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Given a groupoid $G,$ one can consider the canonical epimorphism $$G_0 \to G.$$ Since it is an epimorphism in the $2$-topos of groupoids, $G$ is the weak colimit of the corresponding Cech diagram formed by iterative (2-categorical) fibered products of this morphism against itself. Direct inspection shows that vertices of the $2$-cartesian cube arising from these fibered products can be identified with the (objects of) the nerve of $G$: $G_0$, $G_1$, and $G_2,$ whereas the edges of the cube can be identified with the face maps of the nerve.

My first question is:

Why is this truncated semi-simplicial nerve popping up here? And is there anyway to see what is going on geometrically? It seems like this has to do with relating the geometry of the corner of a cube to that of a 2-simplex.

Secondly, if I am given a (weak) semi-simplicial (truncated) groupoid, that is, groupoids $H_2$, $H_1$, and $H_0$ together with face maps respecting the simplicial identities up to natural isomorphism, let $H$ denote the (weak) colimit of this diagram. Let $$p:H_\cdot \to \Delta_{H}$$ be a colimiting cocone. What is the relationship between the (semi-simplicial) Cech nerve of $$p_0:H_0 \to H$$ and the original diagram?

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    $\begingroup$ Would you please tell me what $G_0$ is? $\endgroup$
    – Spice the Bird
    Mar 13, 2012 at 4:48
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    $\begingroup$ I guess that $G_0$ is the set of objects of $G$. But you may want to be more specific about what "the 2-cartesian cube" you have in mind is (I don't even know what it means for a cube to be "2-cartesian"). $\endgroup$
    – Mike Shulman
    Mar 13, 2012 at 7:52
  • $\begingroup$ @Spice: $G_n$ is notation borrowed from the nerve of a category (when viewed as a simplicial set), and explicitly is the set of $n$-composable arrows, or, in the case when $n=0$, is the set of objects. $\endgroup$
    – David Carchedi
    Mar 13, 2012 at 12:28
  • $\begingroup$ @Mike: I wish I could just make the diagram on here, but I don't know how. But, if you pullback $G_0 \to G$ against itself, you get a $2$-cartesian diagram with $G_1$ sitting as the weak pullback, and with the arrows out of it being the face maps $d_0$ and $d_1$. Imagine now that this is the bottom square of a soon-to-be cube, and place $$G_0 \to G$$ "above this square" and form all the other (weak) pullback squares you can until you have a cube. Some of these will be pullbacks of sets, so actual $1$-categorical pullbacks, but there will still be $2$-categorical one as well. (cont) $\endgroup$
    – David Carchedi
    Mar 13, 2012 at 12:33
  • $\begingroup$ Then, if $G$ is sitting at the bottom face of this cube, on the right and at the front, then on the top face of the cube, on the left, and in the back, is $G_2$, and all the other vertices are $G_0$ and $G_1$ and the edges are all the face maps. The fact that this diagram $2$-commutes, is a manifestation of the simplicial identities for face maps. $\endgroup$
    – David Carchedi
    Mar 13, 2012 at 12:37

2 Answers 2

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The answer to your second question is that the nerve of $p_0$ admits a map from the original diagram, which is "universal among maps into (2,1)-congruences". This is a categorified version of what happens when you take the coequalizer of a parallel pair of morphisms, then the kernel pair of that coequalizer. If the diagram you started with was already a (2,1)-congruence, then it will be equivalent to the nerve of its colimit, because Gpd is a (2,1)-exact (2,1)-category.

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  • $\begingroup$ Thanks Mike. So, is $p_0$ always epi, so that at least the two (possibly different) diagrams have the same colimit? $\endgroup$
    – David Carchedi
    Mar 13, 2012 at 12:40
  • $\begingroup$ Mike, do you know the answer to my first question? I feel like it should be something elementary. $\endgroup$
    – David Carchedi
    Mar 13, 2012 at 16:48
  • $\begingroup$ Yes, $p_0$ is always essentially surjective (the relevant notion of "epi", or more precisely "effective epi" or "strong epi", in this context). $\endgroup$
    – Mike Shulman
    Mar 13, 2012 at 17:29
  • $\begingroup$ Hi Mike, I just realized I never accepted this! $\endgroup$
    – David Carchedi
    Mar 5, 2013 at 22:25
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A topological $n$-simplex can be defined as

$$\Delta^n = \Big\{ (x_0,\dots,x_n) \;\Big|\; 0 \le x_i \quad\text{and}\quad \sum_i x_i = 1 \Big\}$$

In other words, it's a slice that cuts off the neighborhood of a corner of a cube.

Combinatorially, this corresponds to identifying vertices of an $(n+1)$-cube with subsets of $\{0,\dots,n\}$, and then identifying those subsets with the images of coface maps into $[n]$.

Does this help to answer your first question at all?

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