In HTT(Higher Topos Theory) Remark7.3.1.19, it it sketched that the proper base change theorem for $\infty$-topos implies the usual proper base change theorem in (unbounded) derived category. However, I can't understand its detail.

Let $A=Ch(Ab)$ be the category of chain complexes of abelian groups endowed with a conbitional model structure, and $\mathfrak{C}=N(A^{\circ})$ be the underlying $\infty$-category. Then, for a $\infty$-topos $X$, it is written that the homotopy category of $\mathfrak{C}$-valued sheaves over $X$, $hShv(X,\mathfrak{C})$ is equivalent to the usual unbounded derived category $D(X)$. I don't understand why this holds. Please explain me its detail.


Thank you very much for many comments. As far as I understand these comments, it seems that the unboudned derived category of a $\infty$-topos $X$ is equivalent to the homotopy category of hypercomplete $\mathfrak{C}$-valued sheaves $hShv^{hyp}(X,\mathfrak{C})$, and especially if $X$ is hypercomplete, equivalent to $hShv(X,\mathfrak{C})$.

However, since I don't understand the concept of hypercompleteness, I still don't know why this equivalence holds. Is this explained in HTT, HA, or other DAG series? I'd like to know the referrences which study this subject.

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    $\begingroup$ Lurie does not claim (and I think it might be false) that you get Spaltenstein's unbounded derived category this way. In fact, I think the whole point is that you do not, since base change fails for Spaltenstein's category. Maybe the two agree under some conditions. In any case, you do get a copy of bounded below derived category living inside. $\endgroup$ Dec 6, 2014 at 16:05
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    $\begingroup$ On the other hand, I think the two may agree if the $\infty$-topos is hypercomplete. $\endgroup$ Dec 6, 2014 at 16:11
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    $\begingroup$ Hypercompleteness is defined in HTT 6.5.2. The point is that $Shv^{hyp}(X)$ is the localization of $Shv(X)$ at morphisms which induce isomorphisms on homotopy groups. From this it's easy to show that $Shv^{hyp}(X,\mathfrak{C})$ is the localization of $Shv(X,\mathfrak{C})$ at the quasi-isomorphisms. $\endgroup$ Dec 7, 2014 at 20:33

2 Answers 2


Note: answer corrected thanks to the comments of Dylan Wilson and Marc Hoyois.

Let $X$ be a topological space. Its derived category $D(X)$ is the derived category of the abelian category $Ab(Sh(X))$ of sheaves of abelian groups on $X$. As an infinity-category it may be defined as the infinity-category associated to the model category of chain complexes in $Ab(Sh(X))$, where weak equivalences are quasi-isomorphisms (so either the projective or injective model structure). Let me denote this by $D_\infty(Ab(Sh(X)))$.

On the other hand let $Sh_\infty(X, D_\infty(Ab))$ denote the infinity-category of sheaves on $X$ with coefficients in the infinity-category $D_\infty(Ab)$ (= $\mathfrak{C}$ in your notation). This is defined as the full subcategory of the infinity-category of presheaves on $Ouv(X)$, the site of open subsets of $X$, satisfying Cech descent. Let $Sh_\infty^{hyp}(X, D_\infty(Ab))$ be the hypercompletion, i.e. the full sub-$\infty$-category of sheaves satisfying hyperdescent.

A comparison of these two constructions boils down essentially to the fact that a chain complex of sheaves of abelian groups is the same thing as a sheaf of chain complexes of abelian groups. However there is a subtlety with regards to the descent condition. It turns out that $D_\infty(Ab(Sh(X)))$ is equivalent to $Sh_\infty^{hyp}(X, D_\infty(Ab))$. Hence it embeds fully faithfully into $Sh_\infty(X, D_\infty(Ab))$.

  • $\begingroup$ Surely you mean "a chain complex of abelian group objects in sheaves of sets"? $\endgroup$
    – Zhen Lin
    Dec 6, 2014 at 16:06
  • $\begingroup$ Dear @ZhenLin, yes, thanks for the correction ! $\endgroup$ Dec 6, 2014 at 16:09
  • $\begingroup$ In either case the opening line "the claim is simply..." is false. While it is true that sheaves of chain complexes and chain complexes of sheaves are equivalent, this doesn't answer the OPs question. cf my comment above. $\endgroup$ Dec 6, 2014 at 16:10
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    $\begingroup$ I agree with Dylan: for general $X$, $D_\infty(Ab(Sh(X)))$ and $Sh_\infty(X,D_\infty(Ab))$ are not equivalent. That's the whole point of Lurie's remark: proper base change holds for the latter but not the former. $D_\infty(Ab(Sh(X)))$ is equivalent to the $\infty$-category of hypercomplete $D_\infty(Ab)$-valued sheaves. $\endgroup$ Dec 6, 2014 at 17:38
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    $\begingroup$ nitpick: Sh^hyp(X, D(Ab)) is not a topos, but I think it's clear what you mean. $\endgroup$ Dec 7, 2014 at 3:39

The question also confuse me for a long time as a beginner of infinity category. Thanks for previous answers a lot. Here I would like to address a perfect reference for the question: It is a lecture notes of Peter Scholze regarding on 6 functor formalisms: [https://people.mpim-bonn.mpg.de/scholze/SixFunctors.pdf].

For the question in title, it is clear explained in Proposition 7.1. Hope it helps :)


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