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Let $X$ be a complete metric space and $f:X \to X$ be bijective and a connected preserving map i.e. $f$ carries every connected set of $X$ to a connected set of $X$ ; then is it necessarily true that $f^{-1}$ also carries connected sets to connected sets ?

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Let $$X=\bigl(\{-2\}\cup [-1,0]\cup [1,\infty)\bigr)\times \Bbb N_0$$ with the induced metric as a subspace of $\Bbb R^2$. This space is complete. The connected subspaces are precisely those of the form $I\times\{n\}$ where $n\in\Bbb N_0$ and either $I=\{-2\}$ or $I$ is an interval $\subseteq [-1,0]$ or $I$ is an interval $\subseteq [0,\infty)$.

Consider the map $f\colon X\to X$ given by $$f(x,y)=\begin{cases}(-\frac1x,y)&\text{if $x>0$ and $y=0$}\\ (0,0)&\text{if $x=-2$ and $y=0$}\\ (x,y+1)&\text{if $-1\le x\le 0$}\\ (x,y-1)&\text{otherwise.}\end{cases} $$ This is a continuous bijection and preserves connectedness. But its inverse maps the connected set $[-1,0]\times \{0\}$ to the not connected set $\bigl(\{-2\}\cup [1,\infty)\bigr)\times\{0\}$.

The thing works a bit like a zipper:

enter image description here

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  • $\begingroup$ If I'm not mistaken , the inverse map is $f^{-1}(z)=\sqrt z$ , which is continuous , and hence connected preserving $\endgroup$ – user228168 Aug 15 '15 at 13:44
  • $\begingroup$ your map is not injective ; $f(i)=f(-i)$ $\endgroup$ – user228168 Oct 25 '15 at 8:31
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    $\begingroup$ @SaunDev The bijectivity condition was added to the problem statement a day after my answer was posted ... $\endgroup$ – Hagen von Eitzen Oct 25 '15 at 12:54
  • $\begingroup$ @HagenvonEitzen Should that $\frac{1}{x}$ be $-\frac{1}{x}$? $\endgroup$ – Marc Paul Oct 25 '15 at 14:37
  • $\begingroup$ @MarcPaul Yes, I'll fix $\endgroup$ – Hagen von Eitzen Oct 25 '15 at 15:33

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