14
$\begingroup$

If I have a simplicially enriched model category, then I can take the coherent nerve of the full subcategory of bifibrant opjects to obtain a quasicategory. If I have a model category that is not simplicially enriched, then I could take the hammock localisation first and then take the coherent nerve. I have no technical objection to this, but I find it aesthetically displeasing. I would prefer to have a construction that accepts an arbitrary model category $\mathcal{C}$ and produces a quasicategory $N'(\mathcal{C})$ with the same homotopy theory in a single step. Ideally, all objects of $\mathcal{C}$ would appear as $0$-simplices in $N'(\mathcal{C})$, not just the bifibrant ones. Does such a thing exist in the literature?

One idea (in the spirit of the thesis of James Cranch) is as follows. Given a left proper model category $\mathcal{C}$, an $n$-simplex of $N'(\mathcal{C})$ would be an $n$-fold cospan diagram consisting of objects $X_{ij}$ for $0\leq i\leq j\leq n$. There would be cofibrations $X_{ij}\to X_{i,j+1}$ and weak equivalences $X_{i+1,j}\to X_{ij}$ and the resulting squares would be cocartesian. I have not made any serious attempt to check whether this works.

Alternatively, it may be that an answer is essentially contained in the paper of Barwick and Kan on Relative Categories, but I have not properly digested that.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ The construction you outline, although very appealing, unfortunately does not work - it does not actually invert the weak equivalences. (For instance, the mapping spaces are just groupoids.) $\endgroup$
    – Rune Haugseng
    Commented Aug 11, 2015 at 18:00
  • 1
    $\begingroup$ Just as a remark: After taking hammock localization, you first would have to fibrantly replace your simplicial category (for example, by applying $Ex^\infty$ to all mapping spaces) and then apply the coherent nerve. $\endgroup$
    – Lennart Meier
    Commented Aug 12, 2015 at 7:02

2 Answers 2

Reset to default
14
$\begingroup$

It's not quite in the literature, but there is a fully explicit construction that avoids hammock localisation or any kind of fibrant replacement: by a recent result of Lennart Meier, a certain "double cosubdivision" of the Rezk classification diagram of a model category is a complete Segal space, so (by a result of Joyal and Tierney) we can take degreewise 0-simplices to get a quasicategory. The vertices of the quasicategory constructed above are indeed all the objects of the model category we start with, and the edges are diagrams of the form $$\bullet \rightarrow \bullet \leftarrow \bullet \rightarrow \bullet \leftarrow \bullet$$ where every arrow except possibly the interior $\rightarrow$ is a weak equivalence. Perhaps this is more complicated than you hoped for, but it is unavoidable in the general case because we do not always have a two-arrow calculus.

There is also an explicit construction of Karol Szumiło that makes a quasicategory out of a category of cofibrant objects. The vertices are cofibrant cosemisimplicial resolutions, and similarly, the edges are resolutions of cospans. The homotopical correctness of this construction was recently proved by Chris Kapulkin and Karol Szumiło.

Improve this answer
$\endgroup$
4
  • 4
    $\begingroup$ Just to clarify, my construction can be applied to an arbitrary model category. It only depends on the full subcategory of cofibrant objects, but by results of Dwyer and Kan this subcategory is DK-equivalent to the original model category. Hence we still get the right thing. $\endgroup$
    – Karol Szumiło
    Commented Aug 11, 2015 at 15:00
  • $\begingroup$ Sure, of course. I was thinking of using the model category itself as a category of cofibrant objects, but I guess either way the vertices are still going to be more complicated than just objects of the original category. $\endgroup$
    – Zhen Lin
    Commented Aug 11, 2015 at 15:55
  • $\begingroup$ Oh, I see. Do you mean using something like diagrams that are Reedy cofibrant except that their bottom objects might not be cofibrant? $\endgroup$
    – Karol Szumiło
    Commented Aug 11, 2015 at 16:11
  • 2
    $\begingroup$ Something like that, yes. As you remarked, left proper + weak equivalences closed under coproducts implies there is a maximal/canonical cofibration structure. $\endgroup$
    – Zhen Lin
    Commented Aug 11, 2015 at 17:36
7
$\begingroup$

You can consider the marked simplicial set $(N(\mathcal{C}),\mathcal{W})$, where $N$ is the usual nerve functor and $\mathcal{W}$ is the class of weak equivalences in your model category $\mathcal{C}$. Then take $N'(\mathcal{C})$ to be the underlying simplicial set of any fibrant replacement of $(N(\mathcal{C}),\mathcal{W})$ in the Cartesian model structure. This paper by Hinich shows that this procedure gives the "correct" $\infty$-category associated to $\mathcal{C}$. Perhaps some unwinding of definitions will give you a more explicit construction.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .