# Continuity $\iff$ Preimage is closed whenever set is closed

I'm trying to do the following:

Prove that a function $f : S\rightarrow T$ between two topological spaces is continuous iff $f^{-1}(C)$ is closed whenever $C\subset T$ is closed.

To show that the closed-set relationship implies continuity, notice that $T-C$ is open. The preimage of this open set is $f^{-1}\left(T-C\right)=S-f^{-1}(C)$, which is open by assumption ($f^{-1}(C)$ is closed). Of course, any open set $G\subset T$ can be written in the form $T-C$ by letting $C=T-G$. Thus, for every open set $G\subset T$, the preimage is open.

To show the implication in the other way, consider a closed set $C\subset T$. $T-C$ is open, so $f^{-1}(T-C)=S-f^{-1}(C)$ is open as well ($f$ is continuous), which means that $f^{-1}(C)$ is closed.

• Yes, this is correct. – Brian M. Scott Jun 13 '15 at 4:00