TITLE: 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?
QUESTION [9 upvotes]: I've been trying to find this proof:

If there exists $f \colon A\to B$ injective and $g \colon A \to B$ surjective, prove there exists $h \colon A \to B$ bijective.

I thought of using cardinality, but I think it's possible to prove that without using it. Anyone knows how to?

REPLY [4 votes]: As soon as we know that existence of surjective map $A\to B$ is equivalent implies to existence of injective map $B\to A$, this is the Cantor-Bernstein theorem.
The above claim follows from the following equivalent form of Axiom of Choice:
(S) For every surjective map $g: A\to B$ there exists a map $h: B\to A$ such that $(\forall b\in B) h(g(b))=b$.
https://en.wikipedia.org/wiki/Axiom_of_Choice#Equivalents
The proof of AC $\Rightarrow$ (S) was given in Asaf's post. A map $h$ fulfilling the above property is injective.

Some proofs of Cantor-Bernstein can be found here:
https://www.proofwiki.org/wiki/Cantor-Bernstein-Schroeder_Theorem
I do like the proof which uses Knaster-Tarski theorem. (In fact proof 3 from the above link is the same thing as showing Knaster-Tarski in this special case.)
https://planetmath.org/ProofOfSchroederBernsteinTheoremUsingTarskiKnasterTheorem
Brown, Pearcy: An introduction to analysis, Theorem 4.1

EDIT: GEdgar pointed out an error in my original formulation. Let me address this shortly. (I hope I remember it correctly.) The following two claims work for any sets $A$, $B$:

$g: A\to B$ is surjective $\Leftrightarrow$ there exists $h: B\to A$ such that $g\circ h=id_B$. (I.e., $g$ has right inverse.)
We are given a map $h: B\to A$. If there exists $g: A\to B$ such that $h\circ g=id_A$, then $h$ is injective.

Under the assumption $B\ne\emptyset$ we get

$h: B\to A$ is injective $\Rightarrow$ there exists $g: A\to B$ such that $h\circ g=id_A$. (I.e., $h$ has left inverse.)

The last claim is not true for $B=\emptyset$ and $A\ne\emptyset$, since empty function from $\emptyset$ fo $A$ is injective, but there is no function from a non-empty set to $\emptyset$.