10.4 Induced topologies

中文

Induced topologies and topological products

Given a topological space $(X,\mathcal{O})$ and a map $f:Y\to X$ from an arbitrary set $Y$ into $X$, we can ask for the weakest topology on $Y$ for which this map is continuous – it is useless to ask for the finest such topology since, as shown in Example 10.10, the discrete topology on $Y$ always achieves this end. This is known as the topology induced on $Y$ by the map $f$. Let ${\mathcal{O}}_f$ be the family of all inverse images of open sets of $X$,

$${\mathcal{O}}_f=\{f^{-1}(U)|U\in \mathcal{O}\}.$$

Since $f$ is required to be continuous, all members of this collection must be open in the induced topology. Furthermore, ${\mathcal{O}}_f$ is a topology on $Y$ since (i) property (Top1) is trivial, as $\varnothing =f^{-1}(\varnothing )$ and $Y=f^{-1}(X);$ (ii) the axioms (Top2) and (Top3) follow from the

set-theoretical identities

$$f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V)\text{ and }\bigcup _{i\in I}{f}^{-1}(U_i)=f^{-1}\left(\bigcup _{i\in I}{U}_i\right).$$

Hence ${\mathcal{O}}_f$ is a topology on $Y$ and is included in any other topology such that the map $f$ is continuous. It must be the topology induced on $Y$ by the map $f$ since it is the coarsest possible such topology.

Example 10.12

Let $(X,\mathcal{O})$ be any topological space and $A$ any subset of $X$. In the topology induced on $A$ by the natural inclusion map $i_A:A\to X$ defined by $i_A(x)=x$ for all $x\in A$, a subset $B$ of $A$ is open iff it is the intersection of $A$ with an open set of $X;$ that is, $B=A\cap U$ where $U$ is open in $X$. This is precisely the relative topology on $A$ defined in Section 10.2. The relative topology is thus the coarsest topology on $A$ for which the inclusion map is continuous.

More generally, for a collection of maps $\{f_i:Y\to X_i|i\in I\}$ where $X_i$ are topological spaces, the weakest topology on $Y$ such that all these maps are continuous is said to be the topology induced by these maps. To create this topology it is necessary to consider the set of all inverse images of open sets $\mathcal{U}=\{f_i^{-1}(U_i)|U_i\in {\mathcal{O}}_i\}$ . This collection of sets is not itself a topology in general, the topology generated by these sets will be the coarsest topology on $Y$ such that each function $f_i$ is continuous.

Given two topological spaces $(X,{\mathcal{O}}_X)$ and $(Y,{\mathcal{O}}_Y)$ ), let ${\mathrm{pr}}_1:X\times Y\to X$ and ${\mathrm{pr}}_2:X\times Y\to Y$ be the natural projection maps defined by

$${\mathrm{pr}}_1(x,y)=x\text{ and }{\mathrm{pr}}_2(x,y)=y.$$

The product topology on the set $X\times Y$ is defined as the topology induced by these two maps. The space $X\times Y$ together with the product topology is called the topological product of $X$ and $Y$. It is the coarsest topology such that the projection maps are continuous. The inverse image under ${\mathrm{pr}}_1$ of an open set $U$ in $X$ is a ‘vertical strip $U\times Y$ , while the inverse image of an open set $V$ in $Y$ under ${\mathrm{pr}}_2$ is a ‘horizontal strip $X\times V$ . The intersection of any pair of these strips is a set of the form $U\times V$ where $U$ and $V$ are open sets from $X$ and $Y$ respectively (see Fig. 10.4). Since the topology generated by the vertical and horizontal strips consists of all possible unions of such intersections, it follows that in the product topology a subset $A\subset X\times Y$ is open if for every point $(x,y)\in A$ there exist open sets $U\in {\mathcal{O}}_X$ and $V\in {\mathcal{O}}_Y$ such that $(x,y)\in U\times V\subseteq A$

Figure 10.4 Product of two topological spaces

Given an arbitrary collection of sets $\{X_i\mid i\in I\}$ , their cartesian product $P={\prod }_{i\in I}{X}_i$ is defined as the set of maps $f:I\to {\bigcup }_i{X}_i$ such that $f(i)\in X_i$ for each $i\in I$ . For a finite number of sets, taking $I=\{1,2,\dots ,n\}$ , this concept is identical with the set of $n-$ tuples from $X_1\times X_2\times \cdots \times X_n$ . The product topology on $P$ is the topology induced by the projection maps ${\mathrm{pr}}_i:P\to X_i$ defined by ${\mathrm{pr}}_i(f)=f(i)$ . This topology is coarser than the topology generated by all sets of the form ${\Pi }_{i\in I}{U}_i$ where $U_i$ is an open subset of $X_i$

Example 10.13

Let $S^1$ be the unit circle in ${\mathbb{R}}^2$ defined by $x^2+y^2=1$ , with the relative topology. The product space $S^1\times S^1$ is homeomorphic to the torus $T^2$ or ‘donut’, with topology induced from its embedding as a subset of ${\mathbb{R}}^3$ . This can be seen by embedding $S^1$ in the $z=0$ plane of ${\mathbb{R}}^3$ , and attaching a vertical unit circle facing outwards from each point on $S^1$ . As the vertical circles ‘sweep around’ the horizontal cirle the resulting circle is clearly a torus.

The following is an occasionally useful theorem.

Theorem 10.3 · Continuity of slice inclusions into a product space

If $X$ and $Y$ are topological spaces then for each point $x\in X$ the injection map ${\iota }_x:Y\to X\times Y$ defined by ${\iota }_x(y)=(x,y)$ is continuous. Similarly the map ${\iota }_y^{\prime }:X\to X\times Y$ defined by ${\iota }_y^{\prime }(x)=(x,y)$ is continuous.

Proof

Let $U$ and $V$ be open subsets of $X$ and $Y$ respectively. Then

$$({\iota }_x{)}^{-1}(U\times V)=\{\begin{array}{ll}V& \text{ if }x\in U,\\ \varnothing & \text{ if }x\notin U.\end{array}$$

Since every open subset of $U\times V$ in the product topology is a union of sets oftype $U\times V,$ it follows that the inverse image under ${\iota }_x$ of every open set in $X\times Y$ is an open subset of $Y$. Hence the map ${\iota }_x$ is continuous. Similarly for the map ${\iota }_y^{\prime }$ .

Topology by identification

We may also reverse the above situation. Let $(X,\mathcal{O})$ be a topological space, and $f:X\to Y$ a map from $X$ onto an arbitrary set $Y$. In this case the topology on $Y$ induced by $f$ is defined to be the finest topology such that $f$ is continuous. This topology consists of all subsets $U\subseteq Y$ such that $f^{-1}(U)$ is open in $X;$ that is, ${\mathcal{O}}_Y=\{f^{-1}(U)|U\in \mathcal{O}\}$

Exercise

Show that ${\mathcal{O}}_Y$ is a topology on $Y.$

Exercise

Show that ${\mathcal{O}}_Y$ is the strongest topology such that $f$ is continuous.

Figure 10.5 Construction of a torus by identification of opposite sides of a square

A common instance ofthis type ofinduced topology occurs when there is an equivalence relation $E$ defined on a topological space $X$. Let $[x]=\{y\mid yEx\}$ be the equivalence class containing the point $x\in X$ . In the factor space $X/E=\{[x]|x\in X\}$ define the topology obtained by identification from $X$ to be the topology induced by the natural map $i_E:X\to X/E$ associating each point $x\in X$ with the equivalence class to which it belongs, $i_E(x)=[x]$ . In this topology a subset $A$ is open iff its inverse image $i_E^{-1}(A)=\{x\in X\mid [x]\in \begin{array}{r}A\}={\bigcup }_{[x]\in A}[x]\end{array}$ is an open subset of $X$. That is, a subset of equivalence classes $A=\{[x]\}$ is open in the identification topology on $X/E$ iff the union of the sets $[x]$ that belong to $A$ is an open subset of $X$.

Exercise

Verify directly from the last statement that the axioms (Top1)–(Top3) are satisfied by this topology on $X/E$

Example 10.14

As in Example 1.4, we say two points $(x,y)$ and $(x^{\prime },y^{\prime })$ in the plane ${\mathbb{R}}^2$ are equivalent if their coordinates differ by integral amounts,

$$(x,y)\equiv (x^{\prime }{y}^{\prime })\text{ iff }x-x^{\prime }=n,y-y^{\prime }=m(n,m\in \mathbb{Z}).$$

The topology on the space ${\mathbb{R}}^2/\equiv$ obtained by identification can be pictured as the unit square with opposite sides identified (see Fig. 10.5). To understand that this is a representation of the torus $T^2$ , consider a square rubber sheet. Identifying sides $AD$ and $BC$ is equivalent to joining these two sides together to form a cylinder. The identification of $AB$ and $CD$ is now equivalent to identifying the circular edges at the top and bottom of the cylinder. In three-dimensions this involves bending the cylinder until top and bottom join up to form the inner tube of a tyre – remember, distances or metric properties need not be preserved for a topological transformation. The $n$-torus $T^n$ can similarly be defined as the topological space obtained by identification from the corresponding equivalence relation on $R^n$ , whereby points are equivalent if their coordinates differ by integers.

Example 10.15

Let ${\dot{\mathbb{R}}}^3={\mathbb{R}}^3-\{0\}$ be the set of non-zero 3-triples of real numbers given the relative topology in ${\mathbb{R}}^3$ . Define an equivalence relation on ${\dot{\mathbb{R}}}^3$ whereby $(x,y,z)\equiv (x^{\prime },y^{\prime },z^{\prime })$ iff there exists a real number $\lambda \ne 0$ such that $x=\lambda x^{\prime },y=\lambda y^{\prime }$ and $z=\lambda z^{\prime }$ . The factor space $P^2={\dot{\mathbb{R}}}^3/\equiv$ is known as the real projective plane.

Each equivalence class $[(x,y,z)]$ is a straight line through the origin that meets the unit 2-sphere $S^2$ in two diametrically opposite points. Define an equivalence relation on $S^2$ by identifying diametrically opposite points, $(x,y,z)\sim (-x,-y,-z)$ ) where $x^2+y^2+z^2=1$. The topology on $P^2$ obtained by identification from ${\dot{\mathbb{R}}}^3$ is thus identical with that of the 2-sphere $S^2$ with diametrically opposite points identified.

Generalizing, we define real projective $n$-space $P^n$ to be ${\dot{\mathbb{R}}}^{n+1}/\equiv$ where $(x_1,x_2,\dots ,x_{n+1})\equiv (x_1^{\prime },x_2^{\prime },\dots ,x_{n+1}^{\prime })$ if and only ifthere exists $\lambda \ne 0$ such that $x_1=\lambda x_1^{\prime }{x}_2=\lambda x_2^{\prime },...,x_{n+1}=\lambda x_{n+1}^{\prime }$ . This space can be thought of as the set of all straight lines through the origin in ${\mathbb{R}}^{n+1}$ . The topology of $P^n$ is homeomorphic with that of the $n$-sphere $S^n$ with opposite points identified.

Problems

Problem 10.12

If $f:X\to Y$ is a continuous map between topological spaces, we define its graph to be the set $G=\{(x,f(x))|x\in X\}\subseteq X\times Y$ . Show that if $G$ is given the relative topology induced by the topological product $X\times Y$ then it is homeomorphic to the topological space $X.$

Problem 10.13

Let $X$ and $Y$ be topological spaces and $f:X\times Y\to X$ a continuous map. For each fixed $a\in X$ show that the map $f_a:Y\to X$ defined by $f_a(y)=f(a,y)$ is continuous.