Proving Surjective Functions with Finite Y: No Axiom of Choice Required

[jacobrhcp]

Homework Statement

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.

 

[HallsofIvy]

You don't know "proof by induction"? Surely you are just misunderstanding. Since Y is finite, its cardinality is some positive integer, say n. Prove this statement by induction on n.

If n= 1, that means that Y contains exactly one member, call it "y". There exist some function from X such that f(x)= y for all x in X. Can you find a function from Y to X, s(y) such that f(s(y))= y? That should be pretty easy.

Now, suppose that statement is true when y contains n members and prove that it is true and prove it is true when Y contains n+1 members. (Hint: remove a specific one of the members of Y.)

 

[jacobrhcp]

Ah, thank you, yes I was misunderstanding. I do know proof by induction yes, but 'over the cardinality of Y' was a weird statement to me. This clears things up, I should be fine now.