# Inverse of a continuous increasing function is continuous (solution verification)

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.

• Why does this map even have an inverse? $\Bbb R$ is not compact, while $[0,1]$ is, so $f$ cannot possibly be a homeomorphism... – PrudiiArca Feb 13 at 18:46
• I updated the question to specify that $f$ is onto $(0,1)$ not onto $[0,1]$. Does that resolve your point? – dmh Feb 13 at 18:51
• Kind of. You still won't get a homeomorphism of cause, but you might have a retraction $r:[0,1] \rightarrow \Bbb R$ (I would refrain from calling it $f^{-1}$ though, since it is not an inverse as $fr \neq id_{[0,1]}$). – PrudiiArca Feb 13 at 18:58
• I see, that's right it's not an inverse on the whole range, only on $(0, 1)$. Although in this case that's not essential to the question. – dmh Feb 13 at 19:00
• I think your first argument that $f^{-1}$ is continuous iff $f$ is open does not work, as it is phrased like it holds for arbitrary continuous injections. It would be true, if we had a bijection, but as mentioned this is not the case here. Maybe it holds true for injective continous maps with dense image, but then this should be part of your argument... – PrudiiArca Feb 13 at 19:41

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.
• I see, thanks. I'll need to go through this closely. But the point I took away is that the first claim in the first argument breaks down when $x \neq m(r(x))$. In particular I need to be careful with how I define the retraction function $r$ over the domain. I think if I restrict the range of $f$ to $(0, 1)$ then that handles the issue, right? – dmh Feb 13 at 20:30
• Exactly. If the codomain of $f$ is $(0,1)$ you are speaking of a continuous bijection, in which case the inverse being continuous is equivalent to $f$ being open. – PrudiiArca Feb 13 at 21:40
• Well you got the $\varepsilon-\delta$ semi-correct. A function $g$ is continuous, if forall $\varepsilon>0$ there exists $\delta>0$ st forall $x,y$ it holds that $\vert x-y\vert<\delta$ implies $\vert g(x)-g(y)\vert<\varepsilon$. All of this (in particular the order of the quantifiers!) is important. Now of cause you might use the contrapositive implication $\vert g(x)-g(y)\vert>\varepsilon$ implies $\vert x-y\vert>\delta$. But again, if you want to invoke monotonicity of $f$ you need to know that any $x\in [0,1]$ is given by $x=f(t)$ for some $t\in \Bbb R$, which is not the case – PrudiiArca Feb 14 at 0:01