# How do we prove that strictly monotone continuous functions admit strictly monotone increasing continuous inverse?

Let $$a < b$$ be real numbers, and let $$f:[a,b]\to\textbf{R}$$ be a function which is both continuous and strictly monotone increasing. Then $$f$$ is a bijection from $$[a,b]$$ to $$[f(a),f(b)]$$, and the inverse $$f^{-1}:[f(a),f(b)]\to[a,b]$$ is also continuous and strictly monotone increasing.

My solution

To start with, let us prove that $$f$$ is bijective.

Since $$f$$ is strictly monotone increasing, it is injective.

Indeed, suppose that $$x \neq y$$. Then either $$x > y$$ or $$x < y$$. In the first case, $$f(x) > f(y)$$ and in the second case $$f(x) < f(y)$$. In both case, $$f(x)\neq f(y)$$. Thus $$f$$ is injective.

We do also have that $$f$$ is surjective. Indeed, given $$y\in[f(a),f(b)]$$, due to the intermediate value theorem there corresponds a $$c\in[a,b]$$ such that $$f(c) = y$$. Thus $$f$$ is surjective.

Let us now prove that $$f^{-1}$$ is strictly increasing. Let $$f(a) > f(b)$$. Then $$a > b$$, otherwise we would have $$f(a) < f(b)$$, which contradicts our assumption. Consequently, $$f$$ is strictly increasing.

Now it remains to prove that $$f^{-1}$$ is continuous, but I get stuck.

Can someone prove this last part using only the $$\varepsilon-\delta$$ definition or its sequential characterization?

Any comments on the previous attempts are welcome as well.

• $f$ maps open intervals to open intervals. May 16, 2020 at 22:11

Let $$c=f(a), d=f(b)$$ and $$g=f^{-1}$$. You can directly prove that $$g$$ is continuous on $$[c, d]$$. Let $$p\in(c, d)$$ so that $$g(p) \in(a, b)$$. Thus there exists a positive number $$\epsilon_0$$ such that $$(g(p) - \epsilon_0,g(p)+\epsilon_0)\subseteq (a, b)$$ (in particular you can take $$\epsilon _0=\min(g(p)-a,b-g(p))$$).
Consider an arbitrary $$\epsilon >0$$ and let $$\epsilon'=\min(\epsilon, \epsilon_0)$$. Then we have $$(g(p) - \epsilon ', g(p) +\epsilon') \subseteq (a, b)$$. It follows that $$r=f(g(p)-\epsilon'),s=f(g(p)+\epsilon ')$$ both lie in $$(c, d)$$ and $$r. Let us write $$\delta=\min(p-r,s-p)$$ so that $$(p-\delta, p+\delta) \subseteq (r, s)$$ and therefore $$g((p-\delta, p+\delta)) \subseteq g((r, s)) \subseteq (g(r), g(s)) =(g(p) - \epsilon', g(p)+\epsilon') \subseteq (g(p) - \epsilon, g(p) +\epsilon)$$ And this proves that $$g$$ is continuous at $$p$$. The proof can be easily modified/adapted for the case when $$p=c$$ or $$p=d$$.
The proof uses the fact that both $$f, g$$ are strictly increasing on their domains and you should be able to figure out where this has been used in above proof.
Take any $$x_n \rightarrow x$$ and $$f$$ continuous, then $$f(x_n) \rightarrow f(x)$$. But then $$f^{-1}(f(x_n)) = x_n \rightarrow x$$ by definition. On the other hand, if $$y_n = f(x_n)$$ and $$y = f(x)$$, that means $$f^{-1}(y_n) \rightarrow f^{-1}(y)$$, so by the sequential criterion for continuity, $$f^{-1}$$ is continuous.
• Here you have an arbitrary sequence $x_n\to x$. But by definition of sequential continuity you should start with an arbitrary sequence $y_n\to y$ and define $x_n=f^{-1}(y_n)$. May 17, 2020 at 0:32