A classification of the types of poset extensions that raise local-global principle is equivelent to the axiom of choice?

75 Views Asked by At

EDIT: Completely overhauled the question, remove category theory from it (stated in plain set theory):

The original question Adding relations to a partial order

Let $X$ be a set and let $P: Pow(X) \rightarrow Partialorders(X)$ be a multivalued function ($\leq\in P(A)$ is read "$(A,\leq)$ satisfies property $P$").

The following are equivelent:

(1). The following statements are true:

(1a).For all $\leq\in P(X)$ and for all $A\subseteq X$ completely isolated (by $\leq$) subset there is a $\leq ' \in P(X)$ s.t. $\leq\subseteq\leq '$ (subsumption) and in $(X,\leq ')$: $A$ is continuous.

(1b). If For all $\leq\in P(X)$ and for all $A\subseteq X$ completely isolated (by $\leq$) subsets and for all $\leq ' \in P(A)$ s.t. $\leq|\substack{A}\subseteq\leq '$ (subsumption) and in $(A,\leq ')$ $A$ is continuous --------> then there is a partial order $\leq '' \in P(X)$ that subsumes both $\leq$ and $\leq '$.

(2). The following statements are true:

(2a). For all $A\subseteq X$ and for all $\leq \in P(A)$ if $B\subseteq A$ then $\leq\in P(B)$

(2b). For all $A\subseteq X$ every total order $\leq $ on $A$ satisfies: $\leq\in P(A)$.

Explanation in human language:

(1) basically means that every completely disconnected set can be extended to a continous set that preserves $P$ (this is 1a), and every extension of completely disconnected sets to continuous one that is locally in $P$ is globally in $P$ (this is 1b).

(2) gives a characteristic of extensions that satisfy this condition (the fact that (2)=>(1) is a generalization of the answer to the previous question which uses Zorn's lemma). It means that the inclusion map preserves the property $P$ and total orders satisfy it.

The other way around is much trickier.

Note that the exact question I'm asking is: (1)<=>(2) $\equiv$ axiom of choice

Showing that the axiom of choice implies (1)=>(2) is praticulary important and I belive it's more than enough (prooving the AC from (1)<=>(2) looks like a nightmare to even approach)

I'm going to use this extension with wildly bad behaved properties $P$ so I must know exactly how bad they can be, and I conjectured (2) is an exact requirement.

Thanks in advance for all attempting to help.