Suppose f:A→B and g:B→A are both surjective, does this imply that there is a bijection between A and B.
I was told that the statement above is true only with Axiom of Choice. Can someone provide an example of why that is the case?
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1With the Axiom of Choice you can pick right inverses for both $f$ and $g$, and then use regular Cantor-Bernstein on those inverses. However, this result implies the Axiom of Countable Choice, so it cannot be proven in ZF. See this math.overflow post – Arturo Magidin Nov 20 '20 at 23:48
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See also these previous threads here: one, two, and three – Arturo Magidin Nov 20 '20 at 23:51