The title is pretty self-explanatory, but I'll state the full question.
Let $f: X \rightarrow Y$ be a continuous function between topological spaces. Is $f^{-1}$ an open map?
By definition of continuous function, if $V$ is open in $Y$, $f^{-1}(V)$ is open in $X$. Since the image under $f^{-1}$ of an open set is open, $f^{-1}$ is an open map.
I'm not sure about the converse. Is there something I'm missing?
Best Answer
In order that your problem makes sense, there's one thing that's missing: you must add that $f$ is bijective. Otherwise, there will be no function $f^{-1}$. After adding this, yes, $f$ is continuous if and only if $f^{-1}$ is an open map.