# 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... Feb 13, 2021 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, 2021 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 $fr \neq id_{[0,1]}$). Feb 13, 2021 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, 2021 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... Feb 13, 2021 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, 2021 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. Feb 13, 2021 at 21:40
• Thanks! I'm pretty unsure about the second argument. Were you able to look through it?
– dmh
Feb 13, 2021 at 23:01
• 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 Feb 14, 2021 at 0:01