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 Template:Rp.

Introduction

A topological sum of two topological spaces $X_1$, $X_2$ is a triple of a topological space $X$ and two continuous functions $i_1:X_1\to X$, $i_2:X_2\to X$ that satisfies the following universal property.

• For every topological space $Y$ and continuous functions $f_1:X_1\to Y$, $f_2:X_2\to Y$, there is a unique continuous function $f:X\to Y$ such that $f{i}_1=f_1$ and $f{i}_2=f_2$, i.e. the following diagram commutes.

  $\begin{array}{ccc}X& \stackrel{i_2}{\leftarrow }& X_2\\ i_1\uparrow & \searrow f& \downarrow f_2\\ X_1& \underset{f_1}{\overset{}{\to }}& Y\end{array}$

A topological sum of two spaces always exists, and is unique up to homeomorphism. So we can denote it by $X_1\bigsqcup X_2$. 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 $X_1\bigsqcup X_2$ is as a set the disjoint union of $X_1$ and $X_2$, and a subset $U\subset X_1\bigsqcup X_2$ 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 $i_1:X_1\to X_1\bigsqcup X_2$, $i_2:X_2\to X_1\bigsqcup X_2$, $j_1:X_1\to X$, $j_2:X_2\to X$ be inclusion functions. Then the following two conditions are equivalent.

• (a) $X=X_1\bigsqcup X_2$. That is, the unique function $f:X_1\bigsqcup X_2\to X$ defined by $f{i}_1=j_1$, $f{i}_2=j_2$ 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 $X_1\bigsqcup X_2$, 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 $f:X_1\bigsqcup X_2\to Y$ from the topological sum of topological spaces $X_1$, $X_2$ to another topological space $Y$ is continuous, if both $f\upharpoonright X_1:X_1\to Y$ and $f\upharpoonright X_2:X_2\to Y$ 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 $X\setminus (X_1\cap X_2)=X_1\setminus X_2\bigsqcup X_2\setminus X_1$. Let $Y$ be a topological space. Then a function $f:X\to Y$ 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 $f:X\to Y$. Let $A\in 𝒩(x)$ be a neighbourhood of $x$ in $X$. If $f\upharpoonright A$ is continuous at $x$, then $f$ is continuous at $x$.

Proof. Let $U\in 𝒩(f(x))$ 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 $N\in 𝒩(x)$ such that $f^{-1}(U)\cap A=N\cap A$. This and $A\in 𝒩(x)$ imply that $f^{-1}(U)\cap A\in 𝒩(x)$, and so $f^{-1}(U)\in 𝒩(x)$.

Proof of Theorem 2. We show that $f$ is continuous at every point $x\in X$. If $x\in \mathrm{int}{X}_1$ or $x\in \mathrm{int}{X}_2$, then by Lemma 1 $f$ is continuous at $x$. If $x\in ̸\mathrm{int}{X}_1$ and $x\in ̸\mathrm{int}{X}_2$, 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 $x\in ̸\mathrm{int}{X}_1$.

Now let $U\in 𝒩(f(x))$ be any neighbourhood of $f(x)$. Then there are $M,N\in 𝒩(x)$ such that $f(M\cap X_1)\subset U$ and $f(N\cap X_2)\subset U$. Then we have $M\cap N\in 𝒩(x)$ and $f(M\cap N)\subset U$, from which $f$ is continuous at $x$.