In remark (HTT), Lurie states that

Another possible approach to the problem of homotopy coherence is to restrict our attention to simplicial (or topological) categories C in which every homotopy coherent diagram is equivalent to a strictly commutative diagram. For example, this is always true when C arises from a simplicial model category (Proposition

I am working with homotopy coherent diagrams from $\mathbb{Z}_{\ge 0}$ (thought as a poset) to $\text{Ch}_{\ge 0}(\mathbb{Z})$. Here $\text{Ch}_{\ge 0}(\mathbb{Z})$ stands for nonnegatively graded chain complexes of abelian groups, thought as an $\infty$ category. Since I want to compute the colimit of such a diagram, I am wondering the following:

Is $\text{Ch}_{\ge 0}(\mathbb{Z}) $ a simplicial model category?

In particular, in light of Corollary, it seems like for any diagram $N(\mathbb{Z}_{\ge 0}) \to N(\text{Ch}_{\ge 0}(\mathbb{Z})) $ I can find a simplicial functor $\mathbb{Z}_{\ge 0} \to \text{Ch}_{\ge 0}(\mathbb{Z})$ such that the $\infty$ -colimit of the former is equivalent to the homotopy colimit of the latter. Also, it seems like $\mathbb{Z}_{\ge 0}$ being a discrete category, such hocolimit should be computable from a discrete information. In other words, let $sk_0 : \text{sSet-Cat} \to \text{Cat}$ be the functor that takes the $0$ skeleton hom-wise and $LS: \text{Cat} \to \text{sSet-Cat}$ the functor that takes the trivial simplicial set home wise. Note that $LS$ is left adjoint to $sk_0$. Let $J$ be an ordinary category and $C$ a simplicial model category. Let $F: LS(J) \to C$ be a simplicial functor corresponding to $F': J \to sk_0(C)$. Is it true that the homotopy colimit of $F$ and of $F'$ are equivalent?

I hope I have not said too much stupid things :)


1 Answer 1


The result is not only true for simplicial model categories, but for plain combinatorial model categories too - this is Higher Algebra

In fact, for this you can reduce to the case of simplicial model categories, by noting that a combinatorial model category is always Quillen equivalent to a simplicial one. This is for instance corollary 6.4 in Dugger's Universal homotopy theories

It follows that for any small category $I$ and any combinatorial model category $C$, any homotopy coherent diagram $I\to C[W^{-1}]$ can be lifted to a strict diagram $I\to C$; where $C[W^{-1}]$ is the $\infty$-categorical localization.

Also note that there is a simpler approach in your particular situation : $Ch_{\geq 0}(\mathbb Z)$ is Quillen equivalent (in fact, equivalent ! but the equivalence is compatible with the model structures) to $sAb$, simplicial abelian groups, via the usual Dold-Kan correspondance, and the latter is in fact simplicially enriched.

As for colimits, this will depend on $I$, in fact it's also a theorem in HTT that these can be computed as homotopy colimits in $C$ (HTT in the simplicial case, HA in the non simplicial case), in the usual model category theoretic sense. Now $\mathbb Z_{\geq 0}$ is filtered, so since filtered colimits are exact in abelian groups, it follows that if you have a filtered diagram $I\to Ch_{\geq 0}(\mathbb Z)$, then its colimit and its homotopy colimit agree, and in particular the localization functor $Ch_{\geq 0}(\mathbb Z)\to D_{\geq 0}(\mathbb Z)$ preserves these filtered colimits, where I let $D_{\geq 0}(\mathbb Z)$ denote the localization at quasi-isomorphisms (i.e. the underlying $\infty$-category)

Note that the exactness of filtered colimits is true in abelian groups and more generally $R$-modules, but it is not true in general abelian categories.

This is to some extent related to how $\mathbb Z_{\geq 0}$ is "discrete", because it follows from the previous paragraph that the higher homology groups of the homotopy colimit of a functor $I\to Ab\to Ch_{\geq 0}(\mathbb Z)$ are computed as the left derived functors of the colimit functor; and those are related to the nerve of $I$ as a space. The key-word for this is "higher colimit", or perhaps you'll be luckier with "higher limit"

As for your last paragraph, $LS$ is not left adjoint to $sk_0$, rather to evaluation at $0$ (also known as "limit along $\Delta^{op}$", which has an initial object).

But quite generally nonetheless, you can in fact compare a simplicial model category and its underlying plain model category. This is also somewhere in HTT I believe, where it is stated that the homotopy coherent nerve of the full subcategory of cofibrant-fibrant objects of a (sufficiently nice) simplicial model category is equivalent to the localization of its underlying plain model category at the weak equivalences. So you should be able to compare homotopy colimits in one and homotopy colimits in the other.

(I could not find it in HTT but it is in HA)

Hopefully that answers your questions

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    $\begingroup$ For future reference, let me just note, as a comment, that the naive approach to a simplicial structure on $Ch_{\geq 0}(\mathbb Z)$ does not work, i.e. if you take the complex-enriched hom and then feed that to Dold-Kan, this will not produce a nice simplicial structure. The reason is that Dold-Kan is only lax monoidal, and so you do not have tensors over $sSet$. Specifically, denote by $s\hom$ this version of a simplicial hom, by $Hom$ the internal hom of $Ch_{\geq 0}(\mathbb Z)$, and by $DK$ the appropriate Dold-Kan functor. (1/3) $\endgroup$ Jan 25, 2021 at 21:04
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    $\begingroup$ Assume $X$ is a simplicial abelian group and $C,D$ are complexes, then one should have $\hom(X, s\hom(C,D)) \cong \hom(DK(X), Hom(C,D)) \cong \hom(DK(X) \otimes C,D)$ so that the tensoring over $sSet$ would have to be the naive one (plugging in $\mathbb Z[S]$ instead of $X$, for a simplicial set $X$). But this is not a tensoring, because $DK$ is not strong monoidal, so that the natural isomorphism $S\cdot (T\cdot C)\cong (S\times T)\cdot C$ fails, (2/3) $\endgroup$ Jan 25, 2021 at 21:04
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    $\begingroup$ as it is equivalent (by colimit nonsense) to $DK(X\otimes Y)\otimes C \cong DK(X)\otimes DK(Y)\otimes C$ which is equivalent (by taking $C=\mathbb Z$) to strong monoidality of $DK$. (3/3) $\endgroup$ Jan 25, 2021 at 21:04
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    $\begingroup$ If you do it in $sAb$ it is completely fine, but this construction does not work if you perform it in complexes. Indeed, this $A_\bullet$ will be the chain complex associated to $\mathbb Z[\Delta^\bullet]$ and this will produce the naive Dold-Kan thing by some adjunction computations. The reason is again that Dold-Kan is not so good when it comes to tensoring - but if you do that in $sAb$ with $\mathbb Z[\Delta^\bullet]$ directly it will work fine $\endgroup$ Jan 26, 2021 at 8:11
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    $\begingroup$ Yes in general you need something like a telescopic resolution (e.g. in topological spaces) , but as you say, in chain complexes the two colimits end up being the same, and so the homotopy colimit is just a naive colimit. I don't think the sentence "$colim X'$ is still a resolution of $colim X$" is correct; the point is that each $X'(n) \to X(n)$ is a quasi-isomorphism, and those are preserved under filtered colimits (it is an instructive exercise in linear algebra) $\endgroup$ Jan 26, 2021 at 12:47

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