Prove, without using the Axiom of Choice:
if f: X->Y is surjective and Y is finite, there exists a 'section', a function s:Y->X such that f(s(y))=y for all y in Y
Hint: perform induction over the cardinality of Y
The Attempt at a Solution
Induction over the cardinality of Y? I don't even know what they mean. I'd say this would require the AC because we need to pick one of the elements of f^-1 (y) for all y in Y. since X is infinite, one of these f^-1 (y) has to be infinite as well, and I don't know how to pick a specific element of this if I can't just say 'I pick some'.
And I also don't understand the hint. Some help would be appreciated.