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 $h:A\rightarrow B$?
Existence of a bijection without AC
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set-theory
axiom-of-choice