# Is the geometric realisation of a nerve equivalent to the classifying space of its categorisation?

The nerve functor $N:Cat\to sSet$ has a left adjoint, namely the categorisation $C$. In fact there is a natural isomorphism $\epsilon: CN\to Id$ and $N$ is a full embedding.So if I start with a simplicial set $X$, I can generally not hope for $NCX$ to be isomorphic to $X$, but can I say something about the geometric realisation? In other words, is there a nice relation between $BCX$ and $|X|$?

In my example $X$ is the simplicial ring $Gl(R^\bullet)$ where an $n$ simplex is an element of the general linear group over $R[T_0,\cdots,T_n]/(\sum T_i=1)$. $CX$ should then be the category with objects $Gl(R)$ and morphisms $Gl(R[T])$ (source and target given by $T=0,1$, respectively). I tried to make sense of the nerve of this category in comparison to $X$ but failed.

The underlying motivation is a discussion of the hocolim of a functor $F:K\to Cat$ for a small category $K$ and $Cat$ the category of small categories (I am reading Thomasons paper). If I understand it correctly there we are actually taking the hocolim of the nerves over all $F(k)$, $k\in K$ and see then that this hocolim itself is the nerve of a category. Now this category comes with a universal mapping property into other categories. Eventually we obtain a universal mapping property out of hocolim into nerves of arbitrary categories. This seems somehow unnatural since the whole construction worked simplicially and not categorial. Yet I was unable to go back to arbitrary simplicial sets.

Edit Let me get this straight: Given is a functor $\mathcal C\to CX$ for a small category $\mathcal C$ and a simplicial set $X$. I want to show that it induces a map of simplicial sets $N\mathcal C\to X$. If I take the adjoint functors $CSd^2$ and $Ex^2N$ instead ($Sd$ is subdivision and $Ex$ its adjoint) then I get $$Hom(\mathcal C,CSd^2X)\cong Hom(Ex^2N\mathcal C,X)$$ I have an natural acyclic cofibration $N\mathcal C\to Ex^2N\mathcal C$ and a natural map $CSd^2X\to CX$. So all I need to show is that my initial map $\mathcal C\to CX$ factors over $CSd^2X$. Is this correct, is there a good way to show the last bit?

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$|N C X|$ and $|X|$ are certainly not equivalent in general: if $X$ is a Kan complex, then $C X$ is a groupoid, and so $N C X$ is a 1-type. But Thomason showed that if you compose $N$ and $C$ with 2-fold subdivision and its adjoint, they become a Quillen equivalence for an induced model structure on $\mathrm{Cat}$. –  Mike Shulman Jul 26 '12 at 17:08
@Mike Shulman, thanks for your comment but I am afraid it went straight over my head. So the first answer is no, but there is a way of fixing this? –  Simon Markett Jul 26 '12 at 17:26
Have a look at Thomason's paper "Cat as a closed model category". –  Mike Shulman Jul 26 '12 at 19:45
The quick answer to Is there a nice relation'' is no. For example, if $n\geq 3$ (if my memory is correct), $CX$ doesn't know the difference between $X = \Delta[n]$ and $X = \pa\DE[n]$. –  Peter May Jul 27 '12 at 17:35