Choosing a function $f\subseteq R$ with same domain is equivalent to Global Choice?

Question

Assertion. For every class relation $R$ with $\operatorname{dom}(R)$ being a set, there exists a (set) function $f\subseteq R$ with $\operatorname{dom}(f) = \operatorname{dom}(R)$.

Question. Is the above assertion true without using the global axiom of choice?

This question is similar to the analog with sets, asserting that

Proposition. $\mathsf{AC}$ is equivalent to the statement "for every set $A$ and relation $R\subseteq A \times A$, there exists a function $f\subseteq R$ with $\operatorname{dom}(f)=\operatorname{dom}(R)$". (cf. this question.)

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 $A \in \mathcal{F}$. $$\tag*{$\blacksquare$}$$

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.)

Answer 1

Final edit: no, this axiom does not imply global choice. However, if we remove the restriction that $\operatorname{dom}(R)$ is a set, we do get something strong enough to imply global choice.

Your principle is equivalent to small collection + local choice.

We work without foundation. We have the following 4 principles, each implying the next: $(1)$ the axiom of foundation; $(2)$ there is a function $\operatorname{rank} \colon V \to \operatorname{Ord}$ such that $\operatorname{rank}^{-1}(\alpha)$ is a set for all $\alpha \in \operatorname{Ord}$; $(3)$ large collection; and $(4)$ small collection.

Hopefully, you are aware of how to construct the canonical rank function using foundation. So I will skip $(1) \implies (2)$.

Large collection can be phrased as follows: consider any class $R \subseteq V^2$. There exists a class $S \subseteq R$ with $\operatorname{dom}(S) = \operatorname{dom}(R)$ such that for all $x$, $\{y \mid (x, y) \in S\}$ is a set. To prove large collection from (2), simply define $S = \{(x, y) \in R \mid \forall z ((x, z) \in R \to \operatorname{rank}(y) \leq \operatorname{rank}(z))\}$.

First, consider the small collection theorem, which states that for all classes $R \subseteq V^2$ with $\operatorname{dom}(R)$ a set, there exists a set $s \subseteq R$ with $\operatorname{dom}(s) = \operatorname{dom}(R)$. It can easily be shown that an $S$ chosen by applying large collection to $R$ will, in this case, be a set.

Now let us note that your principle clearly implies both small collection and local choice. This is trivial.

Conversely, suppose we have small collection and local choice. Then pick some $s \subseteq R$ such that $\operatorname{dom}(s) = \operatorname{dom}(R)$. Then for each $x \in \operatorname{dom}(s)$, we can, using local choice, pick some $y \in \operatorname{range}(s)$ such that $(x, y) \in s$. This only requires a choice function for $\operatorname{range}(s)$, which is small. $\square$

Now if we assume foundation, we have small collection as a theorem. Thus, your principle is equivalent only to local choice, which does not logically entail global choice. This is well-known.

We can also get to principle (2) using an alternate axiom; the axiom of anti-foundation. NBG is typically not formulated with anti-foundation, but it can be done in principle. Without foundation or anti-foundation, it is consistent that choice holds, but small collection fails.

I don’t know whether any of the implications $(2) \implies (3) \implies (4)$ can be reversed, but I suspect not.

If you drop the restriction that $\operatorname{dom}(R)$ is small and that $f$ is small, your principle is equivalent to global choice + large collection by essentially the same argument.

Answer 2

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.