Relation Contains Mapping is Equivalent to AoC

Theorem
Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$.

Then:
 * there exists a mapping $f \subseteq \mathcal R$ whose domain is the same as the preimage of $\mathcal R$

iff
 * the axiom of choice holds.