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?

  • 1
    $\begingroup$ You have to assume $f$ is bijective in order for $f^{-1}$ to exist. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – Niki Di Giano Aug 9 '18 at 17:16
  • $\begingroup$ @NikiDiGiano Yes, that's right. $\endgroup$ – José Carlos Santos Aug 9 '18 at 17:16
  • $\begingroup$ All clear. Thank you. $\endgroup$ – Niki Di Giano Aug 9 '18 at 17:18

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