Relationship between different versions of AoC in the context of a specific set (in ZF) We can show that various versions of AoC, restricted to a specific set, are equivalent. For example, the following are equivalent for any set, $X$ (in ZF):


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*[$A_X$] $X$ is well-orderable 

*[$B_X$] Every collection of non-empty subsets of $X$ has a choice function 


Similarly, the following statements are also equivalent for any set, $Z$ (in ZF):


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*[$C_Z$] Every partition of $Z$ has a set of representatives 

*[$D_Z$] $\forall Y$, every surjective $f:Z\rightarrow Y$ has a right inverse 


On the other hand, $C_X$, $D_X$ are strictly weaker than $A_X$, $B_X$ (see this answer from Asaf Karagila). My question is whether $A_X$ is equivalent to $C_{F(X)}$ for some definable class function, $F$ (i.e. $ZF\models \forall X(A_X\leftrightarrow C_{F(X)}$). 
I had hoped that letting $F(X)=\mathcal{P}(X)$ would work...but I now suspect it doesn't. Also, I think I can show that $C_{P(X)\times X}$ implies $A_X$, but I don't know whether the converse is true.
More generally, I'm just interested in understanding the relative strengths of these "localized" variants of AoC (even if the answer is not as nice as I had hoped). Thanks for any insights in advance!
 A: This is a fascinating topic, and there's a lot of room for exploration here. After learning the few basic proofs and counterexamples, one should be able to fill in a lot of gaps.
You are correct that $C_{\mathcal P(X)\times X}$ implies $A_X$. Simply because it implies $B_X$. Just consider the partition of $\mathcal P(X)\times X$ defined by: $\{\{(A, a)\mid a\in A\}\mid A\in\mathcal P(X)\}$, along with the remainder of the set. Clearly, a choice function from this is exactly a choice function witnessing $B_X$.
In the other direction, you are right to think the implication may fail. For example, it is possible for $\Bbb R$, standing in for $\mathcal P(\Bbb N)$, to have the property that it can be mapped on certain sets, but there are no injections in the other direction (let alone any inverse functions for the surjection!). And if any set can be well-ordered, $\Bbb N$ is that set.
Note that since $\Bbb R$ is the same size as $\Bbb{R\times N}$, and all the local principles you suggest depend on the cardinality, rather than the exact set, the result follows.
