Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

Yes, at least if the formula can be interpreted as coming from a map.

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 $[b] \mapsto [\alpha^{-1}(b)]$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.