Proof

By hypothesis, $f:A\mapsto B$ and $g:B\mapsto C$ are bijective functions.
It is clear that both functions are invertible. Let $f^{-1}$ and $g^{-1}$ be their inverse functions.
We want to show that for all $c\in C$ we have $(g\circ f)^{-1}(c)=(f^{-1}\circ g^{-1})(c).$
If $c\in C$ , then there is an $a\in A$ with $(g\circ f)(a)=g(f(a))=c,$ since the composition is a function, and in particular it is surjective.
Since $g\circ f$ is bijective, it has an inverse $(g\circ f)^{-1}$ and we have $(g\circ f)^{-1}(c)=a.$ $(*)$
On the other hand, we have $(f^{-1}\circ g^{-1})(c)=f^{-1}(g^{-1}(c))$ .
Since $c=g(f(a))$ we have further $(f^{-1}\circ g^{-1})(c)=f^{-1}(g^{-1}(g(f(a)))=f^{-1}(f(a))=a$ . $(**)$
Comparing $(*)$ with $(**)$ we get $(g\circ f)^{-1}(c)=(f^{-1}\circ g^{-1})(c)$ for all $c\in C.$
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