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?
The axiom of choice asserts that for any set $X$ there exists a function $f : (2^X − \{\emptyset\}) \rightarrow X$ such that for any nonempty $A \subseteq X$, $f(A) \in A$. Show that this is equivalent to the assertion that every relation includes a function with the same domain. I am given this problem in my assignment. My approach is following. I have a set $X$. Then it has a choice function $g : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Suppose $X$ has a relation $R \subseteq X \times X$. $dom(R) \subseteq X$. Then the required function $f : dom(R) \rightarrow X$ is $$f(x) = g(\{y\ |\ (x, y) \in R\})$$ Conversely, suppose $X \supseteq Y$ is a non-empty subset. Let $$R_Y = \{(x, y)\ |\ y \in Y, x \in Y, (p, q) \in R_Y \Rightarrow p = x\}$$ Then there is a function $f_Y : dom(R_Y) \rightarrow X$. But $dom(R_Y)$ is singleton from the definition of $R_Y$. Then $im(f_Y)$ is also singleton. We define our choice function as $f = \{(Y, y)\ |\ y \in im(f_Y)\}$ Are my arguments correct?  Edit.1: what I tried to do for converse is, for any subset $Y$, I construct a relation $Y_1 \times Y$ where $Y \supseteq Y_1 = \{y\}$. Then there is a function $f_Y : Y_1 \rightarrow Y$. Then if $f$ is my choice function I define $f(Y) = f_Y(y)$. May be I should define $R_Y$ as $$R_Y = \{(x, y)\ |\ y \in Y; x \in Y; (p, q), (p', q) \in R_Y \Rightarrow p = p'\}$$ I understand there may be multiple $R_y$'s. But can't I define a singleton set as $\{s\in S\ |\ a, b \in S \Rightarrow a = b\}$? This can be $\{s\}$ $\forall s \in S$. But isn't it a valid specification?  Edit.2: This is the final solution with the help from @Asaf.  If we assume axiom of choice then for any $X$ there is a choice function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$ such that $f(A) \in A$. Suppose $R$ is a relation between $P$ and $Q$. Then we have a choice function $f_Q$ for $Q$. Let $f_R(p) = f_Q(\{q\ |\ pRq\})$. The function $f_R : P \rightarrow Q$ and $dom(f_R) = dom(R)$. So this is our required function.  Conversely, if every relation includes a function with same domain. Let $R$ be a relation between $\mathcal{P}(X) \setminus \left\{\emptyset\right\}$ and $X$ such that $aRb \Rightarrow b \in a$. Now $R$ includes a function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Then $xRf(x) \Rightarrow f(x) \in x$. So $f$ is the choice function for $X$.
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