Inverse of Composite Bijection

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

 * $\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ 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, it follows that $f^{-1} \circ g^{-1}$ is a bijection.