# Topology/Universal property of topological sum in slightly sharper form

A slightly sharper result than the universal property of topological sum is introduced, which is just an excercise given in :68, Exercise 3.2.5.

## Introduction

A topological sum of two topological spaces $X_{1}$ , $X_{2}$  is a triple of a topological space $X$  and two continuous functions $i_{1}\colon X_{1}\to X$ , $i_{2}\colon X_{2}\to X$  that satisfies the following universal property.

• For every topological space $Y$  and continuous functions $f_{1}\colon X_{1}\to Y$ , $f_{2}\colon X_{2}\to Y$ , there is a unique continuous function $f\colon X\to Y$  such that $fi_{1}=f_{1}$  and $fi_{2}=f_{2}$ , i.e. the following diagram commutes.
${\begin{matrix}X&\xleftarrow {i_{2}} &X_{2}\\{i_{1}}\uparrow &\searrow {f}&\downarrow {f_{2}}\\X_{1}&{\xrightarrow[{f_{1}}]{}}&Y\end{matrix}}$

A topological sum of two spaces always exists, and is unique up to homeomorphism. So we can denote it by $X_{1}\sqcup X_{2}$ . Since the inclusion functions $i_{1}$ , $i_{2}$  are both embeddings we can denote $i_{1}(X_{1})$ , $i_{2}(X_{2})$  simply by $X_{1}$ , $X_{2}$ . With this notation the topological sum $X_{1}\sqcup X_{2}$  is as a set the disjoint union of $X_{1}$  and $X_{2}$ , and a subset $U\subset X_{1}\sqcup X_{2}$  is open if and only if $U\cap X_{1}$  is open in $X_{1}$  and $U\cap X_{2}$  is open in $X_{2}$ .

The following theorem characterizes when a topological space can be seen as a topological sum of two given subspaces.

Theorem 1. Let $X$  be a topological space and $X_{1},X_{2}\subset X$  two subsets. Let $i_{1}\colon X_{1}\to X_{1}\sqcup X_{2}$ , $i_{2}\colon X_{2}\to X_{1}\sqcup X_{2}$ , $j_{1}\colon X_{1}\to X$ , $j_{2}\colon X_{2}\to X$  be inclusion functions. Then the following two conditions are equivalent.

• (a) $X=X_{1}\sqcup X_{2}$ . That is, the unique function $f\colon X_{1}\sqcup X_{2}\to X$  defined by $fi_{1}=j_{1}$ , $fi_{2}=j_{2}$  is a homeomorphism.
• (b) $X=X_{1}\cup X_{2}$ , $X_{1}\cap X_{2}=\varnothing$ , and both $X_{1}$ , $X_{2}$  are open in $X$

Proof. (a) ⇒ (b). $X_{1}$  is open in $X_{1}\sqcup X_{2}$ , since $X_{1}$  is open in $X_{1}$  and $\varnothing$  is open in $X_{2}$ . $X_{2}$  is open for a similar reason. (b) ⇒ (a). Suppose $U_{1}$  is open in $X_{1}$  and $U_{2}$  is open in $X_{2}$ . Then, because $X_{1}$ , $X_{2}$  are both open in $X$ , $U_{1}$ , $U_{2}$  is open in $X$  and so $U_{1}\cup U_{2}$  is open in $X$ .

## Main result

According to the universal property of topological sum, a function $f\colon X_{1}\sqcup X_{2}\to Y$  from the topological sum of topological spaces $X_{1}$ , $X_{2}$  to another topological space $Y$  is continuous, if both $f\upharpoonright X_{1}\colon X_{1}\to Y$  and $f\upharpoonright X_{2}\colon X_{2}\to Y$  are continuous. This is a special case of the following theorem.

Theorem 2. Let $X$  be a topological space and $X_{1},X_{2}\subset X$  two subsets. Suppose that $X=X_{1}\cup X_{2}$ , and that $X\setminus (X_{1}\cap X_{2})=X_{1}\setminus X_{2}\sqcup X_{2}\setminus X_{1}$ . Let $Y$  be a topological space. Then a function $f\colon X\to Y$  is continuous if both $f\upharpoonright X_{1}$  and $f\upharpoonright X_{2}$  are continuous.

## Proof of the main result

Lemma 1. Let $X$ , $Y$  be topological spaces, $x\in X$  and $f\colon X\to Y$ . Let $A\in {\mathcal {N}}(x)$  be a neighbourhood of $x$  in $X$ . If $f\upharpoonright A$  is continuous at $x$ , then $f$  is continuous at $x$ .

Proof. Let $U\in {\mathcal {N}}(f(x))$  be any neighbourhood of $f(x)$  in $Y$ . Then $f^{-1}(U)\cap A$  is a neighbourhood of $x$  in $A$ . So there is an $N\in {\mathcal {N}}(x)$  such that $f^{-1}(U)\cap A=N\cap A$ . This and $A\in {\mathcal {N}}(x)$  imply that $f^{-1}(U)\cap A\in {\mathcal {N}}(x)$ , and so $f^{-1}(U)\in {\mathcal {N}}(x)$ .

Proof of Theorem 2. We show that $f$  is continuous at every point $x\in X$ . If $x\in \operatorname {int} X_{1}$  or $x\in \operatorname {int} X_{2}$ , then by Lemma 1 $f$  is continuous at $x$ . If $x\not \in \operatorname {int} X_{1}$  and $x\not \in \operatorname {int} X_{2}$ , then it is easily verified that $x\in X_{1}\cap X_{2}$ . Suppose otherwise that $x\in X_{1}\setminus X_{2}$ . Then $X_{1}\setminus X_{2}$  is open in $X\setminus (X_{1}\cap X_{2})$  and so there is a $V\subset X$  such that $X_{1}\setminus X_{2}=V\setminus (X_{1}\cap X_{2})$ . Then $x\in V\subset X_{1}$  and this is impossible since $x\not \in \operatorname {int} X_{1}$ .

Now let $U\in {\mathcal {N}}(f(x))$  be any neighbourhood of $f(x)$ . Then there are $M,N\in {\mathcal {N}}(x)$  such that $f(M\cap X_{1})\subset U$  and $f(N\cap X_{2})\subset U$ . Then we have $M\cap N\in {\mathcal {N}}(x)$  and $f(M\cap N)\subset U$ , from which $f$  is continuous at $x$ .