Continuity iff inverse of closure contains closure of inverse I showed that $f: R \rightarrow S$ is continuous between two topological spaces iff for all subsets $E \subset R$,  $f(\overline{E}) \subset \overline{f(E)}$. 
How can I show that $f: R \rightarrow S$ is continuous  iff for all subsets $E \subset R$, $\overline{f^{-1}(E)} \subset f^{-1}(\overline{E})$? And how can I show using these two facts that $f$ is continuous and closed iff $\overline{f(E)}=f(\overline{E})$ (can I combine them to show this, even)? 
 A: Suppose that for all $E\subset S$, $\overline{f^{-1}(E)}\subset f^{-1}(\overline{E})$. Now let $E$ be a closed set so that $\overline{E}=E$.
Then observe that $\overline{f^{-1}(E)}\subset f^{-1}(E)$ so that $f^{-1}(E)$ must contain all of its limit points and is thus closed. Therefore $f$ is continuous.

Now let $f$ be continuous. We know that $f^{-1}(\overline{E})$ is closed and clearly $f^{-1}(\overline{E})\supset f^{-1}(E)$ but then because all limit points of $f^{-1}(E)$ must also be limit points of $f^{-1}(\overline{E})$ and because $f^{-1}(\overline{E})$ is closed, the limit points of $f^{-1}(E)$ must be contained in $f^{-1}(\overline{E})$. Therefore, $\overline{f^{-1}(E)}\subset f^{-1}(\overline{E})$.

Suppose now that $f$ is continuous and closed. Then $f(\overline{E})=\overline{f(\overline{E})}\supset\overline{f(E)}$ because $f(\overline{E})$ is closed and because $f(E)\subset f(\overline{E})$. You already proved that for continuous $f$ that $f(\overline{E})\subset \overline{f(E)}$ so we are done.

Now if $\overline{f(E)}=f(\overline{E})$ then $\overline{f(E)}\supset f(\overline{E})$ so $f$ is continuous. For some closed set $E$, $f(E)=f(\overline{E})=\overline{f(E)}$ which is a closed set so $f$ is closed.
