TITLE: Half Cantor-Bernstein Without Choice
QUESTION [22 upvotes]: 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)

REPLY [3 votes]: To your original question, it seems worth mentioning the point that your hypothesis implies the following:

If $|A| = |A \times 2|$ and there is a surjection from $A$ onto $B$, then $B$ injects into $A$.

This really just generalizes the example given by Ricky, but to see this:
Suppose we have a bijection $b: A \rightarrow A \times \{0, 1\}$ and a surjection $g: A \rightarrow B$.  $A$ injects into $A \cup B$ by the inclusion map, and we can get a surjection from $A$ onto $A \cup B$ by mapping an $a \in A$ to itself if the second coordinate of $b(a)$ is $0$ or to $g(a)$ if the second coordinate of $b(a)$ is $1$.  In other words, we split $A$ into two copies of itself, mapping the first copy surjectively (bijectively) onto itself via the identity and the second copy surjectively onto $B$ via $g$.  By your stated principle, this would then mean that we have a bijection $h$ from $A$ onto $A \cup B$ whereby $h^{-1} \upharpoonright B$ would be an injection from $B$ into $A$.
From this observation, I think it becomes intuitively clear how your principle is a form of choice in disguise.  In particular, your principle would imply that $\alpha^+$ injects into $\mathcal{P}(\alpha)$ for every infinite ordinal $\alpha$ since we have a surjection from $\mathcal{P}(\alpha)$ onto $\alpha^+$ (even without choice by virtue of the fact that every ordinal of size $|\alpha|$ is encoded by a subset of $\alpha$).  But as I'm sure you're already aware, models of ZF have been constructed where this will not be the case.