Can Injective Functions Imply Surjective Ones Without the Axiom of Choice?

[jacobrhcp]

Homework Statement

if A is not empty and s:A->B is injective, then there is a surjective function f:B->A such that f(s(a))=a for all a in A. Do not use the Axiom of Choice

The Attempt at a Solution

for all b', b'' in B s(b')=s(b'') means b'=b''. So s^-1 (c) is unique in A. Because any point in A is sent to at most one point in B, we can just let f send every point in B of the form s(c) to c.

now we only need to send all the other points somewhere. Here I need to 'pick' some point once again. Why do I not need the AC here?

 

[tiny-tim]

now we only need to send all the other points somewhere. Here I need to 'pick' some point once again. Why do I not need the AC here?

Hi jacobrhcp!

Because the axiom of choice is only for choosing the whole set … here, you only need to choose one element.

For revision, see http://en.wikipedia.org/wiki/Axiom_of_choice