Is the inverse of a continuous bijective function also continuous? How to prove it?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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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