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Lete $f:A\to B$ be a surjective map.

Show that $f$ is right invertible.

To prove this we need to construct a map $h:B\to A$ such that $f(h)=i_B$. Now as $f$ surjective so for each $b\in B$ there exist $a\in A$ such that $f(a)=b;$Let $A_b:=${$a\in A: f(a)=b$} Now choose any arbitrary element of $A_b$ for each $b$ in $B$. Now clearly it is the right inverse of $f$. But I want a verification that such a map from $B$ to $A$ is well defined.

user128956
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1 Answers1

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You are right that you want to verify that this is a well-defined function. Let $a_b\in A_b$ the chosen element. Then we want to check that $g=\{\langle b,a_b\rangle\mid b\in B\}$ is an injective function from $B$ into $A$.

To see that it is a function, note that if $\langle b,x\rangle$ and $\langle b,y\rangle$ are both in $g$, then by the definition of $g$, $x=a_b$ and $y=a_b$. While our choice was arbitrary, we only chose one element from each $A_b$. Therefore $x=y$ and so $g$ is indeed a function.

The rest, as you note, is quite simple.

It should be noted that (1) the axiom of choice is definitely needed for making these arbitrary choices; (2) depending on how you formulate the axiom of choice, $g$ may just be taken as a choice function for the family $\{A_b\mid b\in B\}$, which you have shown to be a family of non-empty sets; and (3) in either way, this follows from the fact that the axiom of choice (in any of its infinitely many formulations) gives arbitrary choices, but well-defined choices.

Asaf Karagila
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  • Just by curiosity : to prove that $g$ is well defined, why you prove that $g$ is injective ? Don't we only have to prove that if $b,c\in B$, are such that $g(b)\neq g(c)$ then $b\neq c$ ? – user380364 Aug 30 '18 at 15:55
  • Injectivity is left to the reader as an exercise. Indeed it is a simple application of the fact a function is a function. – Asaf Karagila Aug 30 '18 at 15:59
  • sorry ? I'm not asking how prove injectivity... Just "why do you prove injectivity of $g$ to prove that $g$ is well defined ?" (sorry if it was not clear from the beginning). Because for me $g$ is well defined if $g(b)$ has at most one image for all $b\in B$. Here, if $g(b)\neq g(c)$ then $b\neq c$ because $f^{-1}(i)$ are disjoint. – user380364 Aug 30 '18 at 16:03
  • what I said doesn't make sense ? – user380364 Aug 30 '18 at 17:18
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    Err, sorry. Right. Note that I wrote "we want to check that $g=\dots$ is an injective *function* from $B$ into $A$". I am not saying it is enough to verify it is injective to see that it is well-defined. In fact the next sentence, literally, checks that it is a function. Which in your words, is to check that it is well-defined. – Asaf Karagila Aug 30 '18 at 17:24
  • I got it, thanks a lot :) – user380364 Aug 30 '18 at 17:31