Working in "[MK](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory)-Regularity-Limitation of Size + Replacement for sets", call it the Base theory, let's coin the following axiom:


**Axiom of Super-Choice:**$$\forall \ relation \ R \ \exists F \subset R \ (F: dom(R) \to rng(R))$$ Where: $$R \text { is a relation } \equiv\_{df} \forall r \in R \ \exists x,y \ (r=\langle x,y \rangle)\\dom(R)=\{x| \exists y (\langle x,y \rangle \in R)\} \\rng(R)=\{y|\exists x (\langle x,y \rangle \in R)\}$$


I think this would be equivalent to Global Choice over the Base theory.


Now if we weaken the above to:


**Axiom of Proper Class Choice:**$$\forall \ relation \ R \ \text{ with proper class rows } \\\exists F \subset R \ (F: dom(R) \to rng(R))$$


where: $$ z \text{ is a row of }R \iff 
\\ \exists x \in dom(R)[z=\{y| \langle x,y \rangle \in R\}]$$


A relation $R$ is with proper class rows, if every row of it is a proper class.



> 
> Would that still be equivalent to Global Choice over the Base theory?
> 
> 
> If not, then if we also add Choice over sets [i.e., every set has a choice function], would that result in proving Global Choice?
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> 
> 



Is Proper Class Choice equivalent to Global Choice?
Yes, this is still equivalent: given any relation $R$, consider the new relation $$R^{bigrows}=\{\langle x,y\rangle: \exists a,b(y=\langle a,b\rangle\wedge \langle x,a\rangle\in R)\}.$$ Basically, $R^{bigrows}$ just "pads out" the rows of $R$ with a dummy coordinate. 


But from a choice function $F$ for $R^{bigrows}$ we can extract one for $R$ itself: consider $$x\mapsto \pi\_0(F(x))$$ (where $\pi\_0$ denotes projection onto the left coordinate).


