Let G be a braided monoidal groupoid: it does no harm to suppose that the monoidal product on G is *strictly* associative, so I'll do that.

"With inverses" means that for every object $X$ of G, there is an object $Y$ and an isomorphism $X\otimes Y\approx 1$ in G, where $1$ is the unit object.

I'd like to assume, without loss of generality, that inverses exist "on the nose", so that for every object $X$, there is an object $X^{-1}$ such that $X\otimes X^{-1} = 1$. That is, the objects of G form a group.

First question: am I allowed to do this?

Now I can get a 2-category BG by "delooping" G (using the monoidal structure), so that BG has only one object *, and the category of morphisms BG(*,*) is exactly G. This is a 2-groupoid with one object, and it carries some sort of additional structure encoding the braiding.

A connected 2-groupoid is exactly the same thing as a *crossed module*, which consists of data $(H,F, d: H\to F, \phi: F\to \mathrm{Aut}(H))$, where $H$ and $F$ are groups and $d$ and $\phi$ are homomorphisms. In terms of G, F is the group of objects of G, while H is the set of 1-morphisms in G with unit object as domain.

Second question: what extra structure do I put on the crossed module to encode the braiding?

I want to understand such G which are *free* on some set S of objects. In the translation to crossed modules, the group F will have to be the free group on S.

Third question: how do you describe the group H in this crossed module? (That's what I mean by "explicit".)

There's an extensive literature on braided monoidal categories, so I bet someone has thought about this.

(Oh, and I can deloop one more time to get a weak 3-groupoid B^{2}G, and this thing will model a homotopy 3-type X. If G is free, X is a wedge of 2-spheres. Because of this, I know things like the image and kernel of $d: H\to F$. But what is $H$ itself?)

onegenerator, so it amounts to the same thing to give an explicit description of a coproduct of braided monoidal groupoids with inverses. – Charles Rezk Nov 11 '09 at 8:14not2-category) of strict braided monoidal groupoids with strict inverses and (strict?) braided monoidal functors? Otherwise, I don't see why H and F would be uniquely determined. – Reid Barton Nov 11 '09 at 18:13