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.