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 .

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.

  ${\begin{matrix}X&\xleftarrow {i_{2}} &X_{2}\\{\scriptstyle i_{1}}\uparrow &\searrow {\scriptstyle f}&\downarrow {\scriptstyle f_{2}}\\X_{1}&{\xrightarrow[{f_{1}}]{}}&Y\end{matrix}}$

A topological sum of two spaces always exists, and is unique up to homeomorphism. So we can denote it by $X_{1}\sqcup X_{2}$. 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 $x$.