Inverse of Strictly Monotone Function

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

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