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(See Jacob Lurie's "Higher Algebra", section 1.3.5 for context.)

Let $\mathcal{A}$ be a Grothendieck abelian category. Then the stable $\infty$-category $\mathcal{D}(\mathcal{A})$ is a localisation of the dg-nerve $N_{dg}(\mathrm{Ch}(\mathcal{A}))$ by Higher Algebra Proposition 1.3.5.13, and it is presentable according to Higher Algebra Proposition 1.3.5.21, since it is underlying a combinatorial model category.

My question is: what can we say about $N_{dg}(\mathrm{Ch}(\mathcal{A}))$ itself? Is it presentable?

As far as I can understand, the introduction of https://arxiv.org/pdf/1710.11388.pdf seems to suggest that there is an underlying model structure which is not combinatorial but accessible, although I am not sure if this is enough to conclude...

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This fails already with the category of abelian groups. If the dg-nerve of the dg-category of chain complexes of abelian groups were presentable, then the associated triangulated category would be well generated in the sense of Neeman in his book Triangulated categories (Annals of Math. Studies, Vol. 148, 2001); this is easy to deduce from Krause's characterization together with section 1.4.4 of Lurie's Higher Algebra. But in appendix E of his book, Neeman proves that the triangulated category of chain complexes up to chain homotopy equivalences does not have a generating set (Lemma E.3.2 page 438), which provides a definitive obstruction against presentability. He also proves that the opposite category is not well generated, by the way (but this is easier).

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  • $\begingroup$ Is the homotopy category of chain complexes (of abelian groups) accessible? $\endgroup$
    – Z. M
    Commented Jul 14, 2023 at 15:27
  • $\begingroup$ No it is not (since it is cocomplete that would imply it is presentable, which it is not, as I explain above). The adjuncion between the homotopy category and the derived category is accessible though (i.e. the right adjoint of the localization functor by quasi-isomorphisms has a right adjoint that preserves $\kappa$-filtered colimits for $\kappa$ big enough). This is essentially what is proved (in the language of model structures) in the paper of Lyne Moser mentioned in the question. $\endgroup$
    – D.-C. Cisinski
    Commented Jul 16, 2023 at 9:51
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If $\mathcal{A}$ is a small Grothendieck abelian category, one could possibly argue this way: recall that presentable means accessible and closed under small colimits.

Since $\mathcal{A}$ is small, the $\infty$-category $N_{\mathrm{dg}}(\mathrm{Ch}(\mathcal{A}))$ is essentially small by HTT Proposition 5.4.1.2 (1) and admits small colimits, so it is idempotent complete, hence by HTT Corollary 5.4.3.6 it is presentable.

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    $\begingroup$ An abelian Grothendieck category has small colimits. If it is small, by the "uncheatable lemma" from this answer mathoverflow.net/a/365951/1017 it would a partially ordered set and an abelian category, hence equivalent to zero. In particular, it would be presentable, that is true, but for rather trivial reasons! $\endgroup$
    – D.-C. Cisinski
    Commented Apr 28, 2021 at 13:43
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    $\begingroup$ Giulio Lo Monaco has shown "the uncheatable" for infinity categories here: tac.mta.ca/tac/volumes/37/5/37-05.pdf $\endgroup$
    – Ivan Di Liberti
    Commented Apr 28, 2021 at 14:53

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