Composite of Bijections is Bijection/Proof 2

Theorem
Every composite of bijections is also a bijection.

That is:
 * If $f$ and $g$ are both bijections, then so is $f \circ g$.

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

Then from Bijection iff Inverse is Bijection, both $f^{-1}$ and $g^{-1}$ are bijections.

From Inverse of Composite Relation we have that $g^{-1} \circ f^{-1}$ is the inverse of $f \circ g$.

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

Hence the result.