I've got a question to the theorem below:
Let $f: X \rightarrow Y$ be an onto map. If $f$ maps open sets of $X$ to open sets of $Y$, then $f$ is an identification map.
If I got the idea right, we should prove that: $U$ is open in Y $\Leftrightarrow f^{-1}(U)$ open in X.
It's easy to prove from right to left but the reverse is not clear because the open map is not always continuous. Did I miss something or the theorem admits that f is continous?