Inverse of Composite Bijection/Proof 1
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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.
$\blacksquare$