TITLE: Show that the Quotient map is closed.
QUESTION [7 upvotes]: Let $X$ and $Y$ be topological spaces. Let $X \cup Y$ be the space with disjoint union topology.Suppose $A \subset X, A $ closed, $f : A \to Y$ be a continuous closed map.Consider the quotient map $\pi: X \cup Y \to X \cup _{f} Y$.Then $\pi$ is closed map.

I am trying to prove using the definitions directly but its getting bit messy.Is there any 'smart' way to prove $\pi$ is closed?

REPLY [2 votes]: I would just follow the definitions. Suppose that $F\subseteq X\cup Y$ is closed. Then $F$ is the disjoint union of the sets $F_0=(F\cap X)\setminus A$, $F_1=F\cap A$, $F_2=(F\cap Y)\setminus f[A]$, and $F_3=F\cap f[A]$.

Show that $\pi^{-1}[\{\pi(z)\}]=\{z\}$ if $z\in F_0\cup F_2$.  
Show that $\pi^{-1}[\{\pi(z)\}]=f^{-1}[\{f(z)\}]\cup\{f(z)\}$ if $z\in F_1$.  
Show that $\pi^{-1}[\{\pi(z)\}]=f^{-1}[\{z\}]\cup\{z\}$ if $z\in F_3$.  
Conclude that $\pi^{-1}\big[\pi[F]\big]=F\cup f^{-1}\big[f[F\cap X]\big]\cup f^{-1}[F\cap Y]$, which is closed (why?).