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Maxime Ramzi
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The result is not only true for simplicial model categories, but for plain combinatorial model categories too - this is Higher Algebra 1.3.4.25..

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 4.2.4.1 in the simplicial case, HA 1.3.4.24. 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 1.3.4.20. in HA)

Hopefully that answers your questions

The result is not only true for simplicial model categories, but for plain combinatorial model categories too - this is Higher Algebra 1.3.4.25..

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 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 4.2.4.1 in the simplicial case, HA 1.3.4.24. 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 1.3.4.20. in HA)

Hopefully that answers your questions

The result is not only true for simplicial model categories, but for plain combinatorial model categories too - this is Higher Algebra 1.3.4.25..

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 4.2.4.1 in the simplicial case, HA 1.3.4.24. 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 1.3.4.20. in HA)

Hopefully that answers your questions

Source Link
Maxime Ramzi
  • 15.8k
  • 2
  • 40
  • 73

The result is not only true for simplicial model categories, but for plain combinatorial model categories too - this is Higher Algebra 1.3.4.25..

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 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 4.2.4.1 in the simplicial case, HA 1.3.4.24. 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 1.3.4.20. in HA)

Hopefully that answers your questions