# Writing an infinity groupoid as a colimit of sets

If we are given a simplicial set $X:\Delta^{op} \to Set$, we may regard it as a \emph{simplicial} simplicial set, i.e. a bisimplicial set by composing with the "constant" inclusion $Set \to Set^{\Delta^{op}}$. We may choose to view this as a simplicial diagram of infinity groupoids. Its infinity colimit may be computed as the homotopy colimit of this diagram in the Quillen model structure on simplicial sets. A nice model for the homotopy colimit of such a diagram is its diagonal, which in this case is precisely $X$ itself. Taking $X$ to be a Kan complex implies that any infinity groupoid can be written as the colimit of a simplicial diagram of sets (in the infinity category of infinity groupoids). However, in the case of $1$-groupoids, more is true. If $\Delta_+$ denotes the wide subcategory of $\Delta$ consisting of injective maps, if $\mathcal{G}$ is a groupoid, $\mathcal{G}$ is the weak colimit of its $\Delta_+$-nerve regarded as a diagram in the bicategory of groupoids (in fact, we can truncate $\Delta_+$ in this case to contain only $0$, $1$, and $2$, but that is not so important). The proof of this (which I did by pure brute force) uses the Kan condition very clearly. Hence my question is:

If $X:\Delta^{op}\to Set$ is a Kan complex, and $$X^{+}:\left(\Delta_{+}\right)^{op} \to Set^{\Delta^{op}}$$ is the associated diagram of simplicial sets, is $X$ still the homotopy colimit of $X^{+}$? Is this written anywhere?

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I believe that, phrased in a slightly different language, you're asking why the map from the fat geometric realization to the geometric realization is a weak equivalence. Is that what you're asking? –  André Henriques Jan 11 '12 at 14:56
Ah, right, I forgot that the diagonal can be written as a co-end, so, if you take simplicial sets as your target category of realization instead of topological spaces (so that geometric realization is the identity functor), then yes, that is what I am asking. –  David Carchedi Jan 11 '12 at 15:02
But this is subtle, because the difference between simplicial sets and topological spaces is that all topological spaces are fibrant. I'm looking for the link between the Kan condition and the fact that you don't need degeneracies to compute the colimit. –  David Carchedi Jan 11 '12 at 15:03

I just realized that I was stupid: each bisimplicial set is Reedy cofibrant. Btw: The corresponding result for topological spaces can be found in Segals Paper on categories and cohomology theories. For simplicial sets you should be able to deduce it by applying the geometric realization to obtain a diagramm $\Delta^{op} \to Top$ which is Reedy cofibrant. –  Thomas Nikolaus Jan 11 '12 at 16:45