TITLE: Prove that if f is a continuous strictly monotone function defined on an interval, then its inverse is also a continuous function.
QUESTION [13 upvotes]: There is a theorem on continuous function that goes as follow:
If f is a continuous strictly monotone function defined on an interval, then its inverse is also a continuous function.
I have quite an ugly proof on this theorem. My textbook proof doesn't look good either. So I am just wondering if someone can provide me with a more elegant proof. Thanks.
On top of that, I am just wondering why must the function be strictly monotone? Is it to ensure that it is one-to-one so that the inverse exists? Thanks.

REPLY [9 votes]: If a proof with elementary arguments is desired, here's a simple one:
Suppose first that $f$ is strictly increasing on $I=(a,b)$, and $y\in f(I)$, with $f^{-1}(y)=x.$ 
$\forall \epsilon>0$, let $\epsilon'=\min\{\epsilon, \frac{b-x}{2}, \frac{x-a}{2}\}.$ Suppose $f(x+\epsilon')=y+\delta_1$ and $f(x-\epsilon')=y-\delta_2$. Since $f$ is strictly increasing, clearly $f^{-1}(t)<x+\epsilon'$ if $t<y+\delta_1$. Similarly, $f^{-1}(t)>x-\epsilon'$ if $t>y-\delta_2.$  (Obviously $t$ also has to be in $f(I)$.)  Now, with $\delta=\min(\delta_1, \delta_2)$, we have $|f^{-1}(t)-f^{-1}(y)|=|f^{-1}(t)-x|<\epsilon'\le \epsilon,$ if $|t-y|<\delta.$ So $f^{-1}$ is continuous.
Note that $f$ doesn't even need to be continuous in the arguments above, and the extension of the proof to a strictly decreasing $f$, or $I=[a,b], [a,b),$ or $(a,b]$ is straightforward.