# Is the inversion of an invertible map given by a formula again given by the inversion of that formula

This a problem I recently dealt with. It's fairly general.

Is the inversion of an invertible map $f\colon X \to Y$, given representatively by $[a] \mapsto [\alpha(a)]$, where $a$ rsp. $\alpha(a)$ represent elements in $X$ rsp. $Y$, – is its inversion $f^{-1}$ given by $[b] \mapsto [\alpha^{-1}(b)]$?, e.g. is $[a] \mapsto [a+1]$ inverted by $[b] \mapsto [b-1]$?

I think I have an answer, but I feel unsure about it. So I answered this question myself which always feels weird, although it's encouraged. Please have a look at it and check that I didn't make any mistakes. I'm also unsure about whether there are some special cases where a formula can't be interpreted as a map of suitable sets with suitable projections.

-

Let $\alpha\colon A \to B$ denote the map between representatives giving the formula and $\pi_A\colon A \to X$ and $\pi_B\colon B \to Y$ be the surjections which take representatives to their corresponding classes/elements which they represent. Then to say that $f\colon X \to Y$ is given by the formula $[a] \to [\alpha(a)]$ is saying that the obvious diagram commutes, i.e. $\pi_B \circ \alpha = f \circ \pi_A$. Now if $f$ and $\alpha$ are invertible, then you also have $f^{-1} \circ \pi_B = \pi_A \circ \alpha^{-1}$. And this is saying that $f^{-1}\colon Y \to X$ is given by the formula $[b] \mapsto [\alpha^{-1}(b)]$.