAA}}{\mapsto}& \tilde f \colon z \mapsto (x \mapsto f(x,z)) }

   restricts to a natural bijection between sets of continuous functions

   {(X,τX)×(Z,τZ)⟶cts(Y,τY)}HomTop((X,τX)×(Z,τZ),(Y,τY))⟶AAAA{(Z,τZ)⟶ctsMaps((X,τX),(Y,τY))}HomTop((Z,τZ),Maps((X,τX),(Y,τY))).\array{ \underset {Hom_{Top}((X, \tau_X) \times (Z, \tau_Z), (Y, \tau_Y))} {\underbrace{\left\{ (X,\tau_X) \times (Z, \tau_Z) \overset{cts}{\longrightarrow} (Y, \tau_Y) \right\}}} &\underoverset{}{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow}& \underset {Hom_{Top}\left( (Z, \tau_Z) , Maps((X,\tau_X), (Y, \tau_Y)) \right)} {\underbrace{\left\{ (Z, \tau_Z) \overset{cts}{\longrightarrow} Maps( (X,\tau_X), (Y,\tau_Y) ) \right\}}} } \,.

Here Maps((X,τX),(Y,τY))Maps((X,\tau_X), (Y,\tau_Y)) is the mapping space with compact-open topology from example and ()×()(-)\times (-) denotes forming the product topological space (example , example ).

To see the continuity of the evaluation map:

Let VYV \subset Y be an open subset. It is sufficient to show that ev1(V)={(x,f)|f(x)V}ev^{-1}(V) = \{(x, f) \vert f(x) \in V\} is a union of products of the form U×VKU \times V^K with UXU\subset X open and VKHomSet(K,W)V^K \subset Hom_{Set}(K,W) a basic open according to def. .

For (x,f)ev1(V)(x, f) \in ev^{-1}(V), the preimage f1(V)Xf^{-1}(V) \subset X is an open neighbourhood of xx in XX, by continuity of ff.

By local compactness of XX, there is a compact subset Kf1(V)K \subset f^{-1}(V) which is still a neighbourhood of xx, hence contains an open neighbourhood UKU \subset K. Since ff also still takes that into VV, we have found an open neighbourhood

with respect to the product topology. Since this is still contained in ev1(V)ev^{-1}(V), for all (x,f)(x,f) as above, ev1(V)ev^{-1}(V) is exhibited as a union of opens, and is hence itself open.

Regarding the second point:

In one direction, let f:(X,τX)×(Y,τY)(ZτZ)f \colon (X, \tau_X) \times (Y, \tau_Y) \to (Z \, \tau_Z) be a continuous function, and let UKMaps(X,Y)U^K \subset Maps(X,Y) be a sub-basic open. We need to show that the set

is open. To that end, observe that f(K,z)Uf(K,z) \subset U means that K×{z}f1(U)K \times \{z\} \subset f^{-1}(U), where f1(U)X×Yf^{-1}(U) \subset X \times Y is open by the continuity of ff. Hence in the topological subspace K×ZX×YK \times Z \subset X \times Y the inclusion

is an open neighbourhood. Since KK is compact, the tube lemma (prop. ) gives an open neighbourhood Vz{z}V_z \supset \{z\} in YY, hence an open neighbourhood K×VzK×YK \times V_z \subset K \times Y, which is still contained in the original pre-image:

This shows that with every point zf˜1(UK)z \in \tilde f^{-1}\left(U^K\right) also an open neighbourhood of zz is contained in f˜1(UK)\tilde f^{-1}\left(U^K\right), hence that the latter is a union of open subsets, and hence itself open.

In the other direction, assume that f˜:ZMaps((X,τX),(Y,τY))\tilde f \colon Z \to Maps((X,\tau_X),(Y,\tau_Y)) is continuous: We need to show that ff is continuous. But observe that ff is the composite

Here the first function id×f˜id \times \tilde f is continuous since f˜\tilde f is by assumption since the product of two continuous functions is again continuous (example ). The second function evev is continuous by the first point above. hence ff is continuous.

(topological mapping space is exponential object)

In the language of category theory (remark ), prop. says that the mapping space construction with its compact-open topology from def. is an exponential object or internal hom. This just means that it beahves in all abstract ways just as a function set does for plain functions, but it does so for continuous functions and being itself equipped with a topology.

Moreover, the construction of topological mapping spaces in example extends to a functor (remark )

from the product category of the category Top of all topological spaces (remark ) with the opposite category of the subcategory of locally compact topological spaces.

(topological mapping space construction out of the point space is the identity)

The point space *\ast (example ) is clearly a locally compact topological space. Hence for every topological space (X,τ)(X,\tau) the mapping space Maps(*,(X,τ))Maps(\ast, (X,\tau)) (exmaple ) exists. This is homeomorphic (def. ) to the space (x,τ)(x,\tau) itself:

(loop space and path space)

Let (X,τ)(X,\tau) be any topological space.

1. The circle S1S^1 (example ) is a compact Hausdorff space (example ) hence, by prop. , a locally compact topological space (def. ). Accordingly the mapping space

   XMaps(S1,(X,τ))\mathcal{L} X \coloneqq Maps( S^1, (X,\tau) )

   exists (def. ). This is called the free loop space of (X,τ)(X,\tau).

   If both S1S^1 and XX are equipped with a choice of point (basepoint) s0S1s_0 \in S^1, x0Xx_0 \in X, then the topological subspace

   ΩXX\Omega X \subset \mathcal{L}X

   on those functions which take the basepoint of S1S^1 to that of XX, is called the loop space of XX, or sometimes based loop space, for emphasis.

2. Similarly the closed interval is a compact Hausdorff space (example ) hence, by prop. , a locally compact topological space (def. ). Accordingly the mapping space

   Maps([0,1],(X,τ))Maps( [0,1], (X,\tau) )

   exists (def. ). Again if XX is equipped with a choice of basepoint x0Xx_0 \in X, then the topological subspace of those functions that take 0[0,1]0 \in [0,1] to that chosen basepoint is called the path space of (Xτ)(X\tau):

   PXMaps([0,1],(X,τ)).P X \subset Maps( [0,1], (X,\tau) ) \,.

Notice that we may encode these subspaces more abstractly in terms of universal properties:

The path space and the loop space are characterized, up to homeomorphisms, as being the limiting cones in the following pullback diagrams of topological spaces (example ):

1. loop space:

   ΩX⟶Maps(S1,(X,τ))(pb)Maps(consts0,id(X,τ))*⟶constx0XMaps(*,(X,τ)).\array{ \Omega X &\longrightarrow& Maps(S^1, (X,\tau)) \\ \downarrow &(pb)& \downarrow^{\mathrlap{Maps(const_{s_0}, id_{(X,\tau)})}} \\ \ast &\underset{const_{x_0}}{\longrightarrow}& X \simeq Maps(\ast,(X,\tau)) } \,.

2. path space:

   PX⟶Maps([0,1],(X,τ))(pb)Maps(constx,id(X,τ))*⟶constx0XMaps(*,(X,τ))\array{ P X &\longrightarrow& Maps([0,1], (X,\tau)) \\ \downarrow &(pb)& \downarrow^{\mathrlap{Maps(const_x, id_{(X,\tau)})}} \\ \ast &\underset{const_{x_0}}{\longrightarrow}& X \simeq Maps(\ast,(X,\tau)) }

Here on the right we are using that the mapping space construction is a functor as shown in remark , and we are using example in the identification on the bottom right mapping space out of the point space.

(one-point compactification)

Let XX be any topological space. Its one-point compactification X*X^* is the topological space

• whose underlying set is the disjoint union X{}X \cup \{\infty\}

• and whose open sets are

  1. the open subsets of XX (thought of as subsets of X*X^*);

  2. the complements X*\CK=(X\CK){}X^\ast \backslash CK = (X \backslash CK) \cup \{\infty\} of the closed compact subsets CKXCK \subset X.

If XX is Hausdorff, then it is sufficient to speak of compact subsets in def. , since compact subspaces of Hausdorff spaces are closed.

(one-point compactification is well-defined)

The topology on the one-point compactification in def. is indeed well defined in that the given set of subsets is indeed closed under arbitrary unions and finite intersections.

The unions and finite intersections of the open subsets inherited from XX are closed among themselves by the assumption that XX is a topological space.

It is hence sufficient to see that

1. the unions and finite intersection of the (X\CK){}(X \backslash CK) \cup \{\infty\} are closed among themselves,

2. the union and intersection of a subset of the form UopenXX*U \underset{\text{open}}{\subset} X \subset X^\ast with one of the form (X\CK){}(X \backslash CK) \cup \{\infty\} is again of one of the two kinds.

Regarding the first statement: Under de Morgan duality

and

and so the first statement follows from the fact that finite unions of compact subspaces and arbitrary intersections of closed compact subspaces are themselves again compact (example ).

Regarding the second statement: That UXU \subset X is open means that there exists a closed subset CXC \subset X with U=X\CU = X\backslash C. Now using de Morgan duality we find

1. for intersections:

   U((X\CK){})=(X\C)(X\CK)=X\(CCK)\begin{aligned} U \cap ( (X\backslash CK) \cup \{\infty\} ) & = (X \backslash C) \cap (X \backslash CK) \\ & = X \backslash ( C \cup CK ) \end{aligned}

   Since finite unions of closed subsets are closed, this is again an open subset of XX;

2. for unions:

   U(X\CK){}=(X\C)(X\CU){}=(X\(CCK)){}.\begin{aligned} U \cup (X \backslash CK) \cup \{\infty\} & = (X \backslash C) \cup (X \backslash CU) \cup \{\infty\} \\ & = (X \backslash (C \cap CK)) \cup \{\infty\} \end{aligned} \,.

   For this to be open in X*X^\ast we need that CCKC \cap CK is again compact. This follows because subsets are closed in a closed subspace precisely if they are closed in the ambient space and because closed subsets of compact spaces are compact.

(one-point compactification of Euclidean space is the n-sphere )

For nn \in \mathbb{N} the n-sphere with its standard topology (e.g. as a subspace of the Euclidean space n+1\mathbb{R}^{n+1} with its metric topology) is homeomorphic to the one-point compactification (def. ) of the Euclidean space n\mathbb{R}^n

Pick a point Sn\infty \in S^n. By stereographic projection we have a homeomorphism

With this it only remains to see that for U{}U_\infty \supset \{\infty\} an open neighbourhood of \infty in SnS^n then the complement SnUS^n \setminus U_\infty is compact closed, and cnversely that the complement of every compact closed subset of Sn{}S^n \setminus \{\infty\} is an open neighbourhood of {}\{\infty\}.

Observe that under stereographic projection (example ) the open subspaces U{}Sn{}U_\infty \setminus \{\infty\} \subset S^n \setminus \{\infty\} are identified precisely with the closed and bounded subsets of n\mathbb{R}^n. (Closure is immediate, boundedness follows because an open neighbourhood of {}Sn\{\infty\} \in S^n needs to contain an open ball around 0nSn{}0 \in \mathbb{R}^n \simeq S^n \setminus \{-\infty\} in the other stereographic projection, which under change of chart gives a bounded subset. )

By the Heine-Borel theorem (prop. ) the closed and bounded subsets of n\mathbb{R}^n are precisely the compact, and hence the compact closed, subsets of nSn{}\mathbb{R}^n \simeq S^n \setminus \{\infty\}.

(one-point compactification is compact)

For XX any topological space, then its one-point compactification X*X^\ast (def. ) is a compact topological space.

Let {UiX*}iI\{U_i \subset X^\ast\}_{i \in I} be an open cover. We need to show that this has a finite subcover.

That we have a cover means that

1. there must exist iIi_\infty \in I such that Ui{}U_{i \infty} \supset \{\infty\} is an open neighbourhood of the extra point. But since, by construction, the only open subsets containing that point are of the form (X\CK){}(X \backslash CK) \cup \{\infty\}, it follows that there is a compact closed subset CKXCK \subset X with X\CKUiX \backslash CK \subset U_{i \infty}.

2. {UiX}ii\{U_i \subset X\}_{i \in i} is in particular an open cover of that closed compact subset CKXCK \subset X. This being compact means that there is a finite subset JIJ \subset I so that {UiX}iJX\{U_i \subset X\}_{i \in \J \subset X} is still a cover of CKCK.

Together this implies that

is a finite subcover of the original cover.

(one-point compactification of locally compact space is Hausdorff precisely if original space is)

Let XX be a locally compact topological space. Then its one-point compactification X*X^\ast (def. ) is a Hausdorff topological space precisely if XX is.

It is clear that if XX is not Hausdorff then X*X^\ast is not.

For the converse, assume that XX is Hausdorff.

Since X*=X{}X^\ast = X \cup \{\infty\} as underlying sets, we only need to check that for xXx \in X any point, then there is an open neighbourhood UxXX*U_x \subset X \subset X^\ast and an open neighbourhood VX*V_\infty \subset X^\ast of the extra point which are disjoint.

That XX is locally compact implies by definition that there exists an open neighbourhood Uk{x}U_k \supset \{x\} whose topological closure CKCl(Ux)CK \coloneqq Cl(U_x) is a closed compact neighbourhood CK{x}CK \supset \{x\}. Hence

is an open neighbourhood of {}\{\infty\} and the two are disjoint

by construction.

(inclusion into one-point compactification is open embedding)

Let XX be a topological space. Then the evident inclusion function

into its one-point compactification (def. ) is

1. a continuous function

2. an open map

3. an embedding of topological spaces.

Regarding the first point: For UXU \subset X open and CKXCK \subset X closed and compact, the pre-images of the corresponding open subsets in X*X^\ast are

which are open in XX.

Regarding the second point: The image of an open subset UXU \subset X is i(U)=UX*i(U) = U \subset X^\ast, which is open by definition

Regarding the third point: We need to show that i:Xi(X)X*i \colon X \to i(X) \subset X^\ast is a homeomorphism. This is immediate from the definition of X*X^\ast.

(locally compact Hausdorff spaces are the open subspaces of compact Hausdorff spaces)

The locally compact Hausdorff spaces are, up to homeomorphism precisely the ope subspaces of compact Hausdorff spaces.

That every open subspace of a compact Hausdorff space is locally compact Hausdorff was the statement of example . It remains to see that every locally compact Hausdorff space arises this way.

But if XX is locally compact Hausdorff, then its one-point compactification X*X^\ast is compact Hausdorff by prop. and prop. . Moreover the canonical embedding XX*X \hookrightarrow X^\ast exhibts XX as an open subspace of X*X^\ast by prop. .

(proper maps to locally compact spaces are closed)

Let

1. (X,τX)(X,\tau_X) be a topological space,

2. (Y,τY)(Y,\tau_Y) a locally compact Hausdorff space (def. , def. ),

3. f:XYf \colon X \to Y a proper map (def. ).

Then ff is a closed map (def. ).

Let CXC \subset X be a closed subset. We need to show that f(C)Yf(C) \subset Y is closed. By lemma this means we need to show that every yYf(C)y \in Y \setminus f(C) has an open neighbourhood Uy{y}U_y \supset \{y\} not intersecting f(C)f(C)..

By local compactness of (Y,τY)(Y,\tau_Y) (def. ), yy has an open neighbourhood VyV_y whose topological closure Cl(Vy)Cl(V_y) is compact. Hence since ff is proper, also f1(Cl(Vy))Xf^{-1}(Cl(V_y)) \subset X is compact. Then also the intersection Cf1(Cl(Vy))C \cap f^{-1}(Cl(V_y)) is compact, and since continuous images of compact spaces are compact (prop. ) so is

This is also a closed subset, since compact subspaces of Hausdorff spaces are closed (lemma ). Therefore

is an open neighbourhod of yy not intersecting f(C)f(C).

(injective proper maps to locally compact spaces are equivalently the closed embeddings)

Let

1. (X,τX)(X,\tau_X) be a topological space

2. (Y,τY)(Y,\tau_Y) a locally compact Hausdorff space (def. , def. ),

3. f:XYf \colon X \to Y be a continuous function.

Then the following are equivalent

1. ff is an injective proper map,

2. ff is a closed embedding of topological spaces (def. ).

In one direction, if ff is an injective proper map, then since proper maps to locally compact spaces are closed, it follows that ff is also closed map. The claim then follows since closed injections are embeddings (prop. ), and since the image of a closed map is closed.

Conversely, if ff is a closed embedding, we only need to show that the embedding map is proper. So for CYC \subset Y a compact subspace, we need to show that the pre-image f1(C)Xf^{-1}(C) \subset X is also compact. But since ff is an injection (being an embedding), that pre-image is just the intersection f1(C)Cf(X)f^{-1}(C) \simeq C \cap f(X). By the nature of the subspace topology, this is compact if CC is.

(locally finite cover)

Let (X,τ)(X,\tau) be a topological space.

An open cover {UiX}iI\{U_i \subset X\}_{i \in I} (def. ) of XX is called locally finite if for all points xXx \in X, there exists a neighbourhood Ux{x}U_x \supset \{x\} such that it intersects only finitely many elements of the cover, hence such that UxUiU_x \cap U_i \neq \emptyset for only a finite number of iIi \in I.

(refinement of open covers)

Let (X,τ)(X,\tau) be a topological space, and let {UiX}iI\{U_i \subset X\}_{i \in I} be a open cover (def. ).

Then a refinement of this open cover is a set of open subsets {VjX}jJ\{V_j \subset X\}_{j \in J} which is still an open cover in itself and such that for each jJj \in J there exists an iIi \in I with VjUiV_j \subset U_i.

(paracompact topological space)

A topological space (X,τ)(X,\tau) is called paracompact if every open cover of XX has a refinement (def. ) by a locally finite open cover (def. ).

(compact topological spaces are paracompact)

Every compact topological space (def. ) is paracompact (def. ).

Since a finite subcover is in particular a locally finite refinement.

(disjoint unions of paracompact spaces are paracompact)

Let {(Xi,τi)}iI\{(X_i, \tau_i)\}_{i \in I} be a set of paracompact topological spaces (def. ). Then also their disjoint union space (example )

is paracompact.

In particular, by example a non-finite disjoint union of compact topological spaces is, while no longer compact, still paracompact.

Let 𝒰={UjiI(Xi,τi)}jJ\mathcal{U} = \{ U_j \subset \underset{i \in I}{\sqcup} (X_i, \tau_i) \}_{j \in J} be an open cover. We need to produce a locally finite refinement.

Since each XiX_i is open in the disjoint union, the intersections UiXjU_i \cap X_j are all open, and hence by forming all these intersections we obtain a refinement of the original cover by a disjoint union of open covers 𝒰i\mathcal{U}_i of (Xi,τi)(X_i, \tau_i) for all iIi \in I. By the assumption that each (Xi,τi)(X_i, \tau_i) is paracompact, each 𝒰i\mathcal{U}_i has a locally finite refinement 𝒱i\mathcal{V}_i. Accordingly the disjoint union iI𝒱i\underset{i \in I}{\sqcup}\mathcal{V}_i is a locally finite refinement of 𝒰\mathcal{U}.

(second-countable topological space)

A topological space is called second countable if it admits a base for its topology βX\beta_X (def. ) which is a countable set of open subsets.

(Euclidean space is second-countable)

Let nn \in \mathbb{N}. Consider the Euclidean space n\mathbb{R}^n with its Euclidean metric topology (example , example ). Then n\mathbb{R}^n is second countable (def. ).

A countable set of base open subsets is given by the open balls Bx(ϵ)B^\circ_x(\epsilon) of rational radius ϵ00\epsilon \in \mathbb{Q}_{\geq 0} \subset \mathbb{R}_{\geq 0} and centered at points with rational coordinates: xnnx \in \mathbb{Q}^n \subset \mathbb{R}^n.

To see that this is still a base, it is sufficient to see that every point inside very open ball in n\mathbb{R}^n is contains in an open ball of rational radius with rational coordinates of its center that is still itself contained in the original open ball.

To that end, let xx be a point inside an open ball and let d>0d \in \mathbb{R}_{\gt 0} be its distance from the boundary of the ball. By the fact that the rational numbers are a dense subset of \mathbb{R}, we may find epilon\epilon \in \mathbb{Q} such that 0<ϵ<d/20 \lt \epsilon \lt d/2 and then we may find xnnx' \in \mathbb{Q}^n \subset \mathbb{R}^n such that xBx(d/2)x' \in B_x^\circ(d/2). This open ball contains xx and is contained in the original open ball.

To see that this base is countable, use that

1. the set of rational numbers is countable;

2. the Cartesian product of two countable sets is countable.

(sigma-compact topological space)

A topological space is called sigma-compact if it is the union of a countable set of compact subsets (def. ).

(Euclidean space is sigma-compact)

For nn \in \mathbb{N} then the Euclidean space n\mathbb{R}^n (example ) equipped with its metric topology (example ) is sigma-compact (def. ).

For kk \in \mathbb{N} let

be the closed ball (def. ) of radius kk. By the Heine-Borel theorem (prop. ) these are compact subspaces. Clearly they exhaust n\mathbb{R}^n:

(locally compact and second-countable spaces are sigma-compact)

Let XX be a topological space which is

1. locally compact (def. ),

2. second-contable (def. ).

Then XX is sigma-compact (def. ).

We need to produce a countable cover of XX by compact subspaces.

By second-countability there exists a countable base of open subsets

By local compactness, every point xXx \in X has an open neighbourhood VxV_x whose topological closure Cl(Vx)Cl(V_x) is compact.

By definition of base of a topology (def. ), for each xXx \in X there exists BxβB_x \in \beta such that xBxVx{x} \subset B_x \subset V_x, hence such that Cl(Bx)Cl(Vx)Cl(B_x) \subset Cl(V_x).

Since subsets are closed in a closed subspace precisely if they are closed in the ambient space (lemma ), since Cl(Vx)Cl(V_x) is compact by assumption, and since closed subspaces of compact spaces are compact (lemma ) it follows that BxB_x is compact.

Applying this for each point exhibits XX as a union of compact closures of base opens:

But since there is only a countable set β\beta of base open subsets to begin with, there is a countable subset JXJ \subset X such that

Hence

is a countable cover of XX by compact subspaces.

(locally compact and sigma-compact space admits nested countable cover by coompact subspaces)

Let XX be a topological space which is

1. locally compact (def. );

2. sigma-compact (def. ).

Then there exists a countable open cover {UiX}i\{U_i \subset X\}_{i \in \mathbb{N}} of XX such that for each iIi \in I

1. the topological closure Cl(Ui)Cl(U_i) (def. ) is a compact subspace (def. , example );

2. Cl(Ui)Ui+1Cl(U_i) \subset U_{i +1}.

By sigma-compactness of XX there exists a countable cover {KiX}i\{K_i \subset X\}_{i \in \mathbb{N}} of compact subspaces. We use these to construct the required cover by induction.

For i=0i = 0 set

Then assume that for nn \in \mathbb{N} we have constructed a set {UiX}i{1,,n}\{U_i \subset X\}_{i \in \{1, \cdots, n\}} with the required properties.

In particular this implies that the union

is a compact subspace (by example ). We now construct an open neighbourhood Un+1U_{n+1} of this union as follows:

Let {UxX}xQn\{U_x \subset X\}_{x \in Q_n} be a set of open neighbourhood around each of the points in QnQ_n. By local compactness of XX, for each xx there is a smaller open neighbourhood VxV_x with

So {VxX}xQn\{V_x \subset X\}_{x \in Q_n} is still an open cover of QnQ_n. By compactness of QnQ_n, there exists a finite set JnQnJ_n \subset Q_n such that {VxX}xJn\{V_x \subset X\}_{x \in J_n} is a finite open cover. The union

is an open neighbourhood of QnQ_n, hence in particular of Cl(Un)Cl(U_n). Moreover, since finite unions of compact spaces are compact (example ), and since the closure of a finite union is the union of the closures (prop. ) the closure of Un+1U_{n+1} is compact:

In conclusion, by induction we have produced a set {UnX}i\{U_n \subset X\}_{i \in \mathbb{N}} with Cl(Ui)Cl(U_i) compact and Cl(Ui)Ui+1Cl(U_i) \subset U_{i+1} for all ii \in \mathbb{N}. It remains to see that this is a cover. This follows since by construction each Un+1U_{n+1} is an open neighbourhood not just of Cl(Un)Cl(U_{n}) but in fact of QnQ_n, hence in particular of KnK_n, and since the KnK_n form a cover by assumption:

(locally compact and sigma-compact spaces are paracompact)

Let XX be a topological space which is

1. locally compact;

2. sigma-compact.

Then XX is also paracompact.

Let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover of XX. We need to show that this has a refinement by a locally finite cover.

By lemma there exists a countable open cover {VnX}n\{V_n \subset X\}_{n \in \mathbb{N}} of XX such that for all nn \in \mathbb{N}

1. Cl(Vn)Cl(V_n) is compact;

2. Cl(Vn)Vn+1Cl(V_n) \subset V_{n+1}.

Notice that the complement Cl(Vn+1)VnCl(V_{n+1}) \setminus V_n is compact, since Cl(Vn+1)Cl(V_{n+1}) is compact and VnV_n is open, by example .

By this compactness, the cover {UiX}iI\{U_i \subset X\}_{i \in I} regarded as a cover of the subspace Cl(Vn+1)VnCl(V_{n+1})\setminus V_n has a finite subcover {UiX}iJn\{U_i \subset X\}_{i \in J_n} indexed by a finite set JnIJ_n \subset I, for each nn \in \mathbb{N}.

We consider the sets of intersections

Since Vn+2Cl(Vn1)V_{n+2} \setminus Cl(V_{n-1}) is open, and since Cl(Vn+1)Vn+2Cl(V_{n+1}) \subset V_{n+2} by construction, this 𝒰n\mathcal{U}_n is still an open cover of Cl(Vn+1)VnCl(V_{n+1})\setminus V_n. We claim now that

is a locally finite refinement of the original cover, as required:

1. 𝒰\mathcal{U} is a refinement, since by construction each element in 𝒰n\mathcal{U}_n is contained in one of the UiU_i;

2. 𝒰\mathcal{U} is still a covering because by construction it covers Cl(Vn+1)VnCl(V_{n+1}) \setminus V_n for all nn \in \mathbb{N}, and since by the nested nature of the cover {VnX}n\{V_n \subset X\}_{n \in \mathbb{N}} also {Cl(Vn+1)Vn}n\{Cl(V_{n+1}) \setminus V_n\}_{n \in \mathbb{N}} is a cover of XX.

3. 𝒰\mathcal{U} is locally finite because each point xXx \in X has an open neighbourhood of the form Vn+2Cl(Vn1)V_{n+2} \setminus Cl(V_{n-1}) (since these also form an open cover, by the nestedness) and since by construction this has trivial intersection with 𝒰n+3\mathcal{U}_{\geq n+3} and since all 𝒰n\mathcal{U}_n are finite, so that also k<n+3𝒰k\underset{k \lt n+3}{\cup} \mathcal{U}_k is finite.

(Euclidean space is paracompact)

For nn \in \mathbb{N}, the Euclidean space n\mathbb{R}^n (example ), regarded with its metric topology (example ) is a paracompact topological space (def. ).

The Euclidean space is locally compact by example and sigma-compact by example . Therefore the statement follows since locally compact and sigma-compact spaces are paracompact (prop. ).

(topological group)

A topological group is a group GG equipped with a topology τGP(G)\tau_G \subset P(G) (def. ) such that the group operation ()():G×GG(-)\cdot (-)\;\colon\;G \times G \to G and the assignment of inverse elements ()1:GG(-)^{-1} : G \to G are continuous functions.

(Euclidean space as a topological groups)

For nn \in \mathbb{N} then the Euclidean space n\mathbb{R}^n with its metric topology and equipped with the addition operation from its canonical vector space structure is a topological group (def. ) (n,+)(\mathbb{R}^n, +).

(open subgroups of topological groups are closed)

Every open subgroup HGH \subset G of a topological group (def. ) is closed.

The set of HH-cosets is a cover of GG by disjoint open subsets. One of these cosets is HH itself and hence it is the complement of the union of the other cosets, hence the complement of an open subspace, hence closed.

(locally compact topological groups are paracompact)

A topological group (def. ) which is locally compact (def. ) is paracompact (def. ).

By assumption of local compactness, there exists a compact neighbourhood CeGC_e \subset G of the neutral element. We may assume without restriction of generality that with gCeg \in C_e any element, then also the inverse element g1Ceg^{-1} \in C_e.

For if this is not the case, then we may enlarge CeC_e by including its inverse elements, and the result is still a compact neighbourhood of the neutral element: Since taking inverse elements ()1:GG(-)^{-1} \colon G \to G is a continuous function, and since continuous images of compact spaces are compact, it follows that also the set of inverse elements to elements in CeC_e is compact, and the union of two compact subspaces is still compact (example ).

Now for nn \in \mathbb{N}, write CenGC_e^n \subset G for the image of k{1,n}Cek{1,,n}G\underset{k \in \{1, \cdots n\}}{\prod} C_e \subset \underset{k \in \{1, \cdots, n\}}{\prod} G under the iterated group product operation k{1,,n}G⟶G\underset{k \in \{1, \cdots, n\}}{\prod} G \longrightarrow G.

Then

is clearly a topological subgroup of GG.

Observe that each CenC_e^n is compact. This is because k{1,,n}Ce\underset{k \in \{1, \cdots, n\}}{\prod}C_e is compact by the Tychonoff theorem (prop. ), and since continuous images of compact spaces are compact. Thus

is a countable union of compact subspaces, making it sigma-compact. Since locally compact and sigma-compact spaces are paracompact (prop. ), this implies that HH is paracompact.

Observe also that the subgroup HH is open, because it contains with the interior of CeC_e a non-empty open subset Int(Ce)HInt(C_e) \subset H and we may hence write HH as a union of open subsets

Finally, as indicated in the proof of Lemma , the cosets of the open subgroup HH are all open and partition GG as a disjoint union space (example ) of these open cosets. From this we may draw the following conclusions:

• In the particular case where GG is connected (def. ), there is just one such coset, namely HH itself. The argument above thus shows that a connected locally compact topological group is σ\sigma-compact and (by local compactness) also paracompact.

• In the general case, all the cosets are homeomorphic to HH which we have just shown to be a paracompact group. Thus GG is a disjoint union space of paracompact spaces. This is again paracompact by prop. .

(general linear group)

For nn \in \mathbb{N} the general linear group GL(n,)GL(n, \mathbb{R}) is the group of real n×nn \times n matrices whose determinant is non-vanishing

with group operation given by matrix multiplication.

This becomes a topological group (def. ) by taking the topology on GL(n,)GL_(n,\mathbb{R}) to be the subspace topology (def. ) as a subspace of the Euclidean space (example ) of matrices

with its metric topology (example ).

Since matrix multiplication is a polynomial function and since matrix inversion is a rational function, and since polynomials are continuous and more generally rational functions are continuous on their domain of definition (example ) and since the domain of definition for matrix inversion is precisely GL(n,)Matn×n()GL(n,\mathbb{R}) \subset Mat_{n \times n}(\mathbb{R}), the group operations on GL(n,)GL(n,\mathbb{R}) are indeed continuous functions.

(general linear group is subspace of the mapping space)

The topology induced on the real general linear group when regarded as a topological subspace of Euclidean space with its metric topology

(as in def. ) coincides with the topology induced by regarding the general linear group as a subspace of the mapping space Maps(kn,kn)Maps(k^n, k^n),

i.e. the set of all continuous functions knknk^n \to k^n equipped with the compact-open topology.

On the one hand, the universal property of the mapping space (prop. ) gives that the inclusion

is a continuous function for GL(n,)GL(n,\mathbb{R}) equipped with the Euclidean metric topology, because this is the adjunct of the defining continuous action map

This implies that the Euclidean metric topology on GL(n,)GL(n,\mathbb{R}) is equal to or finer than the subspace topology coming from Map(n,n)Map(\mathbb{R}^n, \mathbb{R}^n).

We conclude by showing that it is also equal to or coarser, together this then implies the claims.

Since we are speaking about a subspace topology, we may consider the open subsets of the ambient Euclidean space Matn×n()(n2)Mat_{n \times n}(\mathbb{R}) \simeq \mathbb{R}^{(n^2)}. Observe that a neighborhood base of a linear map or matrix AA consists of sets of the form

for ϵ(0,)\epsilon \in (0,\infty).

But this is also a base element for the compact-open topology, namely

where Ki{ei}K_i \coloneqq \{e_i\} is a singleton and ViBAei(ϵ)V_i \coloneqq B^\circ_{A e^i}(\epsilon) is the open ball of radius ϵ\epsilon around AeiA e^i.

(general linear group is paracompact Hausdorff)

The topological general linear group GL(n,)GL(n,\mathbb{R}) (def. ) is

1. not compact;

2. locally compact;

3. paracompact Hausdorff.

Observe that

is an open subspace, since it is the pre-image under the determinant function (which is a polynomial and hence continuous, example ) of the of the open subspace {0}\mathbb{R} \setminus \{0\} \subset \mathbb{R}:

As an open subspace of Euclidean space, GL(n,)GL(n,\mathbb{R}) is not compact, by the Heine-Borel theorem (prop. ).

As Euclidean space is Hausdorff (example ), and since every topological subspace of a Hausdorff space is again Hausdorff, so Gl(n,)Gl(n,\mathbb{R}) is Hausdorff.

Similarly, as Euclidean space is locally compact (example ) and since an open subspace of a locally compact space is again locally compact, it follows that GL(n,)GL(n,\mathbb{R}) is locally compact.

From this it follows that GL(n,)GL(n,\mathbb{R}) is paracompact, since locally compact topological groups are paracompact by prop. .

(Michael's theorem)

Let XX be a topological space such that

1. XX is regular;

2. every open cover of XX has a refinement by a union of a countable set of locally finite sets of open subsets (the latter not necessarily covering).

Then XX is paracompact topological space.

(metric spaces are paracompact)

A metric space (def. ) regarded as a topological space via its metric topology (example ) is paracompact (def. ).

(second-countable regular spaces are paracompact)

Let XX be a topological space which is

1. second-countable (def. );

2. regular (def. ).

Then XX is paracompact topological space.

Let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover. By Michael's theorem (prop. ) it is sufficient that we find a refinement by a countable cover (hence a countable union of sets consisting of single open subsets).

But second countability implies precisely that every open cover has a countable subcover:

Every open cover has a refinement by a cover consisting of base elements, and if there is only a countable set of these, then the resulting refinement necessarily contains at most this countable set of distinct open subsets.

(CW-complexes are paracompact Hausdorff spaces)

Let XX be a paracompact Hausdorff space, let nn \in \mathbb{N} and let

be a continuous function from the (n1)(n-1)-sphere (with its subspace topology inherited from Euclidean space, example ). Then also the attachment space (example ) XfDnX \cup_f D^n, i.e. the pushout

is paracompact Hausdorff.

This immediately implies that all finite CW-complexes (def. ) relative to a paracompact Hausdorff space are themselves paracompact Hausdorff. In fact this is true generally: all CW-complexes are paracompact Hausdorff spaces.

(every locally finite refinement induces one with the original index set)

Let (X,τ)(X,\tau) be a topological space, let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover (def. ), and let {VjX}jJ\{V_j \subset X\}_{j \in J}, be a refinement (def. ) to a locally finite cover (def. ).

By definition of refinement we may choose a function

such that

Then {WiX}iI\left\{ W_i \subset X \right\}_{i \in I} with

is still a refinement of {UiX}iI\{U_i \subset X\}_{i \in I} to a locally finite cover.

It is clear by construction that WiUiW_i \subset U_i, hence that we have a refinement. We need to show local finiteness.

Hence consider xXx \in X. By the assumption that {VjX}jJ\{V_j \subset X\}_{j \in J} is locally finite, it follows that there exists an open neighbourhood Ux{x}U_x \supset \{x\} and a finite subset KJK \subset J such that

Hence by construction

Since the image ϕ(K)I\phi(K) \subset I is still a finite set, this shows that {WiX}iI\{W_i \subset X\}_{i \in I} is locally finite.

(paracompact Hausdorff spaces are normal)

Every paracompact Hausdorff space (def. , def. ) is normal (def. ).

In particular compact Hausdorff spaces are normal.

Let (X,τ)(X,\tau) be a paracompact Hausdorff space

We first show that it is regular: To that end, let xXx \in X be a point, and let CXC \subset X be a closed subset not containing xx. We need to find disjoint open neighbourhoods Ux{x}U_x \supset \{x\} and UCCU_C \supset C.

First of all, by the Hausdorff property there exists for each cCc \in C disjoint open neighbourhods Ux,c{x}U_{x,c} \supset \{x\} and Uc{c}U_c \supset \{c\}. As cc ranges, the latter clearly form an open cover {UcX}cC\{U_c \subset X\}_{c \in C} of CC, and so the union

is an open cover of XX. By paracompactness of (X,τ)(X,\tau), there exists a locally finite refinement, and by lemma we may assume its elements to share the original index set and be contained in the original elements of the same index. Hence

is a locally finite collection of subsets, such that

is an open neighbourhood of CC.

Now by definition of local finiteness there exists an open neighbourhood Wx{x}W_x \supset \{x\} and a finite subset KCK \subset C such that

Consider then

which is an open neighbourhood of xx, by the finiteness of KK.

It thus only remains to see that

But this holds because the only VcV_{c} that intersect WxW_x are the VkUkV_{k} \subset U_{k} for kKk \in K and each of these is by construction disjoint from Ux,kU_{x,k} and hence from UxU_x.

This establishes that (X,τ)(X,\tau) is regular. Now we prove that it is normal. For this we use the same approach as before:

Let C,DXC,D \subset X be two disjoint closed subsets. By need to produce disjoint open neighbourhoods for these.

By the previous statement of regularity, we may find for each cCc \in C disjoint open neighbourhoods Uc{c}U_c \subset \{c\} and UD,cDU_{D,c} \supset D. Hence the union

is an open cover of XX, and thus by paracompactness has a locally finite refinement, whose elementes we may, again by lemma , assume to have the same index set as before and be contained in the previous elements with the same index. Hence we obtain a locally finite collection of subsets

such that

is an open neighbourhood of CC.

It is now sufficient to see that every point dDd \in D has an open neighbourhood UdU_d not intersecting UCU_C, for then

is the required open neighbourhood of DD not intersecting UCU_C.

Now by local finiteness of {VcX}cX\{V_c \subset X\}_{c \in X}, every dDd \in D has an open neighbourhood WdW_d such that there is a finite set KdCK_d \subset C so that

Accordingly the intersection

is still open and disjoint from the remaining VkV_k, hence disjoint from all of UCU_C.

(shrinking lemma for locally finite covers)

Let XX be a topological space which is normal (def. ) and let {UiX}iI\{U_i \subset X\}_{i \in I} be a locally finite open cover (def. ).

Then there exists another open cover {ViX}iI\{V_i \subset X\}_{i \in I} such that the topological closure Cl(Vi)Cl(V_i) of its elements is contained in the original patches:

(shrinking lemma for binary covers)

Let (X,τ)(X,\tau) be a normal topological space and let {UX}i{1,2}\{U \subset X\}_{i \in \{1,2\}} an open cover by two open subsets.

Then there exists an open set V1XV_1 \subset X whose topological closure is contained in U1U_1

and such that {V1,U2}\{V_1,U_2\} is still an open cover of XX.

Since U1U2=XU_1 \cup U_2 = X it follows (by de Morgan's law, prop. ) that their complements XUiX \setminus U_i are disjoint closed subsets. Hence by normality of (X,τ)(X,\tau) there exist disjoint open subsets

By their disjointness, we have the following inclusions:

In particular, since XV2X \setminus V_2 is closed, this means that Cl(V1)Cl(XV2)=XV2Cl(V_1) \subset Cl(X \setminus V_2) = X \setminus V_2.

Hence it only remains to observe that V1U2=XV_1 \cup U_2 = X, which is true by definition of V1V_1.

(shrinking lemma for finite covers)

Let (X,τ)(X,\tau) be a normal topological space, and let {UiX}i{1,,n}\{U_i \subset X\}_{i \in \{1, \cdots, n\}} be an open cover with a finite number nn \in \mathbb{N} of patches. Then there exists another open cover {ViX}iI\{V_i \subset X\}_{i \in I} such that Cl(Vi)UiCl(V_i) \subset U_i for all iIi \in I.

By induction, using lemma .

To begin with, consider {U1,i=2nUi}\{ U_1, \underoverset{i = 2}{n}{\cup} U_i\}. This is a binary open cover, and hence lemma gives an open subset V1XV_1 \subset X with V1Cl(V1)U1V_1 \subset Cl(V_1) \subset U_1 such that {V1,i=2nUi}\{V_1, \underoverset{i = 2}{n}{\cup} U_i\} is still an open cover, and accordingly so is

Similarly we next find an open subset V2XV_2 \subset X with V2Cl(V2)U2V_2 \subset Cl(V_2) \subset U_2 and such that

is an open cover. After nn such steps we are left with an open cover {ViX}i{1,,n}\{V_i \subset X\}_{i \in \{1, \cdots, n\}} as required.

Beware the induction in lemma does not give the statement for infinite countable covers. The issue is that it is not guaranteed that iVi\underset{i \in \mathbb{N}}{\cup} V_i is a cover.

And in fact, assuming the axiom of choice, then there exists a counter-example of a countable cover on a normal spaces for which the shrinking lemma fails (a Dowker space due to Beslagic 85).

(shrinking lemma for locally finite countable covers)

Let (X,τ)(X,\tau) be a normal topological space and {UiX}i\{U_i \subset X\}_{i \in \mathbb{N}} a locally finite countable cover. Then there exists open subsets ViXV_i \subset X for ii \in \mathbb{N} such that ViCl(Vi)UiV_i \subset Cl(V_i) \subset U_i and such that {ViX}i\{V_i \subset X\}_{i \in \mathbb{N}} is still a cover.

As in the proof of lemma , there exist ViV_i for ii \in \mathbb{N} such that ViCl(Vi)UiV_i \subset Cl(V_i) \subset U_i and such that for every finite number, hence every nn \in \mathbb{N}, then

Now the extra assumption that {UiX}iI\{U_i \subset X\}_{i \in I} is locally finite implies that every xXx \in X is contained in only finitely many of the UiU_i, hence that for every xXx \in X there exists nxn_x \in \mathbb{N} such that

This implies that for every xx then

hence that {ViX}i\{V_i \subset X\}_{i \in \mathbb{N}} is indeed a cover of XX.

of the general shrinking lemma, lemma .

Let {UiX}iI\{U_i \subset X\}_{i \in I} be the given locally finite cover of the normal space (X,τ)(X,\tau). Consider the set SS of pairs (J,𝒱)(J, \mathcal{V}) consisting of

1. a subset JIJ \subset I;

2. an II-indexed set of open subsets 𝒱={ViX}iI\mathcal{V} = \{V_i \subset X\}_{i \in I}

with the property that

1. (iJI)(Cl(Vi)Ui)(i \in J \subset I) \Rightarrow ( Cl(V_i) \subset U_i );

2. (iIJ)(Vi=Ui)(i \in I \setminus J) \Rightarrow ( V_i = U_i ).

3. {ViX}iI\{V_i \subset X\}_{i \in I} is an open cover of XX.

Equip the set SS with a partial order by setting

By definition, an element of SS with J=IJ = I is an open cover of the required form.

We claim now that a maximal element (J,𝒱)(J, \mathcal{V}) of (S,)(S,\leq) has J=IJ = I.

For assume on the contrary that (J,𝒱)(J,\mathcal{V}) is maximal and there were iIJi \in I \setminus J. Then we could apply the construction in lemma to replace that single ViV_i with a smaller open subset ViV'_i to obtain 𝒱\mathcal{V}' such that Cl(Vi)ViCl(V'_i) \subset V_i and such that 𝒱\mathcal{V}' is still an open cover. But that would mean that (J,𝒱)<(J{i},𝒱)(J,\mathcal{V}) \lt (J \cup \{i\}, \mathcal{V}'), contradicting the assumption that (J,𝒱)(J,\mathcal{V}) is maximal. This proves by contradiction that a maximal element of (S,)(S,\leq) has J=IJ = I and hence is an open cover as required.

We are reduced now to showing that a maximal element of (S,)(S,\leq) exists. To achieve this we invoke Zorn's lemma. Hence we have to check that every chain in (S,)(S,\leq), hence every totally ordered subset has an upper bound.

So let TST \subset S be a totally ordered subset. Consider the union of all the index sets appearing in the pairs in this subset:

Now define open subsets WiW_i for iKi \in K picking any (J,𝒱)(J,\mathcal{V}) in TT with iJi \in J and setting

This is independent of the choice of (J,𝒱)(J,\mathcal{V}), hence well defined, by the assumption that (T,)(T,\leq) is totally ordered.

Moreover, for iIKi \in I\setminus K define

We claim now that {WiX}iI\{W_i \subset X\}_{i \in I} thus defined is a cover of XX. Because by assumption that {UiX}iI\{U_i \subset X\}_{i \in I} is locally finite, so for every point xXx \in X there exists a finite set JxIJ_x \subset I such that (iIJx)(xUi)(i \in I \setminus J_x) \Rightarrow (x \notin U_i). Since (T,)(T,\leq) is a total order, it must contain an element (J,𝒱)(J, \mathcal{V}) such that the finite set JxKJ_x \cap K is contained in its index set JJ, hence JxKJJ_x \cap K \subset J. Since that 𝒱\mathcal{V} is a cover, it follows that xViiJxKiIVix \in \underset{i \in J_x \cap K}{V_i} \subset \underset{i \in I}{\cup} V_i, hence in iIWi\underset{i \in I}{\cup} W_i.

This shows that (K,𝒲)(K,\mathcal{W}) is indeed an element of SS. It is clear by construction that it is an upper bound for (T,)(T ,\leq ). Hence we have shown that every chain in (S,)(S,\leq) has an upper bound, and so Zorns lemma implies the claim.

(partition of unity)

Let (X,τ)(X,\tau) be a topological space, and let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover. Then a partition of unity subordinate to the cover is

• a set {fi}iI\{f_i\}_{i \in I} of continuous functions

  fi:X⟶[0,1]f_i \;\colon\; X \longrightarrow [0,1]

  (where [0,1][0,1] \subset \mathbb{R} is equipped with the subspace topology of the real numbers \mathbb{R} regarded as the 1-dimensional Euclidean space equipped with its metric topology);

such that with

denoting the support of fif_i (the topological closure of the subset of points on which it does not vanish) then

1. iI(Supp(fi)Ui)\underset{i \in I}{\forall} \left( Supp(f_i) \subset U_i \right);

2. {Supp(fi)X}iI\left\{ Supp(f_i) \subset X \right\}_{i \in I} is a locally finite cover (def. );

3. xX(iIfi(x)=1)\underset{x \in X}{\forall} \left( \underset{i \in I}{\sum} f_i(x) = 1 \right).

Regarding the definition of partition of unity (def. ) observe that:

1. Due to the second clause in def. , the sum in the third clause involves only a finite number of elements not equal to zero, and therefore is well defined.

2. Due to the third clause, the interiors of the supports {hi1((0,1])X}iI\left \{ h_i^{-1}( (0,1] ) \subset X \right\}_{i \in I} constitute a locally finite open cover:

   1. they are open, since they are the pre-images under the continuous functions fif_i of the open subset (0,1][0,1](0,1] \subset [0,1],

   2. they cover because, by the third clause, for each xxx \in x there is at least one iIi \in I with hi(X)>0h_i(X) \gt 0, hence xhi1((0,1])x \in h_i^{-1}((0,1])

   3. they are locally finite because by the second clause alreay their closures are locally finite.

Consider \mathbb{R} with its Euclidean metric topology.

Let ϵ(0,)\epsilon \in (0,\infty) and consider the open cover

Then a partition of unity {fn:[0,1]}n\{ f_n \colon \mathbb{R} \to [0,1] \}_{n \in \mathbb{N}} (def. )) subordinate to this cover is given by

(paracompact Hausdorff spaces equivalently admit subordinate partitions of unity)

Let (X,τ)(X,\tau) be a Hausdorff topological space (def. ). Then the following are equivalent:

1. (X,τ)(X,\tau) is a paracompact topological space (def. ).

2. Every open cover of (X,τ)(X,\tau) admits a subordinate partition of unity (def. ).

One direction is immediate: Assume that every open cover {UiX}iI\{U_i \subset X\}_{i \in I} admits a subordinate partition of unity {fi}iI\{f_i\}_{i \in I}. Then by definition (def. ) {Int(Supp(f)i)X}iI\{ Int(Supp(f)_i) \subset X\}_{i \in I} is a locally finite open cover refining the original one (remark ), hence XX is paracompact.

We need to show the converse: If (X,τ)(X,\tau) is a paracompact topological space, then for every open cover there is a subordinate partition of unity (def. ).

By paracompactness of (X,τ)(X,\tau), for every open cover there exists a locally finite refinement {UiX}iI\{U_i \subset X\}_{i \in I}, and by lemma we may assume that this has the same index set. It is now sufficient to show that this locally finite cover {UiX}iI\{U_i \subset X\}_{i \in I} admits a subordinate partition of unity, since this will then also be subordinate to the original cover.

Since paracompact Hausdorff spaces are normal (prop. ) we may apply the shrinking lemma to the given locally finite open cover {UiX}\{U_i \subset X\}, to obtain a smaller locally finite open cover {ViX}iI\{V_i \subset X\}_{i \in I}. Apply the lemma once more to that result to get a yet smaller open cover {WiX}iI\{W_i \subset X\}_{i \in I}, so that now

It follows that for each iIi \in I we have two disjoint closed subsets, namely the topological closure Cl(Wi)Cl(W_i) and the complement XViX \setminus V_i

Now since paracompact Hausdorff spaces are normal (prop. ), Urysohn's lemma (prop. ) says that there exist continuous functions of the form

with the property that

This means in particular that hi1((0,1])Vih_i^{-1}((0,1]) \subset V_i and hence that the support of the function is contained in UiU_i

By this construction, the set of function {hi}iI\{h_i\}_{i \in I} already satisfies conditions 1) and 2) on a partition of unity subordinate to {UiX}iI\{U_i \subset X\}_{i \in I} from def. . It just remains to normalize these functions so that they indeed sum to unity. To that end, consider the continuous function

defined on xXx \in X by

Notice that the sum on the right has only a finite number of non-zero summands, due to the local finiteness of the cover, so that this is well-defined. Moreover this is again a continuous function, since polynomials are continuous (example ).

Moreover, notice that

because {Cl(Wi)X}iI\{Cl(W_i) \subset X\}_{i \in I} is a cover so that there is ixIi_x \in I with xCl(Wix)x \in Cl(W_{i_x}), and since hi(Cl(Wix))={1}h_i(Cl(W_{i_x})) = \{1\}, by the above, and since all contributions to the sum are non-negative.

Hence it makes sense to define the ratios

Since Supp(fi)=Supp(hi)Supp(f_i) = Supp(h_i) this still satisfies conditions 1) and 2) on a partition of unity (def. ), but by construction this now also satisfies

and hence the remaining condition 3). Therefore

is a partition of unity as required.

(topological ring and topological field)

A topological ring is

1. a ring (R,+,)(R,+,\cdot),

2. a topology τRP(R)\tau_R \subset P(R) on the underlying set of the ring, making it a topological space (R,τR)(R,\tau_R) (def. )

such that

1. (R,+)(R,+) is a topological group with respect to τR\tau_R (def. );

2. also the multiplication ()():R×R⟶R(-)\cdot (-) \;\colon\; R \times R \longrightarrow R is a continuous function with respect to τR\tau_R and the product topology (example ).

A topological ring ((R,τR),+,)((R, \tau_R),+,\cdot) is a topological field if

1. (R,+,)(R,+,\cdot) is a field;

2. the function assigning multiplicative inverses ()1:R{0}R{0}(-)^{-1} \;\colon\; R \setminus \{0\} \to R \setminus \{0\} is a continuous function with respect to the subspace topology.

There is a redundancy in def. : For a topological ring the continuity of the assignment of additive inverses is already implied by the continuity of the multiplication operation, since

(real and complex numbers are topological fields)

The fields of real numbers \mathbb{R} and of complex numbers 2\mathbb{C} \simeq \mathbb{R}^2 are topological fields (def. ) with respect to their Euclidean metric topology (example , example )

That the operations on these fields are all continuous with respect to the Euclidean topology is the statement that rational functions are continuous on the domain of definition inside Euclidean space (example .)

(topological vector bundle)

Let

1. kk be a topological field (def. )

2. XX be a topological space.

Then a topological kk-vector bundle over XXis

1. a topological space EE;

2. a continuous function EπXE \overset{\pi}{\to} X

3. for each xXx \in X the stucture of a finite-dimensional kk-vector space on the pre-image

   Exπ1({x})EE_x \coloneqq \pi^{-1}(\{x\}) \subset E

   called the fiber of the bundle over xx

such that this is locally trivial in that there exists:

1. an open cover {UiX}iI\{U_i \subset X\}_{i \in I},

2. for each iIi \in I an nin_i \in \mathbb{N} and a homeomorphism

   ϕi:Ui×kni⟶π1(Ui)E\phi_i \;\colon\; U_i \times k^{n_i} \overset{\simeq}{\longrightarrow} \pi^{-1}(U_i) \subset E

   from the product topological space of UiU_i with the real numbers (equipped with their Euclidean space metric topology) to the restriction of EE over UiU_i, such that

   1. ϕi\phi_i is a function over UiU_i in that πϕi=pr1\pi \circ \phi_i = pr_1, hence in that ϕi({x}×kn)π1({x})\phi_i(\{x\} \times k^n) \subset \pi^{-1}(\{x\})

   2. ϕi\phi_i is a linear map in each fiber in that

      xUi(ϕi(x):kni⟶linearEx=π1({x})).\underset{x \in U_i}{\forall} \left( \phi_i(x) \;\colon\; k^{n_i} \overset{\text{linear}}{\longrightarrow} E_x = \pi^{-1}(\{x\}) \right) \,.

Here is the diagram of continuous functions that illustartes these conditions:

Often, but not always, it is required that the numbers nin_i are all equal to some nn \in \mathbb{N}, for all iIi \in I, hence that the vector space fibers all have the same dimension. In this case one says that the vector bundle has rank nn. (Over a connected topological space this is automatic, but the fiber dimension may be distinct over distinct connected components.)

For [E1π1X][E_1 \overset{\pi_1}{\to} X] and [E2ϕ2X][E_2 \overset{\phi_2}{\to} X] two topological vector bundles over the same base space, then a homomorphism between them is

• a continuous function f:E1⟶E2f \colon E_1 \longrightarrow E_2

such that

1. ff respects the projections: π2f=π1\pi_2 \circ f = \pi_1;

2. for each xXx \in X we have that f|x:(E1)x(E2)xf|_x \colon (E_1)_x \to (E_2)_x is a linear map.

(category of topological vector bundles)

For XX a topological space, there is the category whose

• objects are the topological vector bundles over XX,

• morphisms are the topological vector bundle homomorphisms

according to def. . This category is usually denoted Vect(X).

The set of isomorphism classes in this category (topological vector bundles modulo invertible homomorphism between them) we denote by Vect(X)/Vect(X)_{/\sim}.

There is a larger category, where we allow the morphisms to involve a continuous function f˜:X⟶Y\tilde f \colon X \longrightarrow Y between base spaces, so that the continuous functions on total spaces f:E1⟶E2f \colon E_1 \longrightarrow E_2 are, besides being fiberwise linear, required to make the follwoing diagram commute:

(some terminology)

Let kk and nn be as in def. . Then:

For k=k = \mathbb{R} one speaks of real vector bundles.

For k=k = \mathbb{C} one speaks of complex vector bundles.

For n=1n = 1 one speaks of line bundles, in particular of real line bundles and of complex line bundles.

(any two topologial vector bundles have local trivialization over a common open cover)

Let [E1X][E_1 \to X] and [E2X][E_2 \to X] be two topological vector bundles (def. ). Then there always exists an open cover {UiX}iI\{U_i \subset X\}_{i \in I} such that both bundles have a local trivialization over this cover.

By definition we may find two possibly different open covers {Ui11X}i1I1\{U^1_{i_1} \subset X\}_{{i_1} \in I_1} and {Ui22X}i2I2\{U^2_{i_2} \subset X\}_{i_2 \in I_2} with local tivializations {Ui11ϕi11E1|Ui11}i1I1\{ U^1_{i_1} \underoverset{\simeq}{\phi^1_{i_1}}{\to} E_1\vert_{U^1_{i_1}} \}_{i_1 \in I_1} and {Ui22ϕi22E2|Ui22}i2I2\{ U^2_{i_2} \underoverset{\simeq}{\phi^2_{i_2}}{\to} E_2\vert_{U^2_{i_2}} \}_{i_2 \in I_2}.

The joint refinement of these two covers is the open cover given by the intersections of the original patches:

The original local trivializations restrict to local trivializations on this finer cover

and

(topological trivial vector bundle and (local) trivialization)

For XX any topological space, and nn \in \mathbb{N}, we have that the product topological space

canonically becomes a topological vector bundle over XX (def. ). A local trivialization is given over the trivial cover {XX}\{X \subset X\} by the identity function ϕ\phi.

This is called the trivial vector bundle of rank nn over XX.

Given any topological vector bundle EXE \to X, then a choice of isomorphism to a trivial bundle (if it exists)

is called a trivialization of EE. A vector bundle for which a trivialization exists is called trivializable.

Accordingly, the local triviality condition in the definition of topological vector bundles (def. ) says that they are locally isomorphic to the trivial vector bundle. One also says that the data consisting of an open cover {UiX}iI\{U_i \subset X\}_{i \in I} and the homeomorphisms

as in def. constitute a local trivialization of EE.

(section of a topological vector bundle)

Let EπXE \overset{\pi}{\to} X be a topological vector bundle (def. ).

Then a homomorphism of vector bundles from the trivial line bundle (example , remark )

is, by fiberwise linearity, equivalently a continuous function

such that πσ=idX\pi \circ \sigma = id_X

Such functions σ:XE\sigma \colon X \to E are called sections (or cross-sections) of the vector bundle EE.

Namely ff by is necessarily of the form

for a unique such section σ\sigma.

(topological vector sub-bundle)

Given a topological vector bundel EXE \to X (def. ), then a sub-bundle is a homomorphism of topological vector bundles over XX

such that for each point xXx \in X this is a linear embedding of fibers

(This is a monomorphism in the category Vect(X)Vect(X) of topological vector bundles over XX (remark ).)

(topological projective space)

Let kk be a topological field (def. ) and nn \in \mathbb{N}. Consider the product topological space kn+1{1,,n+1}kk^{n+1} \coloneqq \underset{\{1,\cdots, n+1\}}{\prod} k, let kn+1{0}kn+1k^{n+1} \setminus \{0\} \subset k^{n+1} be the topological subspace which is the complement of the origin, and consider on its underlying set the equivalence relation which identifies two points if they differ by multiplication with some ckc \in k (necessarily non-zero):

The equivalence class [x][\vec x] is traditionally denoted

Then the projective space kPnk P^n is the corresponding quotient topological space

For k=k = \mathbb{R} this is called real projective space Pn\mathbb{R}P^n;

for k=k = \mathbb{C} this is called complex projective space Pn\mathbb{C}P^n.

(Riemann sphere)

The first complex projective space (def. ) is homeomorphic to the Euclidean 2-sphere (example )

Under this identification one also speaks of the Riemann sphere.

(standard open cover of topological projective space)

For nn \in \mathbb{N} the standard open cover of the projective space kPnk P^n (def. ) is

with

To see that this is an open cover:

1. This is a cover because with the orgin removed in kn{0}k^n \setminus \{0\} at every point [x1::xn+1][x_1: \cdots : x_{n+1}] at least one of the xix_i has to be non-vanishing.

2. These subsets are open in the quotient topology kPn=(kn{0})/kP^n = (k^n \setminus \{0\})/\sim, since their pre-image under the quotient co-projection kn+1{0}kPnk^{n+1} \setminus \{0\} \to k P^n coincides with the pre-image pri1(k{0})pr_i^{-1}( k \setminus \{0\} ) under the projection onto the iith coordinate in the product topological space kn+1=i{1,,n+1}kk^{n+1} = \underset{i \in \{1,\cdots, n+1\}}{\prod} k.

(canonical cover of Riemann sphere is the stereographic projection)

Under the identification P1S2\mathbb{C}P^1 \simeq S^2 of the first complex projective space as the Riemann sphere, from example , the canonical cover from def. is the cover by the two stereographic projections from example .

(topological tautological line bundle)

For kk a topological field (def. ) and nn \in \mathbb{N}, the tautological line bundle over the projective space kPnk P^n is topological kk-line bundle (remark ) whose total space is the following subspace of the product space (example ) of the projective space kPnk P^n (def. ) with knk^n:

where ⟨x⟩kkn+1\langle \vec x\rangle_k \subset k^{n+1} is the kk-linear span of x\vec x.

(The space TT is the space of pairs consisting of the name of a kk-line in kn+1k^{n+1} together with an element of that kk-line)

This is a bundle over projective space by the projection function

(tautological topological line bundle is well defined)

The tautological line bundle in def. is well defined in that it indeed admits a local trivialization.

We claim that there is a local trivialization over the canonical cover of def. . This is given for i{1,,n}i \in \{1, \cdots, n\} by

This is clearly a bijection of underlying sets.

To see that this function and its inverse function are continuous, hence that this is a homeomorphism notice that this map is the extension to the quotient topological space of the analogous map

This is a polynomial function on Euclidean space and since polynomials are continuous, this is continuous. Similarly the inverse function lifts to a rational function on a subspace of Euclidean space, and since rational functions are continuous on their domain of definition, also this lift is continuous.

Therefore by the universal property of the quotient topology, also the original functions are continuous.

(continuous functions on open subsets with values in the general linear group)

For nn \in \mathbb{N}, regard the general linear group GL(n,k)GL(n,k) as a topological group with its standard topology, given as the Euclidean subspace topology via GL(n,k)Matn×n(k)k(n2)GL(n,k) \subset Mat_{n \times n}(k) \simeq k^{(n^2)} or as the subspace topology GL(n,k)Maps(kn,kn)GL(n,k) \subset Maps(k^n, k^n) of the compact-open topology on the mapping space. (That these topologies coincide is the statement of this prop..

For XX a topological space, we write

for the assignment that sends an open subset UXU \subset X to the set of continuous functions g:UGL(n,k)g \colon U \to GL(n,k) (for UXU \subset X equipped with its subspace topology), regarded as a group via the pointwise group operation in GL(n,k)GL(n,k):

Moreover, for UUXU' \subset U \subset X an inclusion of open subsets, and for gGL(n,k)̲(U)g \in \underline{GL(n,k)}(U), we write

for the restriction of the continuous function from UU to UU'.

(sheaf of groups)

In the language of category theory the assignment GL(n,k)̲\underline{GL(n,k)} from def. of sets continuous functions to open subsets and the restriction operations between these is called a sheaf of groups on the site of open subsets of XX.

(transition functions)

Given a topological vector bundle EXE \to X as in def. and a choice of local trivialization {ϕi:Ui×knE|Ui}\{\phi_i \colon U_i \times k^n \overset{\simeq}{\to} E|_{U_i}\} (example ) there are for i,jIi,j \in I induced continuous functions

to the general linear group (as in def. ) given by composing the local trivialization isomorphisms:

These are called the transition functions for the given local trivialization.

(Cech cocycles)

Let XX be a topological space.

A normalized Cech cocycle of degree 1 with coefficients in GL(n,k)̲\underline{GL(n,k)} (def. ) is

1. an open cover {UiX}iI\{U_i \subset X\}_{i \in I}

2. for all i,jIi,j \in I a continuous function gij:UiUjGL(n,k)g_{i j} \colon U_i \cap U_j \to GL(n,k) as in def.

such that

1. (normalization) iI(gii=const1)\underset{i \in I}{\forall}\left( g_{i i} = const_1 \right) (the constant function on the neutral element in GL(n,k)GL(n,k)),

2. (cocycle condition) i,jI(gjkgij=gikonUiUjUk)\underset{i,j \in I}{\forall}\left( g_{j k} \cdot g_{i j} = g_{i k}\;\;\text{on}\, U_i \cap U_j \cap U_k\right).

Write

for the set of all such cocycles for given nn \in \mathbb{N} and write

for the disjoint union of all these cocycles as nn varies.

(transition functions are Cech cocycles)

Let EXE \to X be a topological vector bundle (def. ) and let {UiX}iI\{U_i \subset X\}_{i \in I}, {ϕi:Ui×knE|Ui}iI\{\phi_i \colon U_i \times k^n \overset{\simeq}{\to} E|_{U_{i}}\}_{i \in I} be a local trivialization (example ).

Then the set of induced transition functions {gij:UiUjGL(n)}\{g_{i j} \colon U_i \cap U_j \to GL(n)\} according to def. is a normalized Cech cocycle on XX with coefficients in GL(k)̲\underline{GL(k)}, according to def. .

This is immediate from the definition:

and

(topological vector bundle constructed from a Cech cocycle)

Let XX be a topological space and let cC1(X,GL(k)̲)c \in C^1(X, \underline{GL(k)}) a Cech cocycle on XX according to def. , with open cover {UiX}iI\{U_i \subset X\}_{i \in I} and component functions {gij}i,jI\{g_{i j}\}_{i,j \in I}.

This induces an equivalence relation on the product topological space

(of the disjoint union space of the patches UiXU_i \subset X regarded as topological subspaces with the product space kn={1,,n}kk^n = \underset{\{1,\cdots, n\}}{\prod} k) given by

Write

for the resulting quotient topological space. This comes with the evident projection

which is a continuous function (by the universal property of the quotient topological space construction, since the corresponding continuous function on the un-quotientd disjoint union space respects the equivalence relation). Moreover, each fiber of this map is identified with knk^n, and hence canonicaly carries the structure of a vector space.

Finally, the quotient co-projections constitute a local trivialization of this vector bundle over the given open cover.

Therefore E(c)XE(c) \to X is a topological vector bundle (def. ). We say it is the topological vector bundle glued from the transition functions.

(bundle glued from Cech cocycle is a coequalizer)

Stated more category theoretically, the constructure of a topological vector bundle from Cech cocycle data in example is a universal construction in topological spaces, namely the coequalizer of the two morphisms

in the category of vector space objects in the slice category Top/XTop/X. Here the restriction of ii to the coproduct summands is induced by inclusion:

and the restriction of μ\mu to the coproduct summands is via the action of the transition functions:

(topological vector bundle reconstructed from its transition functions)

Let [EπX][E \overset{\pi}{\to} X] be a topological vector bundle (def. ), let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover of the base space, and let {Ui×kn⟶ϕiE|Ui}iI\left\{ U_i \times k^n \underoverset{\simeq}{\phi_i}{\longrightarrow} E|_{U_i} \right\}_{i \in I} be a local trivialization.

Write

for the corresponding transition functions (def. ). Then there is an isomorphism of vector bundles over XX

from the vector bundle glued from the transition functions according to def. to the original bundle EE, whose components are the original local trivialization isomorphisms.

By the universal property of the disjoint union space (coproduct in Top), continuous functions out of them are equivalently sets of continuous functions out of every summand space. Hence the set of local trivializations {Ui×knϕiE|UiE}iI\{U_i \times k^n \underoverset{\simeq}{\phi_i}{\to} E|_{U_i} \subset E\}_{i \in I} may be collected into a single continuous function

By construction this function respects the equivalence relation on the disjoint union space given by the transition functions, in that for each xUiUjx \in U_i \cap U_j we have

By the universal property of the quotient space coprojection this means that (ϕi)iI(\phi_i)_{i \in I} uniquely extends to a continuous function on the quotient space such that the following diagram commutes

It is clear that this continuous function is a bijection. Hence to show that it is a homeomorphism, it is now sufficient to show that this is an open map (by prop. ).

So let OO be a subset in the quotient space which is open. By definition of the quotient topology this means equivalently that its restriction OiO_i to Ui×knU_i \times k^n is open for each iIi \in I. Since the ϕi\phi_i are homeomorphisms, it follows that the images ϕi(Oi)E|Ui\phi_i(O_i) \subset E\vert_{U_ i} are open. By the nature of the subspace topology, this means that these images are open also in EE. Therefore also the union f(O)=iIϕi(Oi)f(O) = \underset{i \in I}{\cup} \phi_i(O_i) is open.

(Moebius strip)

Let

be the circle with its Euclidean subspace metric topology. Consider the open cover

with

for any ϵ(0,2π/6)\epsilon \in (0,2\pi/6).

Define a Cech cohomology cocycle (remark ) on this cover by

Since there are no non-trivial triple intersections, all cocycle conditions are evidently satisfied.

Accordingly by example these functions define a vector bundle.

The total space of this bundle is homeomorphic to (the interior, def. of) the Moebius strip from example .

(basic complex line bundle on the 2-sphere)

Let

be the 2-sphere with its Euclidean subspace metric topology. Let

be the two complements of antipodal points

Define continuous functions

Since there are no non-trivial triple intersections, the only cocycle condition is

which is clearly satisfied.

The complex line bundle this defined is called the basic complex line bundle on the 2-sphere.

With the 2-sphere identified with the complex projective space P1\mathbb{C} P^1 (the Riemann sphere), the basic complex line bundle is the tautological line bundle (example ) on P1\mathbb{C}P^1.

(clutching construction)

Generally, for nn \in \mathbb{N}, n1n \geq 1 then the n-sphere SnS^n may be covered by two open hemispheres intersecting in an equator of the form Sn1×(ϵ,ϵ)S^{n-1} \times (-\epsilon, \epsilon). A vector bundle is then defined by specifying a single function

This is called the clutching construction of vector bundles over n-spheres.

(direct sum of vector bundles)

Let XX be a topological space, and let E1XE_1 \to X and E2XE_2 \to X be two topological vector bundles over XX.

Let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover with respect to which both vector bundles locally trivialize (this always exists: pick a local trivialization of either bundle and form the joint refinement of the respective open covers by intersection of their patches). Let

be the transition functions of these two bundles with respect to this cover.

For i,jIi, j \in I write

be the pointwise direct sum of these transition functions

Then the direct sum bundle E1E2E_1 \oplus E_2 is the one glued from this direct sum of the transition functions (by this construction):

(tensor product of vector bundles)

Let XX be a topological space, and let E1XE_1 \to X and E2XE_2 \to X be two topological vector bundles over XX.

Let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover with respect to which both vector bundles locally trivialize (this always exists: pick a local trivialization of either bundle and form the joint refinement of the respective open covers by intersection of their patches). Let

be the transition functions of these two bundles with respect to this cover.

For i,jIi, j \in I write

be the pointwise tensor product of vector spaces of these transition functions

Then the tensor product bundle E1E2E_1 \otimes E_2 is the one glued from this tensor product of the transition functions (by this construction):

(inner product on vector bundles)

Let

1. kk be a topological field (such as the real numbers or complex numbers with their Euclidean metric topology ),

2. XX be a topological space,

3. EXE \to X a topological vector bundle over XX (over \mathbb{R}, say).

Then an inner product on EE is

• a vector bundle homomorphism

  ⟨,⟩:EXE⟶X×\langle -,-\rangle \;\colon\; E \otimes_X E \longrightarrow X \times \mathbb{R}

  from the tensor product of vector bundles of EE with itself to the trivial line bundle

such that

• for each point xXx \in X the function

  ⟨,⟩|x:ExEx\langle -,-\rangle|_x \colon E_x \otimes E_x \to \mathbb{R}

  is an inner product on the fiber vector space, hence a positive-definite symmetric bilinear form.

(coboundary between Cech cocycles )

Let XX be a topological space and let c1,c2C1(X,GL(k)̲)c_1, c_2 \in C^1(X, \underline{GL(k)}) be two Cech cocycles (def. ), given by

1. {UiX}iI\{U_i \subset X\}_{i \in I} and {UiX}iI\{U'_i \subset X\}_{i' \in I'} two open covers,

2. {gij:UiUjGL(n,k)}i,jI\{g_{i j} \colon U_i \cap U_j \to GL(n,k)\}_{i,j \in I} and {gi,j:UiUjGL(n,k)}i,jI\{g'_{i',j'} \colon U'_{i'} \cap U'_{j'} \to GL(n',k) \}_{i', j' \in I'} the corresponding component functions.

Then a coboundary between these two cocycles is

1. the condition that n=nn = n',

2. an open cover {VαX}αA\{V_\alpha \subset X\}_{\alpha \in A}\,,

3. functionsϕ:AI\phi \colon A \to I and ϕ:AJ\phi' \colon A \to J such that αA((VαUϕ(α))and(VαUϕ(α)))\underset{\alpha \in A}{\forall}\left( \left( V_\alpha \subset U_{\phi(\alpha)} \right) \,\text{and}\, \left( V_\alpha \subset U'_{\phi'(\alpha)} \right) \right)

4. a set {κα:VαGL(n,k)}\{ \kappa_\alpha \colon V_\alpha \to GL(n,k) \} of continuous functions as in def.

such that

• α,βA(κβgϕ(α)ϕ(β)=gϕ(α)ϕ(β)καonVαVβ)\underset{ \alpha, \beta \in A }{\forall} \left( \kappa_{\beta} \cdot g_{\phi(\alpha) \phi(\beta)} = g'_{\phi'(\alpha) \phi'(\beta)} \cdot \kappa_{\alpha} \,\, \text{on}\,\, V_\alpha \cap V_\beta \right),

  hence such that the following diagrams of linear maps commute for all α,βA\alpha, \beta \in A and xVαVβx \in V_{\alpha} \cap V_\beta:

  kn⟶gϕ(α)ϕ(β)(x)knκα(x)κβ(x)kn⟶gϕ(α)ϕ(β)(x)kn.\array{ k^n &\overset{ g_{\phi(\alpha) \phi(\beta)}(x) }{\longrightarrow}& k^n \\ {}^{\mathllap{\kappa_{\alpha}(x)} }\downarrow && \downarrow^{\mathrlap{ \kappa_{\beta}(x) }} \\ k^n &\underset{ g'_{\phi'(\alpha) \phi'(\beta)}(x) }{\longrightarrow}& k^n } \,.

Say that two Cech cocycles are cohomologous if there exists a coboundary between them.

(refinement of a Cech cocycle is a coboundary)

Let XX be a topological space and let cC1(X,GL(k)̲)c \in C^1(X, \underline{GL(k)}) be a Cech cocycle as in def. , with respect to some open cover {UiX}iI\{U_i \subset X\}_{i \in I}, given by component functions {gij}i,jI\{g_{i j}\}_{i,j \in I}.

Then for {VαX}αA\{V_\alpha \subset X\}_{\alpha \in A} a refinement of the given open cover, hence an open cover such that there exists a function ϕ:AI\phi \colon A \to I with αA(VαUϕ(α))\underset{\alpha \in A}{\forall}\left( V_\alpha \subset U_{\phi(\alpha)} \right), then

are the components of a Cech cocycle cc' which is cohomologous to cc.

(isomorphism of topological vector bundles induces Cech coboundary between their transition functions)

Let XX be a topological space, and let c1,c2C1(X,GL(n,k)̲)c_1, c_2 \in C^1(X, \underline{GL(n,k)} ) be two Cech cocycles as in def. .

Every isomorphism of topological vector bundles

between the vector bundles glued from these cocycles according to def. induces a coboundary between the two cocycles,

according to def. .

By example we may assume without restriction that the two Cech cocycles are defined with respect to the same open cover {UiX}iI\{U_i \subset X\}_{i \in I} (for if they are not, then by example both are cohomologous to cocycles on a joint refinement of the original covers and we may argue with these).

Accordingly, by example the two bundles E(c1)E(c_1) and E(c2)E(c_2) both have local trivializations of the form

and

over this cover. Consider then for iIi \in I the function

hence the unique function making the following diagram commute:

This induces for all i,jIi,j \in I the following composite commuting diagram

By construction, the two horizonal composites of this diagram are pointwise given by the components gij1g^1_{i j} and gij2g^2_{i j}of the cocycles c1c_1 and c2c_2, respectively. Hence the commutativity of this diagram is equivalently the commutativity of these diagrams:

for all i,jIi,j \in I and xUiUjx \in U_i \cap U_j. By def. this exhibits the required coboundary.

(Cech cohomology)

Let XX be a topological space. The relation \sim on Cech cocycles of being cohomologous (def. ) is an equivalence relation on the set C1(X,GL(k)̲)C^1( X, \underline{GL(k)} ) of Cech cocycles (def. ).

Write

for the resulting set of equivalence classes. This is called the Cech cohomology of XX in degree 1 with coefficients in GL(k)̲\underline{GL(k)}.

(Cech cohomology computes isomorphism classes of topological vector bundle)

Let XX be a topological space.

The construction of gluing a topological vector bundle from a Cech cocycle (example ) constitutes a bijection between the degree-1 Cech cohomology of XX with coefficients in GL(n,k)GL(n,k) (def. ) and the set of isomorphism classes of topological vector bundles on XX (def. , remark ):

First we need to see that the function is well defined, hence that if cocycles c1,c2C1(X,GL(k)̲)c_1, c_2 \in C^1(X,\underline{GL(k)}) are related by a coboundary, c1c2c_1 \sim c_2 (def. ), then the vector bundles E(c1)E(c_1) and E(c2)E(c_2) are related by an isomorphism.

Let {VαX}αA\{V_\alpha \subset X\}_{\alpha \in A} be the open cover with respect to which the coboundary {κα:VαGL(n,k)}α\{\kappa_\alpha \colon V_\alpha \to GL(n,k)\}_{\alpha} is defined, with refining functions ϕ:AI\phi \colon A \to I and ϕ:AI\phi' \colon A \to I'. Let {nψϕ(α)|VαE(c1)|Vα}αA\left\{ \mathbb{R}^n \underoverset{\simeq}{\psi_{\phi(\alpha)}\vert_{V_\alpha} }{\to} E(c_1)\vert_{V_\alpha} \right\}_{\alpha \in A} and {Vα×knψϕ(α)|VαE(c2)|Vα}αA\left\{ V_\alpha \times k^n \underoverset{\simeq}{\psi'_{\phi'(\alpha)}\vert_{V_\alpha} }{\to} E(c_2)\vert_{V_\alpha} \right\}_{\alpha \in A} be the corresponding restrictions of the canonical local trivilizations of the two glued bundles.

For αA\alpha \in A define

Observe that for α,βA\alpha, \beta \in A and xVαVβx \in V_\alpha \cap V_\beta the coboundary condition implies that

because in the diagram

the vertical morphism in the middle on the right is unique, by the fact that all other morphisms in the diagram on the right are invertible.

Therefore by example there is a unique vector bundle homomorphism

given for all αA\alpha \in A by f|Vα=fαf\vert_{V_\alpha} = f_\alpha. Similarly there is a unique vector bundle homomorphism

given for all αA\alpha \in A by f1|Vα=fα1f^{-1}\vert_{V_\alpha} = f^{-1}_\alpha. Hence this is the required vector bundle isomorphism.

Finally to see that the function from Cech cohomology classes to isomorphism classes of vector bundles thus defined is a bijection:

By prop. the function is surjective, and by prop. it is injective.

(homomorphism of vector bundles is isomorphism as soon as it is a fiberwise isomorphism)

Let [E1X][E_1 \to X] and [E2X][E_2 \to X] be two topological vector bundles (def. ).

If a homomorphism of vector bundles f:E1⟶E2f \colon E_1 \longrightarrow E_2 restricts on the fiber over each point to a linear isomorphism

then ff is already an isomorphism of vector bundles.

It is clear that ff has an inverse function

of underlying sets f1:E2E1f^{-1} \colon E_2 \to E_1 which is a function over XX: Over each xXx \in X it it the linear inverse (f|x)1:(E2)x(E1)x(f\vert_x)^{-1} \colon (E_2)_x \to (E_1)_x.

What we need to show is that this is a continuous function.

By remark we find an open cover {UiX}iI\{U_i \subset X\}_{i \in I} over which both bundles have a local trivialization.

Restricted to any patch iIi \in I of this cover, the homomorphism f|Uif|_{U_i} induces a homomorphism of trivial vector bundles

Also the fif_i are fiberwise invertible, hence are continuous bijections. We claim that these are homeomorphisms, hence that their inverse functions (fi)1(f_i)^{-1} are also continuous.

To this end we re-write the fif_i a little. First observe that by the universal property of the product topological space and since they fix the base space UiU_i, the fif_i are equivalently given by a continuous function

as

Moreover since knk^n is locally compact (as every metric space), the mapping space adjunction says (by prop. ) that there is a continuous function

(with Maps(kn,kn)Maps(k^n,k^n) the set of continuous functions knknk^n \to k^n equipped with the compact-open topology) which factors hih_i via the evaluation map as

By assumption of fiberwise linearity the functions h˜i\tilde h_i in fact take values in the general linear group

and this inclusion is a homeomorphism onto its image (by this prop.).

Since passing to inverse matrices

is a rational function on its domain GL(n,k)Matn×n(k)k(n2)GL(n,k) \subset Mat_{n \times n}(k) \simeq k^{(n^2)} inside Euclidean space and since rational functions are continuous on their domain of definition, it follows that the inverse of fif_i

is a continuous function.

To conclude that also f1f^{-1} is a continuous function we make use prop. to find an isomorphism between E2E_2 and a quotient topological space of the form

Hence f1f^{-1} is equivalently a function on this quotient space, and we need to show that as such it is continuous.

By the universal property of the disjoint union space (the coproduct in Top) the set of continuous functions

corresponds to a single continuous function

These functions respect the equivalence relation, since for each xUiUjx \in U_i \cap U_j we have

Therefore by the universal property of the quotient topological space E2E_2, these functions extend to a unique continuous function E2E1E_2 \to E_1 such that the following diagram commutes:

This unique function is clearly f1f^{-1} (by pointwise inspection) and therefore f1f^{-1} is continuous.

(fiberwise linearly independent sections trivialize a vector bundle)

If a topological vector bundle EXE \to X of rank nn admits nn sections (example )

that are linearly independent at each point xXx \in X, then EE is trivializable (example ). In fact, with the sections regarded as vector bundle homomorphisms out of the trivial vector bundle of rank nn (according to example ), these sections are the trivialization

This is because their linear independence at each point means precisely that this morphism of vector bundles is a fiber-wise linear isomorphism and therefore an isomorphism of vector bundles by lemma .

(locally Euclidean topological space)

A topological space XX is locally Euclidean if every point xXx \in X has an open neighbourhood Ux{x}U_x \supset \{x\} which, as a subspace (example ), is homeomorphic (def. ) to the Euclidean space n\mathbb{R}^n (example ) with its metric topology (def. ):

(locally Euclidean spaces are T1T_1-separated, sober, locally compact, locally connected and locally path-connected topological spaces)

Let XX be a locally Euclidean space (def. ). Then

1. XX satisfies the T1T_1 separation axiom (def. );

2. XX is sober (def. );

3. XX is locally compact in the weak sense of def. .

4. XX is locally connected (def. ),

5. XX is locally path-connected (def. ).

Regarding the first statement:

Let xyx \neq y be two distinct points in the locally Euclidean space. We need to show that there is an open neighbourhood UxU_x around xx that does not contain yy.

By definition, there is a Euclidean open neighbourhood nϕUxX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U_x \subset X around xx. If UxU_x does not contain yy, then it already is an open neighbourhood as required. If UxU_x does contain yy, then ϕ1(x)ϕ1(y)\phi^{-1}(x) \neq \phi^{-1}(y) are equivalently two distinct points in n\mathbb{R}^n. But Euclidean space, as every metric space, is T1T_1 (example , prop. ), and hence we may find an open neighbourhood Vϕ1(x)nV_{\phi^{-1}(x)} \subset \mathbb{R}^n not containing ϕ1(y)\phi^{-1}(y). By the nature of the subspace topology, ϕ(Vϕ1(x))X\phi(V_{\phi^{-1}(x)}) \subset X is an open neighbourhood as required.

Regarding the second statement:

We need to show that the map

that sends points to the topological closure of their singleton sets is a bijection with the set of irreducible closed subsets. By the first statement above the map is injective (via lemma ). Hence it remains to see that every irreducible closed subset is the topological closure of a singleton. We will show something stronger: every irreducible closed subset is a singleton.

Let PXP \subset X be an open proper subset such that if there are two open subsets U1,U2XU_1, U_2 \subset X with U1U2PU_1 \cap U_2 \subset P then U1PU_1 \subset P or U2PU_2 \subset P. By prop ) we need to show that there exists a point xXx \in X such that P=X{x}P = X \setminus \{x\} it its complement.

Now since PXP \subset X is a proper subset, and since the locally Euclidean space XX is covered by Euclidean neighbourhoods, there exists a Euclidean neighbourhood nϕUX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U \subset X such that PUUP \cap U \subset U is a proper subset. In fact this still satisfies the condition that for U1,U2openUU_1, U_2 \underset{\text{open}}{\subset} U then U1U2PUU_1 \cap U_2 \subset P \cap U implies U1PUU_1 \subset P \cap U or U2PUU_2 \subset P \cap U. Accordingly, by prop. , it follows that nϕ1(PU)\mathbb{R}^n \setminus \phi^{-1}(P \cap U) is an irreducible closed subset of Euclidean space. Sine metric spaces are sober topological space as well as T1T_1-separated (example , prop. ), this means that there exists xnx \in \mathbb{R}^n such that ϕ1(PU)=n{x}\phi^{-1}(P \cap U) = \mathbb{R}^n \setminus \{x\}.

In conclusion this means that the restriction of an irreducible closed subset in XX to any Euclidean chart is either empty or a singleton set. This means that the irreducible closed subset must be a disjoint union of singletons that are separated by Euclidean neighbourhoods. But by irreducibiliy, this union has to consist of just one point.

Regarding the third statement:

Let xXx \in X be a point and let Ux{x}U_x \supset \{x\} be an open neighbourhood. We need to find a compact neighbourhood KxUxK_x \subset U_x.

By assumption there exists a Euclidean open neighbourhood nϕVxX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} V_x \subset X. By definition of the subspace topology the intersection UxVxU_x \cap V_x is still open as a subspace of VxV_x and hence ϕ1(UxVx)\phi^{-1}(U_x \cap V_x) is an open neighbourhood of ϕ1(x)n\phi^{-1}(x) \in \mathbb{R}^n.

Since Euclidean spaces are locally compact (example ), there exists a compact neighbourhood Kϕ1(x)nK_{\phi^{-1}(x)} \subset \mathbb{R}^n (for instance a sufficiently small closed ball around xx, which is compact by the Heine-Borel theorem, prop. ). Now since continuous images of compact spaces are compact prop. , it follows that also ϕ(K)X\phi(K) \subset X is a compact neighbourhood.

Regarding the last two statements:

We need to show that for every point xXx \in X and every [neighbourhood there exists a neighbourhood which is and a neighbourhood which is .]

By local Euclideanness there exists a chart nϕVxX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} V_x \subset X. Since Euclidean space is locally connected and locally path-connected (def. ), there is a connected and a path-connected neighbourhood of the pre-image ϕ1(x)\phi^{-1}(x) contained in the pre-image ϕ1(UxVx)\phi^{-1}( U_x \cap V_x ). Since continuous images of connected spaces are connected (prop. ), and since continuous images of path-connected spaces are path-connected (prop. ), the images of these neighbourhoods under ϕ\phi are neighbourhoods of xx as required.

(connected locally Euclidean spaces are path-connected)

For a locally Euclidean space (X,τ)(X,\tau) (def. ) the connected components (def. ) coincide with the path-connected components (def. ).

(locally Euclidean spaces are not necessarily T2T_2)

It might superficially seem that every locally Euclidean space (def. ) is necessarily a Hausdorff topological space, since Euclidean space, like any metric space, is Hausdorff, and since by definition the neighbourhood of every point in a locally Euclidean spaces looks like Euclidean space.

But this is not so, see the counter-example below, Hausdorffness is a non-local condition, as opposed to the T0T_0 and T1T_1 separation axioms.

(non-Hausdorff locally Euclidean spaces)

An example of a locally Euclidean space (def. ) which is a non-Hausdorff topological space, is the line with two origins (example ).

(equivalence of regularity conditions for locally Euclidean Hausdorff spaces)

Let XX be a locally Euclidean space (def. ) which is Hausdorff (def. ).

Then the following are equivalent:

1. XX is sigma-compact (def. ).

2. XX is second-countable (def. ).

3. XX is paracompact (def. ) and has a countable set of connected components (def. ).

First, observe that XX is locally compact in the strong sense of def. : By prop. every locally Euclidean space is locally compact in the weak sense that every neighbourhood contains a compact neighbourhood, but since XX is assumed to be Hausdorff, this implies the stronger statement by prop. .

1) \Rightarrow 2)

Let XX be sigma-compact. We show that then XX is second-countable:

By sigma-compactness there exists a countable set {KiX}iI\{K_i \subset X\}_{i \in I} of compact subspaces. By XX being locally Euclidean, each KiK_i admits an open cover by restrictions of Euclidean spaces. By their compactness, each KiK_i has a subcover

with JiJ_i a finite set. Since countable unions of countable sets are countable, we have obtained a countable cover of XX by Euclidean spaces {nϕi,jX}iI,jJi\{ \mathbb{R}^n \overset{\phi_{i,j}}{\to} X\}_{i \in I, j \in J_i}. Now Euclidean space itself is second countable (by example ), hence admits a countable set βn\beta_{\mathbb{R}^n} of base open sets. As a result the union iIjJiϕi,j(βn)\underset{{i \in I} \atop {j \in J_i}}{\cup} \phi_{i,j}(\beta_{\mathbb{R}^n}) is a base of opens for XX. But this is a countable union of countable sets, and since countable unions of countable sets are countable we have obtained a countable base for the topology of XX. This means that XX is second-countable.

1) \Rightarrow 3)

Let XX be sigma-compact. We show that then XX is paracompact with a countable set of connected components:

Since locally compact and sigma-compact spaces are paracompact (prop. ), it follows that XX is paracompact. By local connectivity (prop. ) XX is the disjoint union space of its connected components (def. , lemma ). Since, by the previous statement, XX is also second-countable it cannot have an uncountable set of connected components. (Because there must be at least one base open contained in every connected component.)

2)\Rightarrow 1) Let XX be second-countable, we need to show that it is sigma-compact.

This follows since locally compact and second-countable spaces are sigma-compact (lemma ).

3) \Rightarrow 1)

Now let XX be paracompact with countably many connected components. We show that XX is sigma-compact.

By local compactness, there exists an open cover {UiX}iI\{U_i \subset X\}_{i \in I} such that the topological closures {KiCl(Ui)X}iI\{K_i \coloneqq Cl(U_i) \subset X\}_{i \in I} constitute a cover by compact subspaces. By paracompactness there is a locally finite refinement of this cover. Since paracompact Hausdorff spaces are normal (prop. ), the shrinking lemma applies (lemma ) to this refinement and yields a locally finite open cover

as well as a locally finite cover {Cl(Vj)X}jJ\{Cl(V_j) \subset X\}_{j \in J} by closed subsets. Since this is a refinement of the orignal cover, all the Cl(Vj)Cl(V_j) are contained in one of the compact subspaces KiK_i. Since subsets are closed in a closed subspace precisely if they are closed in the ambient space (lemma ), the Cl(Vj)Cl(V_j) are also closed as subsets of the KiK_i. Since closed subsets of compact spaces are compact (lemma ) it follows that the Cl(Vj)Cl(V_j) are themselves compact and hence form a locally finite cover by compact subspaces.

Now fix any j0Jj_0 \in J.

We claim that for every jJj \in J there is a finite sequence of indices (j0,j1,,jn,jn=j)(j_0, j_1, \cdots, j_n, j_{n} = j) with the property that VjkVjk+1V_{j_k} \cap V_{j_{k+1}} \neq \emptyset for all k{0,,n}k \in \{0, \cdots, n\}.

To see this, first observe that it is sufficient to show sigma-compactness for the case that XX is connected. From this the general statement follows since countable unions of countable sets are countable. Hence assume that XX is connected. It follows from prop. that XX is path-connected.

Hence for any xVj0x \in V_{j_0} and yVjy \in V_{j} there is a path γ:[0,1]X\gamma \colon [0,1] \to X (def. ) connecting xx with yy. Since the closed interval is compact (example ) and since continuous images of compact spaces are compact (prop. ), it follows that there is a finite subset of the ViV_i that covers the image of this path. This proves the claim.

It follows that there is a function

which sends each VjV_j to the minimum natural number nn as above.

We claim now that for all nn \in \mathbb{N} the preimage of {0,1,,n}\{0,1, \cdots, n\} under this function is a finite set. Since countable unions of countable sets are countable this means that ff serves as a countable enumeration of the set JJ and hence implies that {Cl(Vj)X}jJ\{ Cl(V_j) \subset X\}_{j \in J} is a countable cover of XX by compact subspaces, hence that XX is sigma-compact.

We prove this last claim by induction. It is true for n=0n = 0 by construction, since f1({0})=Vj0f^{-1}(\{0\}) = V_{j_0}. Assume it is true for some nn \in \mathbb{N}, hence that f1({0,1,,n})f^{-1}(\{0,1, \cdots, n\}) is a finite set. Since finite unions of compact subspaces are again compact (example ) it follows that

is compact. By local finiteness of the {Cl(Vj)}jJ\{Cl(V_j)\}_{j \in J}, every point xKnx \in K_n has an open neighbourhood WxW_x that intersects only a finite set of the Cl(Vj)Cl(V_j). By compactness of KnK_n, the cover {WxKnfGiKn}xKn\{W_x \cap K_nfGi \subset K_n\}_{x \in K_n} has a finite subcover. In conclusion this implies that only a finite number of the VjV_j intersect KnK_n.

Now by definition f1({0,1,,n+1})f^{-1}(\{0,1,\cdots, n+1\}) is a subset of those VjV_j which intersect KnK_n, and hence itself finite.

(topological manifold)

A topological manifold is a topological space which is

1. locally Euclidean (def. ),

2. paracompact Hausdorff (def. , def. ).

If the Euclidean neighbourhoods nUxX\mathbb{R}^n \overset{\simeq}{\to} U_x \subset X of the points xXx \in X are all of dimension nn for some nn \in \mathbb{N}, then the topological manifold is said to be of dimension nn, too. Sometimes one also speaks of an nn-fold in this case.

(varying terminology regarding topological manifold)

Often a topological manifold (def. ) is required to be second-countable (def. ) or sigma-compact (def. ). But by prop. both conditions are implied by def. as soon as there is a countable set of connected components. Manifolds with uncountably many connected components are rarely considered in practice. The restriction to countably many connected components is strictly necessary for several important theorems (beyond the scope of our discussion here) such as:

1. the Whitney embedding theorem;

2. the embedding of smooth manifolds into formal duals of R-algebras.

(local chart, atlas and gluing function)

Given an nn-dimensional topological manifold XX (def. ), then

1. An open subset UXU \subset X and a homeomorphism ϕ:nAAU\phi \colon \mathbb{R}^n \overset{\phantom{A}\simeq\phantom{A}}{\to} U from the nn-dimensional Euclidean space is also called a local coordinate chart of XX.

2. An open cover of XX by local charts {nϕiUX}iI\left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U \subset X \right\}_{i \in I} is called an atlas of the topological manifold.

3. Given such an atlas then for each i,jIi,j \in I the induced homeomorphism

   nAAϕi1(UiUj)⟶AϕiAUiUj⟶Aϕj1Aϕj1(UiUj)AAn\mathbb{R}^n \supset \phantom{AA} \phi_i^{-1}(U_i \cap U_j) \overset{\phantom{A}\phi_i\phantom{A}}{\longrightarrow} U_i \cap U_j \overset{\phantom{A}\phi_j^{-1}\phantom{A}}{\longrightarrow} \phi_j^{-1}(U_i \cap U_j) \phantom{AA} \subset \mathbb{R}^n

   is called the coordinate transformation or gluing function from chart ii to chart jj.

graphics grabbed from Frankel

(differentiable functions between Euclidean spaces)

Let nn \in \mathbb{N} and let UnU \subset \mathbb{R}^n be an open subset of Euclidean space (example ).

Then a function f:U⟶f \;\colon\; U \longrightarrow \mathbb{R} is called differentiable at xUx\in U if there exists a linear map dfx:nd f_x : \mathbb{R}^n \to \mathbb{R} such that the following limit exists and vanishes as hh approaches zero from all directions at once:

This means that for all ϵ(0,)\epsilon \in (0,\infty) there exists an open neighbourhood VxUV_x\subseteq U of xx such that whenever x+hVx+h\in V we have f(x+h)f(x)dfx(h)h<ϵ\frac{f(x+h)-f(x) - d f_x(h)}{\Vert h\Vert} \lt \epsilon.

We say that ff is differentiable on a subset SS of UU if ff is differentiable at every xSx\in S, and we say that ff is differentiable if ff is differentiable on all of UU. We say that ff is continuously differentiable if it is differentiable and dfd f is a continuous function.

The function dfxd f_x is called the derivative or differential of ff at xx.

More generally, let n1,n2n_1, n_2 \in \mathbb{N} and let Un1U\subseteq \mathbb{R}^{n_1} be an open subset.

Then a function f:U⟶n2f \;\colon\; U \longrightarrow \mathbb{R}^{n_2} is differentiable if for all i{1,,n2}i \in \{1, \cdots, n_2\} the component function

is differentiable in the previous sense

In this case, the derivatives dfi:nd f_i \colon \mathbb{R}^n \to \mathbb{R} of the fif_i assemble into a linear map of the form

If the derivative exists at each xUx \in U, then it defines itself a function

to the space of linear maps from n1\mathbb{R}^{n_1} to n2\mathbb{R}^{n_2}, which is canonically itself a Euclidean space. We say that ff is twice continuously differentiable if dfd f is continuously differentiable.

Generally then, for kk \in \mathbb{N} the function ff is called kk-fold continuously differentiable or of class CkC^kif for all jkj \leq k the jj-fold differential djfd^j f exists and is a continuous function.

Finally, if ff is kk-fold continuously differentiable for all kk \in \mathbb{N} then it is called a smooth function or of class CC^\infty.

(chain rule for differentiable functions between Euclidean spaces)

Let n1,n2,n3n_1, n_2, n_3 \in \mathbb{N} and let

be two differentiable functions (def. ). Then the derivative of their composite is the composite of their derivatives:

(differentiable manifold and smooth manifold)

For p{}p \in \mathbb{N} \cup \{\infty\} then a pp-fold differentiable manifold or CpC^p-manifold for short is

1. a topological manifold XX (def. );

2. an atlas {nϕiX}\{\mathbb{R}^n \overset{\phi_i}{\to} X\} (def. ) all whose gluing functions are pp times continuously differentiable.

A pp-fold differentiable function between pp-fold differentiable manifolds

is

• a continuous function f:XYf \colon X \to Y

such that

• for all iIi \in I and jJj \in J then

  nAA(fϕi)1(Vj)⟶ϕif1(Vj)⟶fVj⟶ψj1n\mathbb{R}^n \supset \phantom{AA} (f\circ \phi_i)^{-1}(V_j) \overset{\phi_i}{\longrightarrow} f^{-1}(V_j) \overset{f}{\longrightarrow} V_j \overset{\psi_j^{-1}}{\longrightarrow} \mathbb{R}^{n'}

  is a pp-fold differentiable function between open subsets of Euclidean space.

(Notice that this in in general a non-trivial condition even if X=YX = Y and ff is the identity function. In this case the above exhibits a passage to a different, but equivalent, differentiable atlas.)

If a manifold is CpC^p differentiable for all pp, then it is called a smooth manifold. Accordingly a continuous function between differentiable manifolds which is pp-fold differentiable for all pp is called a smooth function,

(category Diff of differentiable manifolds)

In analogy to remark there is a category called Diffp{}_p (or similar) whose objects are CpC^p-differentiable manifolds and whose morphisms are CpC^p-differentiable functions, for given p{}p \in \mathbb{N} \cup \{\infty\}.

(diffeomorphism)

Given smooth manifolds XX and YY (def. ), then a smooth function

is called a diffeomorphism, if there is an inverse function

which is also a smooth function (hence if ff is an isomorphism in the category Diff{}_\infty from remark ).

(basic properties of diffeomorphisms)

Let X,YX,Y be differentiable manifolds (def. ). Let

be a diffeomorphisms (def. ) with inverse differentiable function

Then:

1. ff is in particular a homeomorphism (def. ) between the underlying topological spaces.

   Because, by definition, ff is in particular a continuous function, as is its inverse function gg.

2. The derivative (def. ) dfd f of takes values in invertible linear maps, i.e. dfx:TxXTf(x)Xd f_x \colon T_x X \overset{\simeq}{\to} T_{f(x)} X is a linear isomorphsm for all xXx \in X.

   This is because by the chain rule (prop. ) the defining equations

   gf=idXAAAAAfg=idYg \circ f = id_X \phantom{AAAAA} f \circ g = id_{Y}

   imply that

   dgf(x)dfx=d(idX)x=idTxXAAAAAdfxdgf(x)=d(idY)f(x)=idTf(x)Y.d g_{f(x)} \circ d f_x = d (id_X)_x = id_{T_x X} \phantom{AAAAA} d f_{x} \circ d g_{f(x)} = d (id_Y)_{f(x)} = id_{T_{f(x)} Y} \,.

(smooth structure)

Let XX be a topological manifold (def. ) and let

be two atlases (def. ), both making XX into a smooth manifold (def. ).

Then there is a diffeomorphism (def. ) of the form

precisely if the identity function on the underlying set of XX constitutes such a diffeomorphism. (Because if ff is a diffeomorphism, then also f1f=idXf^{-1}\circ f = id_X is a diffeomorphism.)

That the identity function is a diffeomorphism between XX equipped with these two atlases means, by definition , that

Notice that the functions on the right may equivalently be written as

showing their analogy to the gluing functions within a single atlas (def. ).

Hence diffeomorphism induces an equivalence relation on the set of smooth atlases that exist on a given topological manifold XX. An equivalence class with respect to this equivalence relation is called a smooth structure on XX.

(Cartesian space as a smooth manifold)

For nn \in \mathbb{N} then the Cartesian space n\mathbb{R}^n equipped with the atlas consisting of the single chart

is a smooth manifold, in particularly a pp-fold differentiable manifold for every pp \in \mathbb{N} according to def. .

Similarly the open disk DnD^n becomes a smooth manifold when equipped with the atlas whose single chart is the homeomorphism nDn\mathbb{R}^n \to D^n. from example .

(bijective smooth function which is not a diffeomorphism)

Regard the real line 1\mathbb{R}^1 as a smooth manifold via example . Consider the function

This is clearly a smooth function and its underlying function of sets is a bijection.

But it is not a diffeomorphism (def. ): The derivative vanishes at x=0x = 0, and hence it cannot be a diffeomorphism by remark .

(n-sphere as a smooth manifold)

For all nn \in \mathbb{N}, the n-sphere SnS^n becomes a smooth manfold, with atlas consisting of the two local charts that are given by the inverse functions of the stereographic projection from the two poles of the sphere onto the equatorial hyperplane

By the formula given in the proof of prop. the induced gluing function n{0}n{0}\mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n \setminus \{0\} are rational functions and hence smooth functions.

Finally the nn-sphere is a paracompact Hausdorff topological space. Ways to see this include:

1. Snn+1S^n \subset \mathbb{R}^{n+1} is a compact subspace by the Heine-Borel theorem (prop. ). Compact spaces are also paracompact (example ). Moreover, Euclidean space, like any metric space, is Hausdorff (example ), and subspaces of Hausdorff spaces are Hausdorff;

2. The nn-sphere has the structure of a CW-complex (example ) and CW-complexes are paracompact Hausdorff spaces (example ).

(exotic smooth structure)

The constructions in example and define smooth structures (def. ) on the topological spaces underlying the Euclidean spaces n\mathbb{R}^n and the n-spheres SnS^n. These are clearly the standard smooth structures that are used by default whenever these spaces are used in differential geometry, since the beginning of the topic in the work by Gauss 1827.

But since being a smooth manifold is extra structure on a topological space (as opposed to being a topological manifold, which is just extra property) it makes sense to ask whether n\mathbb{R}^n and SnS^n admit other smooth structures besides these standard ones. Remarkably, they do, for special values of the dimension nn. These are called exotic smooth structures. Here are some results:

In dimension 3\leq 3 there are no exotic smooth structures: Two smooth manifolds of dimension 3\leq 3 are diffeomorphic (def. ) as soon as their underlying topological space are homeomorphic (def. ).

For nn \in \mathbb{N} with n4n \neq 4 there is a unique smooth structure on the Euclidean space n\mathbb{R}^n (the standard one from example ).

There are uncountably many exotic smooth structures on Euclidean 4-space 4\mathbb{R}^4.

For each nn \in \mathbb{N}, n5n \geq 5 there is a finite set of smooth structures on the n-sphere SnS^n.

It is still unknown whether there is an exotic smooth structure on the 4-sphere S4S^4.

The only n-spheres with no exotic smooth structure in the range 5n615 \leq n \leq 61 are S5S^5, S6S^6, S12S^{12}, S56S^{56} and S61S^{61}.

For more on all of this see at exotic smooth structure.

(open subsets of differentiable manifolds are again differentiable manifolds)

Let XX be a kk-fold differentiable manifold (def. ) and let SXS \subset X be an open subset of the underlying topological space (X,τ)(X,\tau).

Then SS carries the structure of a kk-fold differentiable manifold such that the inclusion map SXS \hookrightarrow X is an open embedding of differentiable manifolds.

Since the underlying topological space of XX is locally connected (prop. ) it is the disjoint union space of its connected components (def. , lemma ).

Therefore we are reduced to showing the statement for the case that XX has a single connected component. By prop. this implies that XX is second-countable topological space.

Now a subspace of a second-countable Hausdorff space is clearly itself second countable and Hausdorff.

Similarly it is immediate that SS is still locally Euclidean: since XX is locally Euclidean every point xSXx \in S \subset X has a Euclidean neighbourhood in XX and since SS is open there exists an open ball in that (itself homeomorphic to Euclidean space) which is a Euclidean neighbourhood of xx contained in SS.

For the differentiable structure we pick these Euclidean neighbourhoods from the given atlas. Then the gluing functions for the Euclidean charts on SS are kk-fold differentiable follows since these are restrictions of the gluing functions for the atlas of XX.

(coordinate transformations are diffeomorphisms)

Let (X,{nϕiUiX}iI)\left( X, \left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X \right\}_{i \in I} \right) be a differentiable manifold (def. ). By example for all i,jIi,j \in I the open subsets

canonically are diffrentiable manifolds themselves. By definition of differentiable manifolds, the coordinate transformation functions

and

both are differentiable functions. Moreover they are bijective functions by assumption and by construction. This means that they are diffeomorphisms (def. ).

(general linear group as a smooth manifold)

For nn \in \mathbb{N}, the general linear group Gl(n,)Gl(n,\mathbb{R}) (example ) is a smooth manifold (as an open subspace of Euclidean space GL(n,)Matn×n()(n2)GL(n,\mathbb{R}) \subset Mat_{n \times n}(\mathbb{R}) \simeq \mathbb{R}^{(n^2)}, via example and example ).

The group operations, being polynomial functions, are clearly smooth functions with respect to this smooth manifold structure, and thus GL(n,)GL(n,\mathbb{R}) is a Lie group.

(standard open cover of projective space is atlas)

The charts of the standard open cover of projective space, from def. are homeomorphic to Euclidean space knk^n.

If xi0x_i \neq 0 then

and the representatives of the form on the right are unique.

This means that

is a bijection of sets.

To see that this is a continuous function, notice that it is the composite

of the function

with the quotient projection. Now ϕ^i\hat \phi_i is a polynomial function and since polynomials are continuous, and since the projection to a quotient topological space is continuous, and since composites of continuous functions are continuous, it follows that ϕi\phi_i is continuous.

It remains to see that also the inverse function ϕi1\phi_i^{-1} is continuous. Since

is a rational function, and since rational functions are continuous, it follows, by nature of the quotient topology, that ϕi\phi_i takes open subsets to open subsets, hence that ϕi1\phi_i^{-1} is continuous.

(real/complex projective space is smooth manifold)

For k{,}k \in \{\mathbb{R}, \mathbb{C}\} the topological projective space kPnk P^n (def. ) is a topological manifold (def. ).

Equipped with the standard open cover of def. regarded as an atlas by lemma , it is a differentiable manifold, in fact a smooth manifold (def. ).

By lemma kPnk P^n is a locally Euclidean space. Moreover, kPnk P^n admits the structure of a CW-complex (this prop. and this prop.) and therefore it is a paracompact Hausdorff space since CW-complexes are paracompact Hausdorff spaces. This means that it is a topological manifold.

It remains to see that the gluing functions of this atlas are differentiable functions and in fact smooth functions. But by lemma they are even rational functions.

(differentiable vector bundle)

Let kk be a differentiable field, specifically k{,}k \in \{\mathbb{R}, \mathbb{C}\} so that knk^n is equipped with the canonical differentiable sructure from example .

Given a differentiable manifold XX (def. ), then a differentiable k-vector bundle over XX of rank kk is

1. a differentiable manifold EE;

2. a differentiable function E⟶πXE \overset{\pi}{\longrightarrow} X (def. )

3. the structure of a kk-vector space on the fiber π1({x})\pi^{-1}(\{x\}) for all xXx \in X

such that there exists

1. an open cover of XX {dϕi}iI\{\mathbb{R}^d \underoverset{\simeq}{\phi_i}{\to}\}_{i \in I} by open subsets diffeomorphic to Euclidean space with its canonica smooth structure from example (hence an atlas exhibiting the smooth structure of XX)

2. for each iIi \in I a differentiable function as on the top of this diagram

   d×kn⟶ψiE||Uiϕipr1π||UiUi\array{ \mathbb{R}^d \times k^n && \overset{\psi_i}{\longrightarrow} && E\vert|_{U_i} \\ & {}_{\mathllap{\phi_i \circ pr_1}}\searrow && \swarrow_{\mathrlap{ \pi|\vert_{U_i} }} \\ && U_i }

   which makes this diagram commute and which is fiber-wise a linear map.

A homomorphism between differentiabe vector bundles [E1π1X][E_1 \overset{\pi_1}{\to}X] and [E2π2X][E_2 \overset{\pi_2}{\to} X] over the same base differentiable manifolds is a differentiable function as in the top of the following diagram

which makes this diagram commute and which restricts to a linear map

on the fiber over each point xXx \in X.

More generally, if [E1X1][E_1 \overset{}{\to} X_1] and [E2π2X2][E_2 \overset{\pi_2}{\to} X_2] are differentable vector bundles over possibly different differentiable base manifolds, then a homomorphism is a differentiable function f:E1E2f \colon E_1 \to E_2 together with a differentiable function f:XYf \colon X \to Y that make the diagram

commute and such that

is a linear map for all xXx \in X.

This yields a category (remark ) whose

• objects are the differentiable vector bundles;

• morphisms are the homomorphisms between these.

We write Vect(Diff)Vect(Diff) for this category.

Underlying a differentiable vector bundle, is a topological vector bundle (def. ). This yields a forgetful functor

to the category of topological vector bundles from remark .

(tangency relation on differentiable curves)

Let XX be a differentiable manifold of dimension nn (def. ) and let xXx \in X be a point. On the set of smooth functions of the form

such that

define the relations

and

saying that two such functions are related precisely if either there exists a chart (def. ) around xx such that (or else for all charts around xx it is true that) the first derivative of the two functions regarded via the given chart as functions 1n\mathbb{R}^1 \to \mathbb{R}^n, coincide at t=0t = 0 (with tt denoting the canonical coordinate function on \mathbb{R}).

(tangency is equivalence relation)

The two relations in def. are equivalence relations and they coincide.

First to see that they coincide, we need to show that if the derivatives in question coincide in one chart nϕiUiX\mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X, that then they coincide also in any other chart nϕjUjX\mathbb{R}^n \underoverset{\simeq}{\phi_j}{\to} U_j \subset X.

For brevity, write

for the intersection of the two charts.

First of all, since the derivative may be computed in any open neighbourhood around t=0t = 0, and since the differentiable functions γi\gamma_i are in particular continuous functions, we may restrict to the open neighbourhood

of 00 \in \mathbb{R} and consider the derivatives of the functions

and

But then by definition of the differentiable atlas, there is the differentiable gluing function

such that

for n{1,2}n \in \{1,2\}. The chain rule (prop. ) now relates the derivatives of these functions as

Since α\alpha is a diffeomorphism and since derivatives of diffeomorphisms are linear isomorphisms (by remark ), this says that the derivative of γnj\gamma_n^j is related to that of γni\gamma_n^i by a linear isomorphism , and hence

Finally, that either relation is an equivalence relation is immediate.

(tangent vector)

Let XX be a differentiable manifold and xXx \in X a point. Then a tangent vector on XX at xx is an equivalence class of the the tangency equivalence relation (def. , lemma ).

The set of all tangent vectors at xXx \in X is denoted TxXT_x X.

(real vector space structure on tangent vectors)

For XX a differentiable manifold of dimension nn and xXx \in X any point, let nϕUX\mathbb{R}^n \underoverset{\simeq}{\phi}{\to} U \subset X be a chart (def. ) of the given atlas, with xUXx \in U \subset X.

Then there is induced a bijection of sets

from the nn-dimensional Cartesian space to the set of tangent vectors at xx (def. ) given by sending vn\vec v \in \mathbb{R}^n to the equivalence class of the following differentiable curve:

For n⟶ϕUX\mathbb{R}^n \underoverset{\simeq}{\phi'}{\longrightarrow} U' \subset X another chart of the atlas with xUXx \in U' \subset X, then the linear isomorphism relating these two identifications is the derivative

of the gluing function of the two charts at the point xx:

This is also called the transition function between the two local identifications of the tangent space.

If {nϕiUiX}iI\left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X \right\}_{i \in I} is an atlas of the differentiable manifold XX, then the set of transition functions

defined this way satisfies the normalized Cech cocycle conditions (def. ) in that for all i,jIi,j \in I, xUiUjx \in U_i \cap U_j

1. gii(x)=idng_{i i}(x) = id_{\mathbb{R}^n};

2. gjkgij(x)=gik(x)g_{j k}\circ g_{i j}(x) = g_{i k}(x).

The bijectivity of the map is immediate from the fact that the first derivative of ϕ1γvϕ\phi^{-1}\circ \gamma^\phi_{\vec v} at ϕ1(x)\phi^{-1}(x) is

The formula for the transition function now follows with the chain rule (prop. ):

Similarly the Cech cocycle condition follows by the chain rule:

and the normalization simply by the fact that the derivative of the identity function at any point is the identity linear isomorphism:

(tangent space)

For XX a differentiable manifold and xXx \in X a point, then the tangent space of XX at xx is the set TxXT_x X of tangent vectors at xx (def. ) regarded as a real vector space via lemma .

(tangent bundle of Euclidean space)

If X=nX = \mathbb{R}^n is itself a Euclidean space, then for any two points x,yXx,y \in X the tangent spaces TxXT_x X and TyXT_y X (def. ) are canonically identified with each other:

Using the vector space (or just affine space) structure of X=nX = \mathbb{R}^n we may send every smooth function γ:X\gamma \colon \mathbb{R} \to X to the smooth function

This gives a linear bijection

and these linear bijections are compatible, in that for x,y,znx,y,z \in \mathbb{R}^n any three points, then

Moreover, by lemma , each tangent space is identified with n\mathbb{R}^n itself, and this identification in turn is compatible with all the above identifications:

Therefore it makes sense to canonically identify all the tangent spaces of Euclidean space with that Euclidean space itself.

In words, what this identification does is to use the additive group structure on n\mathbb{R}^n to translate any tangent vector at any point xnx \in \mathbb{R}^n to the corresponding tangent vector at 00. (Side remark: Hence this construction is not specific to n\mathbb{R}^n but applies to every Lie group and it fact to every coset space of a Lie group.)

As a result, the collection of all the tangent spaces of Euclidean space is naturally identified with the Cartesian product

equipped with the projection on the first factor

because then the pre-image of a singleton {x}n\{x\} \subset \mathbb{R}^n under this projection are canonically identified with the above tangent spaces:

This way, if we equip Tn=n×nT \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n with the product space topology, then Tn⟶πnT \mathbb{R}^n \overset{\pi}{\longrightarrow} \mathbb{R}^n becomes a trivial topological vector bundle (def. ).

This is called the tangent bundle of the Euclidean space n\mathbb{R}^n regarded as a differentiable manifold.

(chain rule is functoriality of tangent space construction on Euclidean spaces)

Consider the assignment that sends

1. every Euclidean space n\mathbb{R}^n to its tangent bundle TnT \mathbb{R}^n according to def. ;

2. every differentiable function f:n1⟶n2f \colon \mathbb{R}^{n_1} \longrightarrow \mathbb{R}^{n_2} (def. ) to the function on tangent vectors (def. ) induced by postcomposition with ff

   Tn1⟶f()Tn2[1⟶γn1][1⟶fγn2]\array{ T \mathbb{R}^{n_1} &\overset{f \circ (-)}{\longrightarrow}& T \mathbb{R}^{n_2} \\ \left[ \mathbb{R}^1 \overset{\gamma}{\longrightarrow} \mathbb{R}^{n_1} \right] &\mapsto& \left[ \mathbb{R}^1 \overset{ f \circ \gamma}{\longrightarrow} \mathbb{R}^{n_2} \right] }

By the chain rule (prop. ) we have that the derivative of the composite curve fγf \circ \gamma is

and hence that under the identification Tnn×nT \mathbb{R}^n \simeq \mathbb{R}^n \times \mathbb{R}^n of example this assignment takes ff to its derivative

Conversely, in the first form above the assignment ff()f \mapsto f \circ (-) manifestly respects composition (and identity functions). Viewed from the second perspective this respect for composition is once again the chain rule (prop. ) d(gf)=(df)(dg)d(g \circ f) = (d f)\circ (d g):

In the language of category theory this says that the assignment

is an endofunctor on the category CartSp whose

1. objects are the Euclidean spaces n\mathbb{R}^n for nn \in \mathbb{N};

2. morphisms are the differentiable functions between these (for any chosen differentiability class CkC^k with k>0k \gt 0).

In fact more is true: By example TXT X has the structure of a differentiable vector bundle (def. ) and the function TX⟶dfTXT X \overset{ d f}{\longrightarrow} T X is evidently a homomorphism of differentiable vector bundles

Therefore the tangent bundle functor on Euclidean spaces refines to one of the form

to the category of differentiable vector bundles (def. ).

(tangent bundle of a differentiable manifold)

Let XX be a differentiable manifold (def. ) with atlas {nϕiUiX}iI\left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X\right\}_{i \in I}.

Equip the set of all tangent vectors (def. ), i.e. the disjoint union of the sets of tangent vectors

with a topology τTX\tau_{T X} (def. ) by declaring a subset UTXU \subset T X to be an open subset precisely if for all charts nϕiUiX\mathbb{R}^n \underoverset{\simeq}{\phi_i}{\to} U_i \subset X we have that its preimage under

is open in the Euclidean space 2n\mathbb{R}^{2n} (example ) with its metric topology (example ).

Equipped with the function

this is called the tangent bundle of XX.

Equivalently this means that the tangent bundle TXT X is the topological vector bundle (def. ) which is glued (via example ) from the transition functions gijd(ϕj1ϕi)ϕ1()g_{i j} \coloneqq d(\phi_j^{-1} \circ \phi_i)_{\phi^{-1}(-)} from lemma :

(Notice that, by examples , each Ui×nTUiU_i \times \mathbb{R}^n \simeq T U_i is the tangent bundle of the chart UinU_i \simeq \mathbb{R}^n.)

The co-projection maps of this quotient topological space construction constitute an atlas

(tangent bundle is differentiable vector bundle)

If XX is a (p+1)(p+1)-times differentiable manifold, then the total space of the tangent bundle def. is a pp-times differentiable manifold in that

1. (TX,τTX)(T X, \tau_{T X}) is a paracompact Hausdorff space;

2. The gluing functions of the atlas {2ndϕiTUiTX}iI\left\{ \mathbb{R}^{2n} \underoverset{\simeq}{d \phi_i}{\to} T U_i \subset T X \right\}_{i \in I} are pp-times continuously differentiable.

Moreover, the projection π:TXX\pi \colon T X \to X is a pp-times continuously differentiable function.

In summary this makes TXXT X \to X a differentiable vector bundle (def. ).

First to see that TXT X is Hausdorff:

Let (x,v),(x,v)TX(x,\vec v), (x', \vec v') \in T X be two distinct points. We need to produce disjoint openneighbourhoods of these points in TXT X. Since in particular x,xXx,x' \in X are distinct, and since XX is Hausdorff, there exist disjoint open neighbourhoods Ux{x}U_x \supset \{x\} and Ux{x}U_{x'} \supset \{x'\}. Their pre-images π1(Ux)\pi^{-1}(U_x) and π1(Ux)\pi^{-1}(U_{x'}) are disjoint open neighbourhoods of (x,v)(x,\vec v) and (x,vectv)(x',\vect v'), respectively.

Now to see that TXT X is paracompact.

Let {UiTX}iI\{U_i \subset T X\}_{i \in I} be an open cover. We need to find a locally finite refinement. Notice that π:TXX\pi \colon T X \to X is an open map (by example and example ) so that {π(Ui)X}iI\{\pi(U_i) \subset X\}_{i \in I} is an open cover of XX.

Let now {nϕjVjX}jJ\{\mathbb{R}^n \underoverset{\simeq}{\phi_j}{\to} V_j \subset X\}_{j \in J} be an atlas for XX and consider the open common refinement

Since this is still an open cover of XX and since XX is paracompact, this has a locally finite refinement

Notice that for each jJj' \in J' the product topological space Vj×n2nV'_{j'} \times \mathbb{R}^n \subset \mathbb{R}^{2n} is paracompact (as a topological subspace of Euclidean space it is itself locally compact and second countable and since locally compact and second-countable spaces are paracompact, lemma ). Therefore the cover

has a locally finite refinement

We claim now that

is a locally finite refinement of the original cover. That this is an open cover refining the original one is clear. We need to see that it is locally finite.

So let (x,v)TX(x,\vec v) \in T X. By local finiteness of {VjX}jJ\{ V'_{j'} \subset X\}_{j' \in J'} there is an open neighbourhood Vx{x}V_x \supset \{x\} which intersects only finitely many of the VjXV'_{j'} \subset X. Then by local finiteness of {WkjVj}\{ W_{k_{j'}} \subset V'_{j_'}\}, for each such jj' the point (x,v)(x,\vec v) regarded in Vj×nV'_{j'} \times \mathbb{R}^n has an open neighbourhood UjU_{j'} that intersects only finitely many of the WkjW_{k_{j'}}. Hence the intersection π1(Vx)(jUj)\pi^{-1}(V_x) \cap \left( \underset{j'}{\cap} U_{j'} \right) is a finite intersection of open subsets, hence still open, and by construction it intersects still only a finite number of the WkjW_{k_{j'}}.

This shows that TXT X is paracompact.

Finally the statement about the differentiability of the gluing functions and of the projections is immediate from the definitions

(differentials of differentiable functions between differentiable manifolds)

Let XX and YY be differentiable manifolds and let f:X⟶Yf \;\colon\; X \longrightarrow Y be a differentiable function. Then the operation of postcomposition, which takes differentiable curves in XX to differentiable curves in YY,

descends at each point xXx \in X to the tangency equivalence relation (def. , lemma ) to yield a function on sets of tangent vectors (def. ), called the differential dfxd f_x of ff at xx

Moreover:

1. (linear dependence on the tangent vector) these differentials are linear functions with respect to the vector space structure on the tangent spaces from lemma , def. ;

2. (differentiable dependence on the base point) globally they yield a homomorphism of real differentiable vector bundles between the tangent bundles (def. , lemma ), called the global differential dfd f of ff

   df:TX⟶TY.d f \;\colon\; T X \longrightarrow T Y \,.

3. (chain rule) The assignment fdff \mapsto d f respects composition in that for XX, YY, ZZ three differentiable manifolds and for

   X⟶AfAY⟶AgAZX \overset{\phantom{A}f\phantom{A}}{\longrightarrow} Y \overset{\phantom{A}g\phantom{A}}{\longrightarrow} Z

   two composable differentiable functions then their differentials satisfy

   d(gf)=(dg)(df).d(g \circ f) = (d g) \circ (d f) \,.

All statements are to be tested on charts of an atlas for XX and for YY. On these charts the statement reduces to that of example .

(tangent functor)

In the language of category theory (remark ) the statement of prop. says that forming tangent bundles TXT X of differentiable manifolds XX and differentials dfd f of differentiable functions f:XYf \colon X \to Y constitutes a functor

from the category Diff of differentiable manifolds to the category of differentiable real vector bundles.

(vector field)

Let XX be a differentiable manifold with differentiable tangent bundle TXXT X \to X (def. ).

A differentiable section v:XTXv \colon X \to T X of the tangent bundle is called a (differentiable) vector field on XX. We write Γ(TX)\Gamma(T X) for the real vector space of tangent vector fields on XX.

(notation for tangent vectors in a chart)

Under the bijection of lemma one often denotes the tangent vector corresponding to the the ii-th canonical basis vector of n\mathbb{R}^n by

because under the identification of tangent vectors with derivations on the algebra of differentiable functions on XX as above then it acts as the operation of taking the iith partial derivative. The general tangent vector corresponding to vnv \in \mathbb{R}^n is then denoted by

Notice that this identification depends on the choice of chart, which is left implicit in this notation.

Sometimes, notably in texts on thermodynamics, one augments this notation to indicate the chart being used by listing the remaining coordinate functions as subscripts. For instance if two functions f,gf,g on a 2-dimensional manifold are used as coordinate functions for a local chart (i.e. so that x1=fx^1 = f and x2=gx^2 = g ), then one writes

for the tangent vectors x1\frac{\partial}{\partial x^1} and x2\frac{\partial}{\partial x^2}, respectively.

(immersion and submersion of differentiable manifolds)

Let f:X⟶Yf \colon X \longrightarrow Y be a differentiable function between differentiable manifolds.

If for each xXx \in X the differential (prop. )

is

1. an injective function, then ff is called an immersion of differentiable manifolds

2. a surjective function, then ff is called a submersion of differentiable manifolds.

3. a bijective function, then ff is called a local diffeomorphism.

(embedding of smooth manifolds)

An embedding of smooth manifolds is a smooth function f:XYf : X \hookrightarrow Y between smooth manifolds XX and YY (def. ) such that

1. the underlying continuous function is an embedding of topological spaces (def. );

2. ff is an immersion (def. ).

A closed embedding is an embedding such that the image f(X)Yf(X) \subset Y is a closed subset.

If XYX \hookrightarrow Y is an embedding of smooth manifolds, then XX is also called a submanifold of YY.

(immersions that are not embeddings)

Consider an immersion f:(a,b)2f \;\colon\; (a,b) \to \mathbb{R}^2 of an open interval into the Euclidean plane (or the 2-sphere) as shown on the right. This is not a embedding of smooth manifolds: around the points where the image crosses itself, the function is not even injective, but even at the points where it just touches itself, the pre-images under ff of open subsets of 2\mathbb{R}^2 do not exhaust the open subsets of (a,b)(a,b), hence do not yield the subspace topology.

As a concrete examples, consider the function

While this is an immersion and injective, it fails to be an embedding due to the points at t=±πt = \pm \pi touching the point at t=0t = 0:

Every open neighbourhood in 2\mathbb{R}^2 which contains the origin (0,0)(0,0) also contains the image ϕ((π,π+ϵ)(πϵ,π))\phi( (-\pi , -\pi + \epsilon) \sqcup (\pi-\epsilon, \pi) ) for some ϵ\epsilon and hence in the subspace topology on (π,π)athbbR2(-\pi,\pi) \hookrightarrow \athbb{R}^2 none of the intervals (δ,δ)(π,π)(-\delta, \delta) \subset (-\pi,\pi) is open, contrary to the actual Euclidean topolgy on (π,π)(-\pi,\pi).

graphics grabbed from Lee

(proper injective immersions are equivalently the closed embeddings)

Let XX and YY be smooth manifolds (def. ), and let f:XYf \colon X \to Y be a smooth function. Then the following are equivalent

1. ff is a proper injective immersion (def. , def. );

2. ff is a closed embedding of smooth manifolds (def. ).

Since topological manifolds are locally compact topological spaces (prop. ), this follows directly since injective proper maps to locally compact spaces are equivalently the closed embeddings by prop. .

(bump function)

A bump function is a function on Cartesian space n\mathbb{R}^n, for some nn \in \mathbb{R} with values in the real numbers \mathbb{R}

such that

1. bb is smooth (def. );

2. bb has compact support (def. ).

(a class of bump functions)

For every closed ball Bx0(ϵ)={xn|xx0ϵ}nB_{x_0}(\epsilon) = \{x \in \mathbb{R}^n \,\vert\, {\Vert x - x_0 \Vert} \leq \epsilon\} \subset \mathbb{R}^n (def. ) there exists a bump function b:nb \colon \mathbb{R}^n \to \mathbb{R} (def. ) with support Supp(b)Cl(b1((0,)))Supp(b) \coloneqq Cl\left( b^{-1}( (0,\infty) ) \right) being that closed ball:

Consider the function

given by

By construction the support of this function is the closed unit ball at the origin, Supp(ϕ)=B0(1)Supp(\phi) = B_0(1).

We claim that ϕ\phi is smooth. That it is smooth away from rx=0r \coloneqq {\Vert x \Vert} = 0 is clear, hence smoothness only need to be checked at r=0r = 0, where it amounts to demanding that all the derivatives of the exponential function vanish as r0r \to 0.

But that is the case since

This clearly tends to zero as r1r \to 1. A quick way to see this is to consider the inverse function and expand the exponential to see that this tends to \infty as r1r \to 1:

The form of the higher derivatives is the same but with higher inverse powers of (r21)(r^2 -1) and so this conclusion remains the same for all derivatives. Hence ϕ\phi is smooth.

Now for arbitrary radii ε>0\varepsilon \gt 0 define

This is clearly still smooth and Supp(ϕε)=B0(ϵ)Supp(\phi_{\varepsilon}) = B_0(\epsilon).

Finally the function xϕε(xx0)x \mapsto \phi_\varepsilon(x-x_0) has support the closed ball Bx0(ε)B_{x_0}(\varepsilon).

(open cover of smooth manifold admits locally finite refinement by closed balls)

Let XX be a smooth manifold (def. ) and let {UiX}iI\{U_i \subset X\}_{i \in I} be an open cover. Then there exists cover

which is a locally finite refinement of {UiX}iI\{U_i \subset X\}_{i \in I} with each patch diffeomorphic to a closed ball (def. ) regarded as a subspace of Euclidean space.

First consider the special case that XX is compact topological space (def. ).

Let

be a smooth atlas representing the smooth structure on XX (def. ) (hence an open cover by patches which are diffeomorphic to standard Euclidean space). The intersections

still form an open cover of XX. Hence for each point xXx \in X there is iIi \in I and jJj \in J with xUiVjx \in U_i \cap V_j. By the nature of the Euclidean space metric topology, there exists a closed ball BxB_x around ϕj1(x)\phi_j^{-1}(x) in ϕj1(UiVj)n\phi_j^{-1}(U_i \cap V_j) \subset \mathbb{R}^n. Its image ϕj(Bx)X\phi_j(B_x) \subset X is a neighbourhood of xXx \in X diffeomorphic to a closed ball.

The interiors of these balls form an open cover

of XX which, by construction, is a refinement of {UiX}iI\{U_i \subset X\}_{i \in I}. By the assumption that XX is compact, this has a finite subcover

for LL a finite set. Hence

is a finite cover by closed balls, hence in particular locally finite, and by construction it is still a refinement of the orignal cover. This shows the statement for XX compact.

Now for general XX, notice that without restriction we may assume that XX is connected (def. ), for if it is not, then we obtain the required refinement on all of XX by finding one on each connected component (def. ), and so we are immediately reduced to the connected case.

But, by the proof of prop. , if a locally Euclidean paracompact Hausdorff space XX is connected, then it is sigma-compact and in fact admits a countable increasing exhaustion

by open subsets whose topological closures

exhaust XX by compact subspaces KnK_n.

For nn \in \mathbb{N}, consider the open subspace

which canonically inherits the structure of a smooth manifold by example . As above we find a refinement of the restriction of {UiX}iI\{U_i \subset X\}_{i \in I} to this open subset by closed balls and since the further subspace Kn+1KnK_{n+1}\setminus K_n is still compact (example ) there is a finite set LnL_n such that

is a finite cover of Kn+1KnK_{n+1} \setminus K_n by closed balls refining the original cover.

It follows that the union of all these

is a refinement by closed balls as required. Its local finiteness follows by the fact that each BlnB_{l_n} is contained in the strip Vn+2Kn1V_{n+2} \setminus K_{n-1}, each strip contains only a finite set of BlnB_{l_n}-s and each strip intersects only a finite number of other strips. (Hence an open subset around a point xx which intersects only a finite number of elements of the refined cover is given by any one of the balls BlnB_{l_n} that contain xx.)

(smooth manifolds admit smooth partitions of unity)

Let XX be a smooth manifold (def. ). Then every open cover {UiX}iI\{U_i \subset X\}_{i \in I} has a subordinate partition of unity (def. ) by functions {fi:Ui}iI\{f_i \colon U_i \to \mathbb{R}\}_{i \in I} which are smooth functions.

By lemma the given cover has a locally finite refinement by closed subsets diffeomorphic to closed balls:

Given this, let

be the function which on VjV_j is given by a smooth bump function (def. , example )

with support supp(bj)=B0(ϵj)supp(b_j) = B_0(\epsilon_j):

By the nature of bump functions this is indeed a smooth function on all of XX. By local finiteness of the cover by closed balls, the function

given by

is well defined (the sum involves only a finite number of non-vanishing contributions), non vanishing (since every point is contained in the support of one of the hih_i) and is smooth (since finite sums of smooth functions are smooth). Therefore if we set

for all jJj \in J then

is a subordinate partition of unity by smooth functions as required.

(weak embedding theorem)

For every compact (def. ) smooth manifold XX of finite dimension (def ), there exists some kk \in \mathbb{N} such that XX has an embedding of smooth manifolds (def. ) into the Euclidean space of dimension kk, regarded as a smooth manifold via example :

Let

be an atlas exhibiting the smooth structure of XX (def. ), hence an open cover by pathces diffeomorphic to Euclidean space. By compactness there exists a finite subset JIJ \subset I such that

is still an open cover.

Since XX is a smooth manifold, there exists a partition of unity {fiC(X,)}iJ\{f_i \in C^\infty(X,\mathbb{R})\}_{i \in J } subordinate to this cover (def. ) with smooth functions fif_i (by prop. ).

This we may use to extend the inverse chart identifications

to smooth functions on all of XX

by setting

The idea now is to use the universal property of the product topological space to combine all these functions to obtain an injective function of the form

This function is an immersion: On the interior of the support of the bump functions the product functions fiψif_i\cdot \psi_i have smooth inverses ψi1fi\tfrac{\psi_i^{-1}}{f_i} and therefore their differentials have vanishing kernel.

Hence it remains to see that the function is also an embedding of topological spaces.

Observe that it is an injective function: If two points x,yXx,y \in X have the same image, this means that they have the same image under all the fiψif_i \cdot \psi_i. But where these are non-vanishing, they are bijective. Moreover, since their supports cover XX, not all of them vanish on xx and yy. Therefore xx and yy must be the same.

Hence we have an injective immersion. With this prop. says that it is now sufficient to show that we also have a closed map. But this follows generally since XX is a compact topological space by assumption, and since Euclidean metric space is a Hausdorff topological space (example ), and since maps from compact spaces to Hausdorff spaces are closed and proper (prop. ).

(Whitney embedding theorem)

The Whitney embedding theorem (which we do not prove here) strengthens the statement of prop. in two ways:

1. it applies to non-compact smooth manifolds with a countable set of connected components;

2. it gives the upper bound of 2n2n on the dimension of the ambient Euclidean space (for embedding of nn-dimensional manifolds) which turns out to be the minimal ambient dimension such that all nn-manifolds have an embedding.