User blog comment:Vel!/A medley of countability proofs/@comment-1605058-20130516053008/@comment-5529393-20130516230352

If there is a injection from A to B and an injection from B to A, then there is a bijection from A to B, and this is provable without the Axiom of Choice. The proof is by direct construction, and is quite nice. (This is know as the Cantor-Schroeder-Bernstein theorem, by the way.)

If there is a surjection both ways, then there is a bijection between the two, but this time the Axiom of Choice is required. I'm not sure about when there is an injection and a surjection from A to B, whether the Axiom of Choice is required.