Existence of a bijection without AC

Question

Let $A$ and $B$ be subsets of a set $X$ and let $f:A\rightarrow X\setminus A$ and $g:B\rightarrow X\setminus B$ be bijections. Is it possible to show without AC that there is a bijection $h:A\rightarrow B$?

Answer 1

Yes. Let $\sim$ denote the "of the same cardinality" equivalence relation.

In the paper of Conway and Doyle, Division by Three they prove that if $A\times 2\sim B\times 2$ then $A\sim B$. Of course they argue only in ZF, without using the axiom of choice.

In fact they prove a more general theorem, but your question is a particular case:

If $A\sim X\setminus A$ then $X\sim A\times 2$; similarly for $B$. Therefore using Conway-Doyle we obtain that $A\sim B$.