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Let $C$ be a small site with fibre products. The (injective) Čech model structure on simplicial presheaves $\operatorname{sPre}(C)$ on $C$ presents an $(\infty,1)$-topos and one may ask if this $(\infty,1)$-topos is hypercomplete or not.

If I understand correctly, hypercompleteness means in this setting precisely that a Bousfield localization of the Čech model structure on $\operatorname{sPre}(C)$ at hypercovers (which results in Jardine's local (injective) model structure) doesn't change anything. In other words, the Čech model structure is already Jardine's local model structure.

Let $X$ be a scheme and $\mathcal{F}$ a quasi-coherent sheaf on $X$ with values in abelian groups. Sometimes, the Zariski-sheaf cohomology $H^n(X,\mathcal{F})$ of $X$ with values in $\mathcal {F}$ can be calculated by Čech cohomology since a certain spectral sequence degenerates. This is the case for example when $X$ is separated.

I hope that I recall it correctly but there seem to be results (stacks project, TAG 01H0) that this is true for a general $X$, if one considers ''Hypercover-Čech cohomology'', i.e. Čech cohomology with respect to hypercovers instead of just ordinary covers. For example, if $X$ is not separated, the intersection of two open affines does not have to be affine - but it is covered by affines, and so on.

Let $X_{Zar}$ denote the small Zariski-site on a scheme $X$.

What is the relation between the $(\infty,1)$-topos associated to $X_{Zar}$ being hypercomplete and the property that sheaf cohomology can be calculated by ordinary Čech cohomology?

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  • $\begingroup$ The question is appealing. Do you know Hypercovers and simplicial presheaves of Dugger et alii? They show in example A.10 that in general there isn't hypercompleteness. But the base site is pathological, so it doesn't relate directly with your question. $\endgroup$
    – Mauro Porta
    Commented Nov 30, 2013 at 20:36
  • $\begingroup$ However, Example A.11 seems more related to your question: for every noetherian scheme S of finite type, the (big, I suppose) Zariski site over $S$ is hypercomplete. $\endgroup$
    – Mauro Porta
    Commented Nov 30, 2013 at 20:39
  • $\begingroup$ Yes (Example A.11 is for a noetherian scheme of finite Krull dimension). Unfortunately, I don't see the relation to the equality of sheaf and Čech cohomology clearly. $\endgroup$
    – sopot
    Commented Nov 30, 2013 at 21:26

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There is no relation between hypercompleteness and the property that Čech cohomology agrees with genuine cohomology, i.e., there is no implication either way. For example, étale cohomology of nice schemes can be computed using Čech cohomology even though the small étale (∞,1)-topos is typically not hypercomplete, and, conversely, Zariski cohomology cannot always be computed as Čech cohomology, even though the small Zariski (∞,1)-topos is often hypercomplete (for any noetherian scheme of finite Krull dimension).

Verdier's hypercovering theorem, which says that hypercovers can be used to compute cohomology, follows from the existence of a category of fibrant objects on locally fibrant simplicial presheaves in which hypercovers are the acyclic fibrations. There is no such thing for Čech covers, or even, as far as I know, for bounded hypercovers. In fact, since an (∞,1)-topos and its hypercompletion have the same cohomology, hypercovers also compute the cohomology in the Čech localization of $sPre(C)$.

Update: The reason that $sPre(C)_{Cech}$ and $sPre(C)_{hyper}$ have the same cohomology is that the Eilenberg–Mac Lane object $K(A,n)∈ sPre(C)_{Cech}$, which is a fibrant replacement of the presheaf $U\mapsto K(A(U),n)$, is already local with respect to all hypercovers, so it’s already fibrant in $sPre(C)_{hyper}$. This is because $K(A,n)$ is $n$-truncated, i.e., its sheaves of homotopy groups vanish in degree $>n$. So the mapping space from a hypercover $U_\bullet$ into $K(A,n)$ is the same as the mapping space from the $n$-bounded hypercover $cosk_n U_\bullet$ into $K(A,n)$, but every fibrant object in $sPre(C)_{Cech}$ is already local with respect to bounded hypercovers (see Dugger–Hollander–Isaksen).

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  • $\begingroup$ Thank you very much! Is 'cohomology of an $(\infty,1)$-topos' presented by a model category $\operatorname{sPre(C)}_{Cech}$ the same as this?: There is a functor$$F:Fun(C^{op},sSet)=Fun(C^{op}\times\Delta^{op},Set)\to Fun(C^{op}\times\Delta^{op},Ab)=Fun(C^{op},sAb)$$ induced by the free abelian group functor $Set\to Ab$. Taking first the (levelwise) injective model structures and then Čech localizing each side turns $F$ into a left Quillen functor. Then the (co)homology is $H_n(X)=\pi_n(F(X))$? Why does the hypercompletion have the same cohomology? (ie. why does $F$ get rid of the 'hyper'?) $\endgroup$
    – sopot
    Commented Dec 1, 2013 at 0:16
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    $\begingroup$ Your $\pi_n F(X)$ is going to be $Hom(\pi_0(X),\mathbb{Z})$ in degree $0$ and trivial in positive degrees, so that's not the right way to define cohomology. To compute $H^n(C,A)$, where $C$ is a site with final object $X$ and $A$ is a sheaf of abelian group on $X$, you form the Eilenberg-Mac Lane simplicial presheaf $K^{pre}(A,n)$ and choose a fibrant replacement $K(A,n)$ (i.e. sheafify). Then $H^n(C,A)=\pi_0(K(A,n)(X))$. $\endgroup$
    – Marc Hoyois
    Commented Dec 1, 2013 at 1:00
  • $\begingroup$ Dear Marc, I guess you mean $sk_n U_\bullet$ as the left-adjoint. $\endgroup$
    – sopot
    Commented Dec 1, 2013 at 23:44
  • $\begingroup$ Actually if $X$ is a fibrant simplicial set, the map $X\to cosk_nX$ in $Ho(sSet)$ is the initial map to a simplicial set with no homotopy groups in degrees $>n$. So $cosk_n$ is left adjoint to the inclusion of that subcategory. $\endgroup$
    – Marc Hoyois
    Commented Dec 2, 2013 at 3:23
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    $\begingroup$ The point of a category of fibrant objects is that you can compute a mapping space $Map(A,B)$ by resolving only the source $A$ and not the target $B$ (this is explained on the nLab page). Verdier's hypercovering is an instance of this for $B=K^{pre}(A,n)$. To compute $Map(A,B)$ using a model structure on $sPre(C)$ you need to replace $B$ by a fibrant object (for $B=K^{pre}(A,n)$ this means taking an injective resolution of $A$). So the hypercovering theorem is not a simple application of model category/∞-category techniques. $\endgroup$
    – Marc Hoyois
    Commented Dec 4, 2013 at 4:34

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