# $X$ be a complete metric space and $f:X \to X$ be a bijective and connected preserving map ; then is $f^{-1}$ also connected preserving?

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 ?

• So are you supposing that $f$ is bijective? – Crostul Aug 12 '15 at 16:32
• – Mizar Oct 28 '15 at 13:52

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 $\bigl(\{-2\}\cup [1,\infty)\bigr)\times\{0\}$.
• If I'm not mistaken , the inverse map is $f^{-1}(z)=\sqrt z$ , which is continuous , and hence connected preserving – user228168 Aug 15 '15 at 13:44
• your map is not injective ; $f(i)=f(-i)$ – user228168 Oct 25 '15 at 8:31
• @HagenvonEitzen Should that $\frac{1}{x}$ be $-\frac{1}{x}$? – Marc Paul Oct 25 '15 at 14:37