Show that for a continuous, increasing function $f : \mathbb R \rightarrow [0,1]$ that is onto $(0,1)$ it's inverse $f^{-1}$ is also continuous.
What I've tried
My first approach:
$f^{-1}$ is continuous if $f$ maps open sets to open sets. Since $f$ is increasing it is a one-to-one mapping and it suffices to show that $f$ maps open intervals to open intervals. Let $V \subset [0,1]$ so that $f^{-1}(V)$ is open (since $f$ is continuous) and $f(f^{-1}(V))=V$ which is also open. All that remains to show is that every open interval $U = f^{-1}(V)$ for some open interval $V$. This is true since $f$ is a one-to-one mapping.
My second approach:
$f^{-1}$ is continuous if there exists a $\delta$ such that when:
$$|\alpha – \alpha'| < \delta$$
we have:
$$|f^{-1}(\alpha) – f^{-1}(\alpha')| < \epsilon$$
Call the first event $A$ and the second event $B$. To show $A \implies B$ it is sufficient to show that $B^c \implies A^c$. Let $\alpha = f(x),\, \alpha' = f(x')$ for some $x,x'$ so that we need to show for each $\epsilon$: $|x-x'| \geq \epsilon \implies |f(x) – f(x')| \geq \delta$ for some $\delta$. But this is clear since $f$ is strictly increasing.
I could use some feedback on my reasoning for both approaches.
Best Answer
I think the first argument is broken: it is not true that for a continuous map $m:X\rightarrow Y$ with a retraction function $r:Y \rightarrow X$ the map $r$ is continuous iff $m$ is open. The problem is that if $m$ is not surjective onto $Y$ preimages of open sets $U \subseteq X$ under $r$ might contain elements not in the image of $U$ under $m$. We have canonically $m(U) \subseteq r^{-1}(U)$, but the latter needs not be open.
I can give you a counterexample: Consider the topological spaces $X=\{\circ,\square\}$ and $Y=\{\circ,\square,\bullet\}$, where $\circ,\square$ denote open points, $\bullet$ is a closed point and we consider the topologies generated by this conditions. Then the obvious inclusion is a continous and open injection $m:X\rightarrow Y$ and we can define the retraction $r:Y \rightarrow X$ doing the obvious things and sending $\bullet$ to $\circ$. The latter is not continous though as the preimage $r^{-1}(\{\circ\}) = \{\circ,\bullet\}$ is not open in the topology on $Y$.
Edit I think the statement itself is wrong (if you use $[0,1]$). The function $\operatorname{arctan}:\Bbb R \rightarrow (-\frac{\pi}{2}, \frac{\pi}{2}) \subseteq [-\frac{\pi}{2},\frac{\pi}{2}]$ is a continuous strictly increasing function, which is not surjective. If it had a continuous retraction $r:[-\frac{\pi}{2},\frac{\pi}{2}]\rightarrow \Bbb R$ we would have in particular $r\vert_{(-\frac{\pi}{2},\frac{\pi}{2})} = \tan$ and continuity would force $$r(\frac{\pi}{2})=\lim \limits_{x \rightarrow \frac{\pi}{2}} r(x) = \lim \limits_{x \rightarrow \frac{\pi}{2}} \tan(x) = \infty \in \Bbb R,$$ which is absurd.