Suppose that $f:A \rightarrow B$ is surjective , then there exists a subset of $A$ like $C$ such that $C$ and $B$ are equipotent. I've used this property a couple of times but I've never tried to prove this. I researched a little bit and it was said that we need the axiom of choice. Where do we need this? Sorry if this sounds dumb. If an explanation of this is available somewhere I'd like to see a link. Thank you very much!
2026-04-08 13:11:55.1775653915
Why do we need axiom of choice to prove that if $f\colon A\to B$ is surjective, then there is $C\subseteq A$ such that $C$ and $B$ are equipotent?
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Yes, you need to use the Axiom of Choice for this proof. Or rather the Partition Principle, which says that if $A$ can be mapped onto $B$, then $B$ can be injected into $A$. The range of such an injection is this set $C$ you're looking for.
It is currently open whether or not the Partition Principle implies the Axiom of Choice. But it is certainly not provable in $\sf ZF$.
For example, it is always true that $\Bbb R$ can be mapped onto $\{A\subseteq\Bbb R\mid A\text{ is countable}\}$. But it is not provable that there is an injection in the other direction. Similarly, it is always true that $\Bbb R$ can be mapped onto $\omega_1$, but again, it is not provable that there is an injection from $\omega_1$ into $\Bbb R$.