Is composition of reversible functions reversible too?

Question

This sounds obvious. Too obvious for me to be able to prove it formally. Can you please help with a formal proof of this?

Answer 1

Assuming that by "reversible" you mean "invertible," suppose $f,g$ are invertible with inverses $f^{-1},g^{-1}$ respectively. We claim the inverse of $f\circ g$ is $g^{-1}\circ f^{-1}$. Indeed, $$(f\circ g)\circ (g^{-1}\circ f^{-1})=f\circ \text{Id}\circ f^{-1}=\text{Id}$$ $$(g^{-1}\circ f^{-1})\circ (f\circ g)=g^{-1}\circ \text{Id}\circ g=\text{Id}.$$ Therefore $f\circ g$ is invertible with inverse $g^{-1}\circ f^{-1}$,

Answer 2

It is required to show that, given $y$ in the range of $f$ composition $g$, there is precisely $1$ $x$ such that $f(g(x))=y$. Firstly, since f is reversible or bijective, there is only one $g(x)$ such that $f(g(x))=y$. Since $g$ is also reversible, we find that there is only one $x$ such that $g(x)=g(x)$. Therefore, $f$ composition $g$ is also reversible.