This seems like it should be a "standard" thing, and I think I remember even seeing it somewhere, but I can't remember where.

Let $C$ be a small category. Is there a category $\Lambda C$ whose nerve (or classifying space) is a model for the free loop space of the nerve (or classifying space) of $C$? Probably the objects of $\Lambda C$ should be something like zigzags of morphisms in $C$. A reference would be the best thing to hear.

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    $\begingroup$ Limits in the category of categories exist, so why not take the equalizer of the two maps Fun([1], C) evaluating at 0 and 1? The nerve respects limits so at least you'll get a good looking simplicial set. If I recall May correctly, geometric realization preserves pullbacks... So this seems like a good candidate. $\endgroup$ Nov 26 '12 at 16:56
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    $\begingroup$ @Dylan: this category has wrong homotopy type in general. It is actually isomorphic to the category of functors from the monoid $\mathbb{N}$ to $C$. If you take $C$ to be a category with two objects and two parallel arrows between them, then $C$ is homotopy equivalent to a circle, so its loop space should be countably infinite discrete (up to homotopy), but your construction gives a finite category. $\endgroup$ Nov 26 '12 at 20:37
  • $\begingroup$ I don't know whether this contains an answer or not: books.google.com/books/about/… but it does have some connection I think. $\endgroup$ Nov 27 '12 at 18:11
  • $\begingroup$ @Bob: Well, I can't tell since Google books won't even let me read the table of contents. The title doesn't make me hopeful, though. $\endgroup$ Nov 28 '12 at 1:29
  • $\begingroup$ @Mike: the abstract says that: A method for associating to each topological category G, a principal top cat BXG is discussed. The first step is to associate with each G, a top 2-cat XG. The second is to apply the classifying space functor B. It is shown that there is a homomorphism BXG->G, which when restricted to the morphism spaces is a htpy equivalence of spaces over Ob G x Ob G. The particular example BX\Gamma^0 is considered and is shown to act on the disjoint union of (BA)^n, where A is a permutative category. $\endgroup$ Nov 29 '12 at 11:49

My French is not good enough to be sure about it, but it seems that this paper has the definition you are after.

Evrard, Marcel Fibrations de petites catégories. Bull. Soc. Math. France 103 (1975), no. 3, 241–265. (Numdam)

  • $\begingroup$ It certainly has a likely-looking definition of $\Lambda C$, but I can't find in it a statement of the theorem I want. He asserts that the homotopy groups of $\Lambda C$ are isomorphic to those of the nerve of $C$ (referring to his thesis, which is apparently not available online, for the proofs). But I don't see the stronger statement I want, that the nerve of $\Lambda C$ is weak homotopy equivalent to the loop space of the nerve of $C$. $\endgroup$ Nov 27 '12 at 5:49
  • $\begingroup$ Thanks, though. Maybe if I could get a copy of his thesis, it would be in there. $\endgroup$ Nov 27 '12 at 5:49

I will try to answer the question. As I said in a comment, the Thomason model structure on $Cat$ is not simplicial model structure. Let $C$ be a small category, we will view it as a topological category. Denote by $C[C^{-1}]$ the topological category where we invert all maps of $C$ such that $C\rightarrow C[C^{-1}]$ is a cofibration of topological categories, then the coherent nerve $N_{\bullet}C\rightarrow N_{\bullet}C[C^{-1}]$ induces a weak equivalence of simplicial sets. Notice that $C[C^{-1}]$ is an infinity groupoid.

Let $C$ be a cofibrant topological category. The mapping space $map(C,D)$ in the model category of topological categories is given by the (standard) nerve of the following $HOM(C,D)$ category :

$\underline{Objects}$ are topological functor $F:C^{op}\times D:\rightarrow Top$ such that for any $c\in C,$ $F(c)$ is equivalent to a representable functor $D(d,-)$ for some $d\in D$.

$\underline{Morphisms}$ in this category are natural transformation $H:F\rightarrow G$ such that $F(c,d)\rightarrow G(c,d)$ is a weak equivalence for all $c\in C$ and $d\in D$.

Let $S^{1}$ a simplicial model for a circle. Let $k: sSet\rightarrow sSet$ the cocontinues Joyal functor which take $\Delta^{n}$ to the nerve of the groupoid with $n+1$ objects and only one isomorphism between any two objects.

Recall that $\mathfrak{C}: sSet\rightarrow Cat_{\Delta}$ is the left quillen adjoint to the coherent $N_{\bullet}$ betwen the joyal model structre on $sSet$ and the Bergner model structure on $Cat_{\Delta}$

Now $k(S^{1})$ is a simplicial set, and the cofibrant topological category $|\mathfrak{C}[k(S^{1})]|$ is an infinity groupoid and its cohenrent nerve is equivalent to $S^{1}$.
The finial result is that $HOM(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])$ is a model for $\Lambda C$, since the nerve of $HOM(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])$ is equivalent to $ map(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])\sim Map(S^{1},N_{\bullet}C[C^{-1}])\sim Map(S^{1}, N_{\bullet} C)=\Lambda N_{\bullet}C$

N.B. The only point that I did not explained is the construction of $C[C^{-1}]$. Let 1 be the category with two objets a and b and a unique morphisms from $: f:a\rightarrow b$. Let $\widehat{1}=|\mathfrak{C}[k(\Delta^{1})]|$, then $ C[C^{-1}]$ is the pushout $colim (\sqcup_{mor C} \widehat{1}\leftarrow \sqcup_{mor C} 1\rightarrow C )$ i.e., for each porphism of $C$ there is a map $1\rightarrow C$.

  • $\begingroup$ But the construction of $C[C^{-1}]$ seems to be where a lot of the meat lies... that's where there'll be zigzags and stuff. Also, this is not really concrete enough for what I want; too many geometric realizations and coherent nerves of very large categories. $\endgroup$ Dec 6 '12 at 1:38
  • $\begingroup$ I put the construction of $C[C^{-1}]$, it does not require zig-zags and it conceptually very easy. I agree that the constructed $\Lambda C$ is not very concrete but still you have a general model for the cotensorisation functor in $Cat$ (with Thomason model structure) by $\sSet$. $\endgroup$
    – Ilias A.
    Dec 6 '12 at 7:21
  • $\begingroup$ Computing pushouts of categories is not what I would call "easy"; that's where the zigzags are going to come in. $\endgroup$ Dec 24 '12 at 14:28

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