[Math] If f is continuous and strictly increasing, then the function $f^{-1}:f(I)\rightarrow I$ is continuous and strictly increasing.

Question

Let $I \subseteq \Bbb R$ be a non-degenerate open interval, and let $f:I\rightarrow \Bbb R$ be a function. Suppose that f is strictly monotone. If f is continuous and strictly increasing (or decreasing), then the function $f^{-1}:f(I)\rightarrow I$ is continuous and strictly increasing (or decreasing).

Strictly monotone means the function is either strictly increasing or strictly decreasing.
Strictly increasing means that $x<y$ implies $f(x)<f(y)$ for all $x,y\in A$.
Strictly decreasing means that $x<y$ implies $f(x)>f(y)$ for all $x,y\in A$.

Now for my attempt:
Suppose that $f^{-1}(x)>0$ for all x in the interior of I. Let $x,y\in I$. Suppose that $x<y$. Then $[x,y]\subseteq I$. The Mean Value Theorem applied to $f\rvert [x,y]$ implies that there is some $c\in (x,y)$ such that $f^{-1}(c)=$${f(y)-f(x)}\over {y-x}$.
Because $c\in (x,y)$ then c is in the interior of I, and it follows that $f(y)-f(x)=f^{-1}(c)(y-x)>0$. Therefore $f(x)<f(y)$, and f is increasing.

Answer 1

First we will show $f^{-1}$ is strictly increasing:

• If $\exists a,b \in f(I)$ with $a < b$ s.t. $f^{-1}(a) = x$, and $\text{ }f^{-1}(b) = y$ and $ y \leq x $, then $$\Rightarrow b = f(y) \leq f(x) = a$$ as $f$ is strictly increasing. This is a contradiction.

Thus $f^{-1}$ is strictly increasing as required.

Now to show $f^{-1}$ is continuous:

• Let $x_0 \in I$ and define $y_0 = f(x_0)$. We will show that $\lim_{y \rightarrow y_0 }$ $f^{-1}(y) = x_0$. For any $\epsilon > 0$ s.t. $(x_0 - \epsilon, x_0 + \epsilon) \subset I$, we have that $$f(x_0 - \epsilon) < f(x_0) < f(x_0 + \epsilon)$$ Take $\delta = \frac{1}{2}min( f(x_0) -f(x_0 - \epsilon), f(x_0 + \epsilon) - f(x_0))$. Then: $$f(x_0 - \epsilon) < f(x_0) - \delta \text{ and } f(x_0) + \delta < f(x_0 + \epsilon) $$ That is to say, if $|y - y_0| < \delta$, then $|f^{-1}(y) - f^{-1}(y_0)| < \epsilon$ as required, and we have continuity.