Invertible functions - proving that $g \circ f$ is invertible

Question

If I were to suppose that $f : A \to B$ and $g : B \to C$ are functions which are both invertible, how would I go about proving that $g\circ f$ is invertible with $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$? Any help is appreciated.

Answer 1

A function $h:A\rightarrow C$ is invertible if a function $k:C\rightarrow A$ exists such that $h\circ k=1_C$ and $k\circ h=1_A$.

Here $1_C:C\rightarrow C$ and $1_A:A\rightarrow A$ are the functions prescribed by $c\mapsto c$ and $a\mapsto a$ respectively.

In this context function $k$ is unique and is the so-called inverse of $f$. Often it is denoted as $h^{-1}$.

The fact that $f$ and $g$ are invertible tells us that there are functions $f^{-1}:B\rightarrow A$ and $g^{-1}:C\rightarrow B$ such that $f\circ f^{-1}=1_B$, $f^{-1}\circ f=1_A$, $g\circ g^{-1}=1_C$ and $g^{-1}\circ g=1_B$.

Based on that we find: $$(g\circ f)\circ (f^{-1}\circ g^{-1})=g\circ(f\circ f^{-1})\circ g^{-1}=g\circ 1_B\circ g^{-1}=g\circ g^{-1}=1_C$$ and:$$(f^{-1}\circ g^{-1})\circ(g\circ f)=f^{-1}\circ(g^{-1}\circ g)\circ f=f^{-1}\circ1_B\circ f=f^{-1}\circ f=1_A$$

Proved is now that $f^{-1}\circ g^{-1}$ is the (unique) inverse of $g\circ f$.

Answer 2

Assume $x,y \in A, x\ne y$. Then since f is invertible : $f(x) \ne f(y)$, and since g is invertible, $g(f(x)) \ne g(f(y))$. So $g \circ f$ is invertible.

Assume $g(f(x))=z$. Then $(g \circ f)^{-1}(z)=x$. But we also have $f(x)=g^{-1}(z)$, and this again gives that: $x = f^{-1}(g^{-1}(z))$. This must hold for any $z$, so $(g \circ f)^{-1}=f^{-1}\circ g^{-1}$.