# 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 [1]:68, Exercise 3.2.5.

## Introduction

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

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

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

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

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

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

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

## Main result

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

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

## Proof of the main result

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

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

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

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

## References

1. Brown, Ronald (2006). Topology and groupoids. A geometric account of general topology, homotopy types and the fundamental groupoid (in en) (3 ed.). ISBN 1-4196-2722-8.