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Is the inverse of a continuous bijective function also continuous? How to prove it?

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    $\begingroup$ You should also visit math.stackexchange.com/questions/68800/… ; $\endgroup$ – Marra Apr 22 '13 at 2:37
  • $\begingroup$ The question to which this is supposedly a duplicate mentions nothing about bijectivity. Certainly, there are answers there which use bijective functions, but this question is not a duplicate, in my opinion. $\endgroup$ – robjohn Oct 27 '17 at 15:31
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Take the function $f(x)=x^2$ for $x\in(-1,0]\cup[1,2]$. Then $f:(-1,0]\cup[1,2]\to[0,4]$ is continuous and bijective, but the inverse is not continuous. We can see the inverse is not continuous since $[0,4]$ is connected but $(-1,0]\cup[1,2]$ is not connected.

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Take any set $S$. Let $X$ be $S$ with the discrete topology and $Y$ be $S$ with the coarse topology. Note that the identity $i:X\to Y$ is continuous, but its inverse, the identity $i:Y\to X$, is not.

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Define $f: [0,1) \cup [2,3] \rightarrow [0,2]$ by

$$f(x)=\begin{cases} x & x \in [0,1) \\ x-1 & x \in [2,3] \end{cases}$$

This is a counter-example from TonyK.

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