Inverse of Composite Bijection

Theorem
Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$.

Then:


 * $\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$

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