Inverse of Composite Bijection/Proof 2

Theorem
Let $f$ and $g$ be bijections.

Then:


 * $\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$

and $f^{-1} \circ g^{-1}$ is itself a bijection.

Proof
Let $g: X \to Y$ and $f: Y \to Z$ be bijections.

Then:

Hence the result.