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 $N(L^H\mathcal M)^f$ ?


1 Answer 1


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.

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    $\begingroup$ 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
    Commented Jul 7, 2018 at 12:00
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    $\begingroup$ @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 $M$. $\endgroup$ Commented Jul 7, 2018 at 17:44

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