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

ferdinandutkin
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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 $g^{-1}\circ f^{-1}$,

Connor Gordon
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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.

George Adams
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    Your argument as written is not quite complete. In particular, bijectivity of $f$ tells you there exists a unique $z$ such that $f(z)=y$. You need to use bijectivity of $g$ now to say that $z=g(x)$ for a unique $x$. – Stefan Dawydiak Apr 19 '22 at 15:54
  • This is what I did by saying "there is only one $x$ such that $g(x)=g(x)$." but perhaps the notation could do with some work. – George Adams Apr 19 '22 at 16:28