When do surjections split in ZF? Two surjections imply bijection?

Question

We have that the Axiom of Choice is equivalent to the principle that every surjection has a right inverse. However, without the Axiom of Choice we can determine for some $X$ that $X\succeq Y\implies Y\preceq X$. Some immediate instances are when $X$ is finite (by the Principle of Finite Choice) and even when $X$ is countable (put in bijection with $\mathbb{N}$ and then choose the least element of the preimages).

I was wondering for what other $X$ does this hold? Clearly it holds for well-orderable sets, but does the principle extend to sets like $\mathcal{P}(\mathbb{N})$ or any other set that ZF can 'sanely' construct (as in be created without Choice)?

That is my first question. My second question is whether the following principle is equivalent to the Axiom of Choice: $$X\succeq Y\wedge Y\succeq X\implies X\simeq Y$$ This principle somewhat mirrors Schroeder-Bernstein, and can be derived with the Axiom of Choice; however, I've never seen a proof (not even a remark) that it is equivalent to or weaker than Choice.

These questions are just out of personal curiosity.

Notation: $X\succeq Y$ means $X$ surjects onto $Y$, and $Y\preceq X$ means $Y$ injects into $X$.

Answer 1

First of all notation. Using the same symbol in reverse for different things is horrible. Let's use the usual notation. $X\leq Y$ means there is an injection from $X$ into $Y$, $X\leq^* Y$ means that either $X$ is empty, or there is a surjection from $Y$ onto $X$, and we'll keep the $X\simeq Y$ as you did for the existence of a bijection.

In $\sf ZF$ the Cantor-Bernstein theorem holds. We don't need $\sf AC$ to prove: $$X\leq Y\land Y\leq X\iff X\simeq Y.$$

On the other hand, the dual notion for $\leq^*$ is not provable in $\sf ZF$. We call this the Dual Cantor-Bernstein Theorem, and it states the very thing we expect it to, $$X\leq^*Y\land Y\leq^*X\iff X\simeq Y.$$

When this statement is true, for example, then every infinite set has a countably infinite set. Something which is consistently false in $\sf ZF$.

The problem on how much choice it implies, and how much choice is needed to prove it is open. The only thing that was proved, to my knowledge, is that if we make a certain strengthening of this theorem then the axiom of choice holds, but we don't know anything else.

Now, for extending the reversibility of surjections, the answer is that in most cases, you can expect it to fail. For example, It is provable in $\sf ZF$ that $\mathcal P(\omega)$ can be mapped onto $\omega_1$, but there are models where there is no injection back. There is exactly one case that I know of where it is provable to have reversible surjections. The following case.

Definition. For an infinite $\aleph$ cardinal $\kappa$ we say that $A$ is $\kappa$-amorphous if every subset of $A$ is of size $<\kappa$, or its complement is of cardinality $<\kappa$. We say that $A$ is strongly $\kappa$-amorphous if for every $S$, a partition of $A$, the set $\{X\in S\mid |X|>1\}$ has size strictly less than $\kappa$.

Theorem. Suppose that $\sf AC_{<\kappa}$ holds, and $A$ is strongly $\kappa$-amorphous then every $f\colon A\to B$ which is surjective can be reversed.

(Where $\sf AC_{<\kappa}$ is the statement that every family of strictly less than $\kappa$ non-empty sets admits a choice function.)

Proof. Left to the interested reader. $\square$

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Further reading.

1. Is there a Cantor-Schroder-Bernstein statement about surjective maps?
2. Cantor-Bernstein-like theorem: If $f\colon A\to B$ is injection and $g\colon A\to B$ is surjective, can we prove there is a bijection as well?
3. Sur- in- bijections and cardinality.
4. Does a injective function $f: A \to B$ and surjective function $g : A\to B$ imply a bijective function exists?
5. Proving existence of a surjection $2^{\aleph_0} \to \aleph_1$ without AC
6. Dual Schroeder-Bernstein theorem (MathOverflow)
7. Half Cantor-Bernstein Without Choice (MathOverflow)
8. Does $2^X=2^Y\Rightarrow |X|=|Y|$ imply the axiom of choice? (MathOverflow)
9. Uncountable subset with uncountable complement, without the Axiom of Choice
10. Infinite Set is Disjoint Union of Two Infinite Sets
11. A question about the Axiom of Choice (MathOverflow)