4
$\begingroup$

Consider a model category $\mathcal M$. Because it is a category with weak equivalences, we can use the following construction to obtain the "underlying quasicategory" of $\mathcal M$:

  • taking the hammock localization $L^H\mathcal M$ makes it a simplicially enriched category,
  • of which we can take the fibrant replacement $(L^H\mathcal M)^f$ in Bergner's model structure,
  • then the underlying quasicategory of $\mathcal M$ is the image $N(L^H\mathcal M)^f$ through the homotopy coherent nerve.

It is conceptually quite nice, but it is kind of cumbersome to actually compute those stuff. And in a sense, we do not use the full structure of the model category $\mathcal M$, we only use it as a relative category.

Is there a (possibly simpler) construction that makes use of $\mathcal M$ as a model category?

I was thinking of something along the lines of the theory of simplicial frames. Recall that given $X$ in $\mathcal M$, one can consider the "diagonal map" $X\to \tilde X$ between the simplicial objects $X:n\mapsto X$ and $\tilde X:n\mapsto X^n$. This map factors as $X \overset j \to \bar X \overset q\to \tilde X$ with $j$ acyclic cofibration and $q$ fibration for the Reedy model structure on $s\mathcal M$. Then the homotopy type of the simplicial set $\mathcal C(A,\bar X)$ is a good candidate for the mapping space between $A$ and $X$ in the underlying quasicategory of $\mathcal M$ (the so-called derived hom-space if I understand correctly this notion). If i understand correctly Hovey's chapter on framings, under mild hypothesis on $\mathcal M$ (functorial factorization systems mostly), we even have a functor $$ \operatorname{Ho}(\mathcal M)^{\rm op} \times \operatorname{Ho}(\mathcal M) \to \operatorname{Ho}(\mathsf{sSet})$$ that equip $\operatorname{Ho}(\mathcal M)$ with the structure of a closed $\operatorname{Ho}(\mathsf{sSet})$-module (with tensor given by cosimplicial frame).

Can we use this construction to cook up a quasicategory (or even a simplicially enriched one) weakly equivalent to $N(L^H\mathcal M)^f$ ?

$\endgroup$

1 Answer 1

2
$\begingroup$

Great question. One can do this, but to the best of my knowledge not via techniques that have been published. Here's the idea. Dwyer and Kan say that if $x$ is cofibrant and $y$ is fibrant, whatever nice model for the mapping space $\mathrm{Map}(x,y)$ you might have in mind lives in a natural zigzag of weak equivalences with $L^H\mathcal M^f(x,y)$. Now of course this isn't nearly enough to transfer the simplicial category structure on the family of Kan complexes $(L^H\mathcal{M}^f(x,y))_{x,y}$ over to such a structure on $(\mathrm{Map}(x,y))_{x,y}$.

However! This is exactly what one needs to construct an $A^\infty$-category $L_{A^\infty}\mathcal M$ whose objects are those of $\mathcal M$ and whose mapping spaces are exactly $\mathrm{Map}(x,y)$. Unfortunately, if you look up $A^\infty$ category you'll mostly find references to a concept involving weak enrichment in chain complexes. I mean an $A^\infty$ category in Kan complexes. You can view this formally as a category over the $A^\infty$ operad in Kan complexes. What such a structure consists of is precisely a family of objects $x,y,...$, Kan complexes $\mathrm{Map}(x,y)$ for each pair of objects, composition morphisms $\mathrm{Map}(x,y)\times \mathrm{Map}(y,z)$ for every triple, an associativity homotopy $\mathrm{Map}(x,y)\times \mathrm{Map}(y,z)\times \mathrm{Map}(z,w)\times \Delta^1\to \mathrm{Map}(x,w)$ witnessing the equivalence of the two induced composition maps for every quadruple, and so on, with higher coherence conditions in every dimension. If you're family with the $A^\infty$ operad, hopefully this seems rather natural. And now we can think of a category enriched in Kan complexes as exactly an $A^\infty$-category in which all the higher-dimensional structure is trivial.

Anyway, one of the foundational motivations for the study of $A^\infty$ spaces (where by "space" I mean "Kan complex") is that the forgetful functor $A^\infty-\mathrm{Spaces}\to\mathrm{Spaces}$ is a homotopy isofibration: given a space $X$ and a homotopy equivalence $X\to Y$, we can transfer $A^\infty$ structures from $Y$ to $X$ and vice versa. The same fact is true for $A^\infty$-categories: given a family of spaces, such as $(\mathrm{Map}(x,y))_{x,y}$, and a family of homotopy equivalences, such as those with $(L^H\mathcal{M}^f(x,y))_{x,y}$, we can transfer $A^\infty$ structures between the two families. In particular, the strict $A^\infty$ structure on $(L^H\mathcal{M}^f(x,y))_{x,y}$ transfers to one on $(\mathrm{Map}(x,y))_{x,y}$ which is almost certainly not strict, i.e. doesn't come from a simplicially enriched category.

If you now want a quasicategory, then you're in luck, as $A^\infty$-categories also admit homotopy coherent nerves. At least, it's obvious that they should. But I don't think anybody has ever written it down! In any case, you can see a proof for the differential graded case here.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks, this is an awesome answer. I will need some time to process that and try to compute something, but it definitively helps. Just to be sure, when you say the $A^\infty$ operad in Kan complexes, you mean a cofibrant replacement of the associative operad $A$ (generated just by $A(0)=A(1)=1$) in the model category of (Kan complexes)-enriched operads? I tried to follow the links of definition in the nLab, but sometimes "resolution" is used without precision about which kind of resolution... $\endgroup$
    – Pece
    Jul 7, 2018 at 12:00
  • 1
    $\begingroup$ @Pece Yes, I mean a cofibrant resolution of the associative opread. There won't be a model category of Kan complex-valued operads as there's no model category of Kan complexes, but you can do a cofibrant-fibrant replacement over simplicial sets. Anyway, the cofibrant resolution story is a bit anachronistic for this example: I'm really thinking of the operad valued in Stasheff's associahedra, which is a decade older than the word "operad" itself, either with the singular simplicial set functor applied to it or applied directly to some topological space model of the mapping spaces in $M$. $\endgroup$ Jul 7, 2018 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.