Let be a surjection.
then
Let
Using the Axiom of Choice, for each we can choose any of the elements to be identified as , and thereby define:
Then we see that
and thus .
Prove: f is surjective iff f has a right inverse.
f is surjective if for all b in B there is some a in A such that f(a) = b.
f has a right inverse if there is a function h: B ---> A such that f(h(b)) = b for every b in B.
i. Suppose f has a right inverse h: B --> A such that f(h(b)) = b for every b in B. Then for each b in B, let a in A such that f(a) = b. Let h(b) = a, then f(h(b)) = f(a) = b. Thus f is surjective.
ii. Suppose f is surjective. That is, for every b in B, there is some a in A such that f(a) = b. How to show that f has a right inverse? Please help. Thank you.
Excellent question.
We know that f is a surjection, but f is not necessarily bijective, thus there could be more than one element of S mapped to one element of T. Thus the inverse of f is not necessarily a mapping, as an element of T does not necessarily map to a single element of S.
So technically speaking you can not talk about the inverse element of y, but you can talk about the set of elements of S that the inverse of f relates to.
What I mean by is actually the preimage of , and in this context I find that considering y as a set (admittedly a singleton one) rather than an element enhances the understanding of what's going on.