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Let me first recall some definition: Let $A$ be a Grothendieck Abelian category. Then, then category $\mathrm{Ch}(A)$ (I am using homological indexing) admits a combinatorial model structure (see for eg. Lurie's Higher Algebra, 1.3.5.3), where:

  • Cofibrations: level-wise injective chain maps.
  • Weak equivalence: quasi-isomorphisms.
  • Fibration: maps with right lifting property with respect to trivial (i.e. acyclic) cofibrations.

Using this model structure, we could form an unbounded ($\infty$-)derived category $D(A)$ (which is a stable $\infty$-category), whose objects are fibrant objects of $\mathrm{Ch}(A)$ (note that all objects in $\mathrm{Ch}(A)$ are automatically cofibrant). This category is also equipped with a $t$-structure defined by the vanishing of the homology groups (see loc. cit. 1.3.5.16).

In loc. cit., 1.3.5.21, it is proved that $D(A)$ with the $t$-structure mentioned above is right complete, using (the dual version of) 1.2.1.19. But in remark 1.3.5.22, it is mentioned that the stable $\infty$-category $D(A)$ is generally not left-complete.

My question is: why can't we use 1.2.1.19 to prove also that $D(A)$ is left-complete? It seems to me that the category admits countable product, and $D(A)_{\geq 0}$ is stable under this operation, since $A$ is a Grothendieck abelian category. Moreover, $D(A)_{\geq \infty} = D(A)_{\leq \infty}$ since both just consist of those fibrant complexes that are acyclic. But $D(A)_{\leq\infty} = 0$, at least because of the claim that $D(A)$ is right-complete.

I would appreciate it very much if someone could point out where my confusion lies!

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    $\begingroup$ Neeman gives examples of Grothendieck categories where $D(A)_{\geq 0}$ is not closed under countable products: arxiv.org/pdf/1103.5539v1.pdf. One such is the category of representations of $\mathbb{G}_a$ over a field of positive characteristic. $\endgroup$
    – Marc Hoyois
    Commented Aug 3, 2014 at 19:39
  • $\begingroup$ @MarcHoyois: Thanks for the reference. There's one small point that I'm confused about, and I would appreciate it very much if you could clarify. Namely, since we are working with a Grothendieck category, taking products is exact, and product in the derived category could be done level-wise. So suppose I have a collection of complexes in $D(A)_{\geq 0}$, then they are all acyclic at levels $<0$. But by what I mentioned, their product should also be acyclic at levels $<0$. In other words, the product lies in $D_{\geq 0}$? $\endgroup$
    – QcH
    Commented Aug 4, 2014 at 16:51
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    $\begingroup$ Infinite products are not necessarily exact in a Grothendieck category, as Neeman's example shows. $\endgroup$
    – Marc Hoyois
    Commented Aug 4, 2014 at 21:20
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    $\begingroup$ @MarcHoyois: Great! I've been mistakenly thinking that Grothendieck Categories satisfy both AB4 and AB4* as part of the definition. Thanks! $\endgroup$
    – QcH
    Commented Aug 4, 2014 at 21:40
  • $\begingroup$ @QcH Is there a section 1.3.5 in Lurie's Higher Algebra? I'm asking because I'm looking at math.harvard.edu/~lurie/papers/higheralgebra.pdf, and I can't find it. Perhaps you could cite the statements that you are referencing? $\endgroup$
    – user62675
    Commented Aug 14, 2014 at 2:55

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