for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.

Let $X$ and $Y$ be two topological spaces.

Prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.

I am stuck on the converse. Suppose $f: X \to Y$ is continuous. Then for every closed set C in Y, $f^{-1}(C)$ is closed in X.

WTS for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.

Since $\overline{f(A)}$ is a closed set in Y, $f^{-1} \circ \overline{f(A)}$ = $ \overline{ f^{-1} \circ \overline{f(A)} }$

Also, $f(\overline{A}) \subseteq \overline{ f(\overline{A})}$

Solution 1:

Suppose than for any $A\subseteq X$, we have $f(\overline A)\subseteq \overline{f(A)}$. Pick a closed set $F\subseteq Y$. We want to show $f^{-1}(F)$ is closed. Using the above, can you show that $\overline {f^{-1}(F)}\subseteq f^{-1}(F)$ must hold true? You need to use $F$ is closed.

ADD (Spoiler) We use $F=\overline F$ and $ff^{-1}(F)\subseteq F$. Then using $f(\overline A)\subseteq \overline{f(A)}$ with $A=f^{-1}(F)$, we get $f\left(\overline{f^{-1}(F)}\right)\subseteq \overline{ff^{-1}(F)}\subseteq \overline F=F$. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}f\left(\overline{f^{-1}(F)}\right)\subseteq f^{-1}(F)$, so $f^{-1}(F)$ is closed and $f$ is continuous.