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If function $\textbf{F}^{-1}(x)$ is an inverse of function $\textbf{F}$ and $\textbf{F}^{-1}(x)$ is continuous. Is it true that $\textbf{F}(x)$ is continuous too?

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Take $f^{-1}(x) = x$ on $[0,1)$ and $f^{-1}(x) = x-1$ on $[2,3]$.

Then $f(x) = x$ on $[0,1)$ and $f(x) = 1+x$ on $[1,2]$.

Here $f^{-1}:[0,1) \cup [2,3] \rightarrow [0,2]$ is continuous, and $f$ is discontinuous at $x=1$.

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I think the map $f:[0,1) \rightarrow S^1$ given by $x \rightarrow (\sin 2\pi x, \cos 2\pi x)$ is a counter.

For a somewhat-trivial example , take $id: (X,\text{discrete} )\rightarrow (X, \text{indiscrete})$ , for any set $X$ .

As a general result, if $f A \rightarrow B$ is a continuous bijection; $A$ is compact and $B$ is Hausdorff, then $f^{-1}$ is necessarily continuous.

For counters from a space to itself with the same topology, see: https://mathoverflow.net/questions/30661/non-homeomorphic-spaces-that-have-continuous-bijections-between-them

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The answer is no: take $f^{-1}(x) = e^{ix}$, defined from $[0,2\pi)$ to $\mathbb{S}^1$ (the unit sphere in the plane) This function is clearly continuous. Unfortunately, its inverse cannot be continuous since otherwise $[0,2\pi)$ would be compact being the image of the compact set $\mathbb{S}^1$ under a continuous function.

I hope it helps, let me know if the details are clear enough :)

EDIT: something easier to check: the identity map from the reals with the lower limit topology ($\mathbb{R}_l$) to the real with the standard topology ($\mathbb{R}$) is another counterexample! This is immediate to check since the inverse of the identity map is of course the identity map itself which now goes from $\mathbb{R}$ to $\mathbb{R}_l$. Clearly the pre image of the open set $[a,b)$ is not open!

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  • $\begingroup$ Don't you mean [0, 2\pi) would be compact? $\endgroup$ – user99680 Aug 13 '14 at 6:47
  • $\begingroup$ yeah, of course I do! :) thank you for pointing it out! $\endgroup$ – user67133 Aug 13 '14 at 6:48

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