Proof
(related to Proposition: The Inverse Of a Composition)
 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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References
Bibliography
 Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994