I am trying to establish the equivalence between the following statement and the Axiom of Choice:
Suppose that $R$ is a relation between two non-empty sets $A$ and $B$. There is a function $F$ with the same domain as $R$ such that $F \subseteq R$.
I have established the direction from the Axiom of Choice to this statement. My proof goes as follows.
For each set $x$, we may define the following set: \begin{equation*} A_{x} = \left\{y \in B: \langle x,y \rangle \in R\right\}. \end{equation*} Specifically, if $x \in \operatorname{dom}\left(R\right)$, then $A_{x} \neq \emptyset$. Further, we may define the following set: \begin{equation*} S = \left\{A_{x}: x \in \operatorname{dom}\left(R\right)\right\}. \end{equation*} It is obvious that $S$ is a set of non-empty sets and $\bigcup{S} \subseteq B$. According to the Axiom of Choice, there is a function $h: S \to \bigcup{S}$ such that \begin{equation*} \left(\forall T \in S\right)\left(h\left(T\right) \in T\right). \end{equation*} We may define a function $F: \operatorname{dom}\left(A\right) \to B$ such that \begin{equation*} F\left(x\right) = h\left(A_{x}\right),\ x \in \operatorname{dom}\left(R\right). \end{equation*} It is obvious that $F \subseteq R$.
I am wondering how to establish the reverse direction. I understand that there are already several equivalent forms of the Axiom of Choice. For instance, the following statement is the axiom used in my proof above:
Suppose that $\mathscr{F}$ is a family of non-empty sets. There is a function $h: \mathscr{F} \to \bigcup{\mathscr{F}}$ such that for each $A \in \mathscr{F}$, $h\left(A\right) \in A$.
We also have the following versions:
Suppose that $\mathscr{F}$ is a family of disjoint non-empty sets. There is a function $h: \mathscr{F} \to \bigcup{\mathscr{F}}$ such that for each $A \in \mathscr{F}$, $h\left(A\right) \in A$.
Suppose that $M$ is a non-empty set. Then there is a function $h: \mathscr{P}\left(M\right)\setminus \left\{\emptyset\right\} \to M$ such that for all non-empty subsets $A$ of $M$, $h\left(A\right) \in A$.
Can anyone help with completing the proof?
