# closed maps on topological spaces

Prove that $f$ is closed if and only if $f(\text{cl}(A)) \supseteq \text{cl}(f(A)).$ I can show the forward directions by saying:
Suppose $f$ is closed. Then, $f(\text{cl}(A)) = f(A) \cup f(\partial A)$ which is closed implying the inclusion of the smallest closed set containing $f(A)$ in $f(\text{cl}(A)).$

But I am struggling going in the other direction. I want to say take a closed set $B \subset X.$ $B = \text{cl}(A)$ for some open $A \subset X.$ But then I am stuck.

For the forward direction there’s no need to deal with the boundary of $A$: just observe that if $f$ is closed, then $f[\operatorname{cl}A]$ is a closed set containing $f[A]$ and must therefore contain $\operatorname{cl}f[A]$, since that’s the intersection of all closed sets containing $f[A]$.
For the other direction it’s probably easiest to suppose that $f$ is not closed and try to find a set $A\subseteq X$ such that $f[\operatorname{cl}A]\nsupseteq\operatorname{cl}f[A]$. If $f$ is not closed, there is a closed set $F\subseteq X$ such that $f[F]$ is not closed. Clearly $f[\operatorname{cl}F]=f[F]$, but $f[F]$ is not closed, so ... ?
• Another way to do the other direction: Let $A$ be closed. $\operatorname{cl} A = A$. Thus $f(A) \supset \operatorname{cl}(f(A))$. Hence $f(A)$ is closed. – Ayman Hourieh Sep 29 '13 at 22:56
• @sergey: It says that the closure of the image of $A$ may be smaller than the image of the closure of $A$, which I see as a kind of shrinkage. – Brian M. Scott Sep 30 '13 at 1:01