# Is the inverse of a continuous function an open map?

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?

• You have to assume $f$ is bijective in order for $f^{-1}$ to exist. – Cheerful Parsnip Aug 9 '18 at 16:58

In order that your problem makes sense, there's one thing that's missing: you must add that $$f$$ is bijective. Otherewise, there will be no function $$f^{-1}$$. After adding this, yes, $$f$$ is continuous if and only if $$f^{-1}$$ is an open map.
• Suppose I consider $f(x)=x^2$. Even if the inverse is not a function, open intervals of $\Bbb{R}$ have a preimage that takes the form of the union of two open intervals (at most) - thus the preimage is open. However, $f^{-1}$ is not a function and thus cannot be an open map or any other kind of map, is that right? – Niki Di Giano Aug 9 '18 at 17:16