Inverse of Composite Bijection/Proof 1

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.

Proof

$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of Inverse of Composite Relation.

As $f$ and $g$ are bijections then by Bijection iff Inverse is Bijection, so are both $f^{-1}$ and $g^{-1}$.

By Composite of Bijections is Bijection, it follows that $f^{-1} \circ g^{-1}$ is a bijection.

$\blacksquare$