TITLE: Relation between hypercompleteness and the property that Cech cohomology calculates sheaf cohomology
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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).