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I had a discussion with one of my teachers the other day, which boiled to the following question:

Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and $g\colon A\to B$ which is surjective.

Is there $h\colon A\to B$ which is bijective?

Of course it is enough to show that there is an injection from $B$ into $A$, and by the Cantor-Bernstein theorem (which does not require choice) we finish the proof.

My intuition says that this is true, his intuition says it is false.

Insights, references and possible solutions will be appreciated!

As Ricky shows below, it depends on $A$ and $B$.

So to a more specific choice of sets (for which I think it is true) we have $A=\mathbb R^\omega$ (i.e. infinite sequences of real numbers) and $B=[\mathbb R]^\omega$ (i.e. countable subsets of real numbers)

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Half Cantor-Berstein Without Choice

I had a discussion with one of my teachers the other day, which boiled to the following question:

Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and $g\colon A\to B$ which is surjective.

Is there $h\colon A\to B$ which is bijective?

Of course it is enough to show that there is an injection from $B$ into $A$, and by the Cantor-Bernstein theorem (which does not require choice) we finish the proof.

My intuition says that this is true, his intuition says it is false.

Insights, references and possible solutions will be appreciated!