Exercise 4.5.8

Define a function $f$ to be strictly increasing (decreasing) on $A$ if $f(x)<f(y)$ ( $f(x)>f(y)$) for all $x<y$ in $A$, and define strictly monotone to mean either strictly increasing or strictly decreasing. It’s easy to show by contradiction using IVT that a one-to-one continuous function $f$ must be strictly monotone, and similarly so is $f^{-1}$. Moreover, if $f$ is strictly increasing then so is $f^{-1}$, and if $f$ is strictly decreasing then so is $f^{-1}$.

We now show that $f^{-1}$ satisfies the intermediate value property. Assume for now that $f$ is strictly increasing - the proof for the case of strictly decreasing is similar. Consider $f(x_1)=y_1<f(x_2)=y_2$ and any $L$ between $f^{-1}(y_1)$ and $f^{-1}(y_2)$, i.e. $x_1<L<x_2$; we need to show there is some $y_3\in [y_1,y_2]$ such that $f^{-1}(y_3)=L$; clearly $y_3=f(L)$ works. By Exercise 4.5.3, this shows $f^{-1}(x)$ is continuous.