TITLE: how to show strictly increasing function on an interval has continuous inverse
QUESTION [5 upvotes]: my question is to show that a strictly increasing function that is defined on an interval has a continuous inverse.
Thanks.

REPLY [14 votes]: First suppose $I:=]a,b[$. Since $f:I\to\mathbb{R}$ is strictly increasing, it is injective, so $f:I\to f(I)$ is a bijection and $f^{-1}$ exists. $f^{-1}$ is continuous iff $f$ is an open map. It suffices to check that $\forall ]\alpha,\beta[\subseteq I: f(]\alpha,\beta[)$ is open in $f(I)$.
Let $]\alpha,\beta[\subseteq I$. We claim that $f(]\alpha,\beta[)=]f(\alpha),f(\beta)[\cap f(I)$. If $\alpha<x<\beta$, then $f(\alpha)< f(x) < f(\beta)$ and since $x\in ]\alpha,\beta[\subseteq I, f(x)\in f(I)$. Conversely, if $y\in ]f(\alpha),f(\beta)[\cap f(I)$, there is an $x\in I: f(x)=y$. Also since $f(\alpha)<f(x)=y<f(\beta), x\in ]\alpha,\beta[$, so that $y\in f(]\alpha,\beta[)$. This means that $f(]\alpha,\beta[)$ is an open subset with respect to the subspace topology of $f(I)\subseteq\mathbb{R}$. Therefore $f$ is open.
Other cases follow analogously.

P.S.: For reference purposes, this is Exercise 2.7.41 from Royden & Fitzpatrick's Real Analysis, 4e (p. 53).