Category = Groupoid x Poset? - MathOverflow

Category = Groupoid x Poset?

2010-03-24T13:29:11Z

Is it possible to split a given category $C$ up into its groupoid of isomorphisms and a category that resembles a poset?

"Splitting up" should be that $C$ can be expressed as some kind of extension of a groupoid $G$ by a poset $P$ (or "directed category" $P$ the only epimorphisms in $P$ are the identities, all isomorphisms in $P$ are identities).

Answer by Mikola for Category = Groupoid x Poset?

2010-03-24T14:57:20Z

Given any locally small category, $C$, the collection of all isomorphisms forms a subgroupoid, $G \subseteq C$, where $Ob(G) = Ob(C)$ and $Hom_G(A,B) = \left ( f \in Hom_C(A,B) : \exists g, h \in Hom_C(B,A) g \circ f = id_A, f \circ h = id_B \right ) $.

Because $G$ is a groupoid, it determines an equivalence relation, $R$ on the objects and morphisms of $C$ such for $A, B \in Ob(C)$:

$A \equiv_R B \Longleftrightarrow Hom_G(A,B) \neq \emptyset$

And for $f, g \in Hom_C(A,B)$:

$f \equiv_{R_{A,B}} g \Longleftrightarrow \exists h_B \in Hom_G(B,B), h_A \in Hom_G(A,A) : h_B \circ f = g \circ h_A$

If I understand what you are asking, then the quotient $C/R$ should be the 'poset' you want.

Answer by Peter Arndt for Category = Groupoid x Poset?

2010-03-24T15:02:38Z

I am also looking forward to answers to your question. Meanwhile here is something pointing roughly into that direction:

One can study a category $C$ through its set-valued functor category $Set^C$. By the Yoneda lemma, $C$ sits as a full subcategory inside this functor category, and from it one can reconstruct something close to $C$ (I think the idempotent completion of $C$). But non-equivalent categories can give rise to equivalent functor categories, e.g. category $C$ in which not every idempotent splits and its idempotent completion, i.e. the category made from $C$ by adjoining objects such that each idempotent becomes a composition of projection to and inclusion of a subobject and thus splits. One calls such categories Morita-equivalent.

Now $Set^C$ is a Grothendieck topos (:=category of sheaves on a site, in this case with trivial topology) and there is the following theorem about those:

A locale is a distributive lattice closed under meets and finite joins, just like the lattice of open sets of a topological space, so it is a particular poset. The theorem of Joyal and Tierney, from their monograph "An extension of the Galois theory of Grothendieck", states that every Grothendieck topos is equivalent to the category of $G$-equivariant sheaves on a groupoid object in locales - see e.g. here.

Well at least it is a statement which separates a category into a groupoid and a poset part. So if you look from very far and take it with a boulder of salt you could read this as saying that every category is "Morita-equivalent" (not really!) to a groupoid internal to posets (it makes some intuitive sense to see this as an extension).

Answer by Tilman for Category = Groupoid x Poset?

2010-03-24T16:17:39Z

Unless I totally misunderstand the question, this doesn't even work for categories with one object, i.e. monoids (which are not groups), does it?

Answer by Ben Wieland for Category = Groupoid x Poset?

2010-03-24T23:05:07Z

One type of category that factors nicely is called an EI category. The definition is that every Endomorphism is an Isomorphism. After taking the quotient by the groupoid, every endomorphism is the identity. But it is still not a poset. It could be something like the category of two parallel arrows, where Mor(A,B) has two elements, Mor(B,A) none, and the endomorphisms of each object are only the identity. This has a further poset quotient $A\to B$, but it isn't there yet.

So groupoid and poset are only two kinds of behavior in categories. Monoids are a third that have been mentioned before. In particular, idempotents, as in the monoid {0,1} under multiplication, do not embed in any group. And the two parallel arrows category is yet a fourth.