Inverse of Strictly Monotone Function

Theorem
Let $f$ be a real function which is defined on $I \subseteq \R$.

Let $f$ be strictly monotone on $I$.

Let the image of $f$ be $J$.

Then $f$ always has an inverse function $f^{-1}$ and:
 * if $f$ is strictly increasing then so is $f^{-1}$;
 * if $f$ is strictly decreasing then so is $f^{-1}$.

Proof
The function $f$ is a bijection from Strictly Monotone Function is Bijective.

Hence from Bijection iff Inverse is Bijection, $f^{-1}$ always exists and is also a bijection.

From the definition of strictly increasing:
 * $x < y \iff f \left({x}\right) < f \left({y}\right)$

Hence:
 * $f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff f^{-1} \left({f \left({x}\right)}\right) < f^{-1} \left({f \left({y}\right)}\right)$

and so:
 * $f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff x < y$.

Similarly, from the definition of strictly decreasing:
 * $x < y \iff f \left({x}\right) > f \left({y}\right)$

Hence:
 * $f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff f^{-1} \left({f \left({x}\right)}\right) > f^{-1} \left({f \left({y}\right)}\right)$

and so:
 * $f^{-1} \left({x}\right) < f^{-1} \left({y}\right) \iff x > y$.