Subspace Topology

Definition (Subspace Topology)

Recall if $(X,{\tau }_X)$ is a topological space and $A\subseteq X$, we can define the subspace topology defined on $A$:

$${\tau }_A=\{U\cap A\mid U\text{ open in }X\}$$

This does form a topology on $A$ because of the canonical inclusion map

$$i:A\hookrightarrow X$$

is continuous. We check open $U\subset X$, such that $i^{-1}(U)=U\cap A$, which is open (relative to $A$). So, we have

$$(A,{\tau }_A)\stackrel{i}{\hookrightarrow }(X,{\tau }_X)$$

which is continuous. In fact, ${\tau }_A$ is the smallest topology on $A$ for which this inclusion map is continuous.

Lemma (Subspace Continuity by Canonical Inclusion)

Given $A\subseteq X$ is a subspace, and $Z$ is a third space, then a map

$$f:Z\to A$$

is continuous iff its composite $Z\to A{\to }^{i}X$ is continuous.

Proof: $(⟹)$ We simply have the composition of continuous functions, which is continuous.

$(⟸)$ Given $U$ open in $A$, WTS $f^{-1}(U)$ is open in $Z$. But we know $U=A\cap V$ for some open $V$ in $X$. Then $f^{-1}(U)=f^{-1}(A\cap V)=f^{-1}{i}^{-1}(V)=(i\circ f{)}^{-1}(V)$ which is open if $i\circ f$ is continuous.

Lemma (Universal Property of the Subspace Topology)

The subspace topology satisfies the property:

$"\\usepackage{tikz-cd}\n\\begin{document}\n\\begin{tikzcd} \n\t{(A, \\tau_A)} & {(X, \\tau)} \\\\ \n\t{(Z, \\tau_Z) } \n\t\\arrow[\"i\", hook, from=1-1, to=1-2] \n\t\\arrow[\"f\", from=2-1, to=1-1] \n\t\\arrow[\"{i \\circ f}\"', dotted, from=2-1, to=1-2] \n\\end{tikzcd}\n\\end{document}"$

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$f$ is continuous $⟺$ $i\circ f$ is continuous.

This means that if you want to construct a continuous map to $A$, it is enough to find a continuous map to $X$ whose image lies in $A$. In fact, this property characterizes the subspace topology in the following sense:

if ${\tau }_1$ and ${\tau }_2$ are two topologies on $A$ which satisfies the property that for all test spaces and maps, then ${\tau }_1={\tau }_2$.

Proof: Suppose we had

$"\\usepackage{tikz-cd}\n\\begin{document}\n\\begin{tikzcd} \n\t{(A, \\tau_1)} & {(X, \\tau)} \\\\ \n\t{(A, \\tau_2)} \n\t\\arrow[\"{i_1}\", hook, from=1-1, to=1-2] \n\t\\arrow[\"{i_2}\", hook, from=2-1, to=1-2]\n\t\\arrow[\"{id_1}\", shift left=2 ,from=1-1, to=2-1]\n\t\\arrow[\"{id_2}\", from=2-1, to=1-1]\n\\end{tikzcd}\n\\end{document}"$

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with ${\tau }_1,{\tau }_2$ satisfying the property above. We observe that

• ${\text{Id}}_1$ is continuous because $i_2\circ {\text{Id}}_1=i_1$ is continuous, and
• ${\text{Id}}_2$ is continuous because $i_1\circ {\text{Id}}_2=i_2$ is continuous Hence the two $\text{Id}$ maps do form a homeomorphsism. Continuity of both implies they have the exact same open sets. Thus, ${\tau }_1\subseteq {\tau }_2$, and ${\tau }_2\subseteq {\tau }_1$.

By test, we mean the conditions are written in some other space $Z$.

Gluing Lemma

If $\{U_i\}$ is an open cover of $X$, then a continuous function $f:X\to Y$ is the same as a collection of continuous functions

$$f_i:U_i\to Y$$

which agree on all overlaps:

$$f_i{|}_{U_i\cap U_j}=f_j{|}_{U_i\cap U_j}$$

Open Cover means open sets where the union covers $X$.

Proof: $(⟹)$ This is straightforward because $f{|}_{U_j}=f\circ i_j:U_j\stackrel{i_j}{\hookrightarrow }X\stackrel{f}{\to }Y$.

$(⟸)$ The question is whether the global function $f$ created by setting

$$f(x)=f_i(x)x\in U_i$$

is continuous or not. Consider $V$ open in $Y$. Then

$$f^{-1}(V)=\bigcup _{i\in I}{f}^{-1}(V)\cap U_i$$

Now,

$$f^{-1}(V)\cap U_i=(f{|}_{U_i}{)}^{-1}(V)=(f_i{)}^{-1}(V)$$

so it is open in $U_i$ because $f_i$ is continuous, using Definition (Topological Characterization of Continuity).

Fortunately, we know that an open subset $f^{-1}(V)\cap U_i$ of a subspace $U_i$ which is itself open on a subset of $X$, is open in $X$. Hence,

$$f^{-1}(V)=\bigcup \text{ sets open relative to }X$$

so by Lemma (Openness and Closure), $f^{-1}(V)$ is open.

This lemma describes how we can glue together multiple continuous functions, each defined on an open patch to create a single continuous function.

Gluing Lemma (But Closed)

If $\{F_i{\}}_{i=1,\dots ,n}$ is a covering by finitely many closed closed subsets, then a continuous function $f:X\to Y$ is the same as a collection of continuous $f_i:F_i\to Y$ which agree on overlaps.

Proof: The same as the Gluing Lemma but using $f^{-1}(\text{closed sets})$. Take note of the finiteness condition.

These lemmas are similar to how we prove a piecewise function is continuous. We break up the cases, just like how we break up $X$ to $U_i$ and prove continuity on each $U_i$.