Question 1. Is the following assertion true without using the global axiom of choice: every class realtion $R$ with $\operatorname{dom}(R)$ being a set, there exists a function $f\subseteq R$ with $\operatorname{dom}(f) = \operatorname{dom}(R)$?
This question is similar to the (true) statement that
Proposition 2. $\mathsf{AC}$ is equivalent to the assertion that for every set $A$ and relation $R\subseteq A \times A$, there exists a function $f\subseteq R$ with $\operatorname{dom}(f)=\operatorname{dom}(R)$.
(This is similar to the problem stated here.)
Proof.
$(\Longrightarrow)$ Assume $\mathsf{AC}$. Let $A_0 = \operatorname{dom}(R)\subseteq A $ (this is a set by separation) and denote $\mathcal{F} = \{R[a_0] \mid a_0 \in A_0\}$ (this is a set by replacement). By definition of $A_0$, the set $\mathcal{F}$ is a family of non-empty sets, therefore it admits a choice function $c:\mathcal{F}\to \bigcup \mathcal{F}$. Define $f:A_0 \to A$ by $f(a_0)=c(R[a_0])\in R[a_0]$, so if $b=f(a_0)$ we have $a_0 R b$, therefore, $f\subseteq R$.
$(\Longleftarrow)$ Assume the assertion above. Let $\mathcal{F}$ be a family of non-empty sets. Define a relation $R$ on $\mathcal{P}(\bigcup \mathcal{F})$ by
$$R =\{ \langle A,\{a\}\rangle \mid A\in \mathcal{F}, a\in A\} .$$
Note that $\operatorname{dom}(R) = \mathcal{F}$. From assumption, there exists a function $f\subseteq R$ with $\operatorname{dom}(f) = \operatorname{dom}(R) = \mathcal{F}$. Define $c:\mathcal{F}\to \bigcup \mathcal F, A\mapsto \bigcup f(A) $. From the definition of $R$, we have $c(A)\in A$ for each $A \in \mathcal{F}$. $\mathrm{~~~~~~} \blacksquare$
We cannot use the same trick for the question above, as $R[a_0]$ may not be a set. Is it true that we have to use global choice, or is there a way to prove it without it?
(I see that this post asks a similar question, but not quite the same.)
Best Answer
Elaborating on my comment, define a class function $F$ from $A := dom(R)$ to the universe, which maps $a \in A$ to the smallest set $V_\alpha$ in the cumulative hierarchy such that $a \in V_\alpha$ (this is indeed definable). By replacement, $F(A)$ is a set. Let $B = \bigcup F(A)$. Then for each $a \in A$, $B$ is a set containing at least one element in $R[a]$. Now we can apply AC.