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I am trying to understand the essential image of the following functor. Given a scheme $X$, we consider the corresponding small Zariski site $X_{zar}$. For a commutative ring $\Lambda$, let $\mathcal D(\Lambda)$ denote the derived $\infty$-category of $\Lambda$, and let $Fun(X_{zar}^{op}, \mathcal D(\Lambda))$ denote the $\infty$-category of the $\mathcal D(\Lambda)$-valued presheaves on $X_{zar}$. Then for any $F$ in $\mathcal D(X, \Lambda)$, the derived $\infty$-category of sheaves of $\Lambda$-modules, we can define $i(F)\in Fun(X_{zar}^{op}, \mathcal D(\Lambda))$ by the formula:$$i(F)(U)=R\Gamma_{zar}(U, F)$$ Here is my question: is $i(F)$ already an $\infty$-categorical sheaf?

An $\infty$-categorical sheaf $G$ here is a presheaf satisfying Čech descent (I am even not sure this "definition" is correct in practice...), i.e. for any covering $U=\cup U_i$ $$G(U)\simeq \lim_{n}G(ČU)$$ where $ČU$ is the corresponding Čech complex. A naive idea is that at least in the bounded below case if we take an injective resolution $I$ of $F$, then the homotopy limit here is just the limit and this becomes the definition of sheaf.

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    $\begingroup$ Welcome to mathoverflow! Your question seems to be closely related to this one: mathoverflow.net/questions/265557/…. You might find there information relevant to what you need. $\endgroup$ Commented Apr 10, 2019 at 19:50
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    $\begingroup$ @YonatanHarpaz Thanks very much for the link! According to your answer to that question, could I think in general $i(F)$ is not already a sheaf? $\endgroup$
    – user662742
    Commented Apr 10, 2019 at 20:14
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    $\begingroup$ The question in the body and the title have different answers, respectively yes and no. As @YonatanHarpaz says, the answer to the title question is no. The derived category of sheaves is the hypersheaves, a subcategory of ∞-sheaves. For CW complexes or finite dimensional Zariski spectra all ∞-sheaves are hypersheaves. But not for the étale topology or the Hilbert cube. The dualizing complex of the Hilbert cube is an ∞-sheaf with zero stalks, hence not a hypersheaf. If you restrict to finite dimensional schemes they're the same, but that's a nontrivial theorem... $\endgroup$ Commented Apr 11, 2019 at 13:42
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    $\begingroup$ ...eg, Quillen's localization theorem pretty much immediately proves that $K$-theory is a ∞-sheaf, but Brown and Gersten have to do a lot of work to prove that ∞-sheaves are hypersheaves and thus that get the BGQ spectral sequence relating stalks to sections. . . 2. Body question: yes, but overkill. What is RΓ? If defined in terms of sheaves, then what will go wrong is that $i$ will be the hypersheafification functor, when you want the ∞-sheafification functor. But it's still a sheaf, so the answer to your question is yes. (But again, if finite dim Zariski, the two functors are the same.) $\endgroup$ Commented Apr 11, 2019 at 13:45

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Let $\tau$ be a Grothendieck topology on the category of schemes (of suitably bounded cardinality if $\tau$ is ‘large’), $X$ a scheme, $\Lambda$ a ring, $\operatorname{Mod}_\Lambda$ the abelian category of $\Lambda$-sheaves on $X_\tau$, $\operatorname{Ch}:=\operatorname{Ch}^{\geq0}(\operatorname{Mod}_\Lambda)$ the abelian category of non-negative cochain complexes, $\mathcal D^{\geq0}(X)$ the derived $\infty$-category, $\mathcal F\in\mathcal D^{\geq0}(X)$ a sheaf (i.e. complex of sheaves). If $U\to X$ is a map of schemes, let $R\Gamma(U_\tau,-):\mathcal D^{\geq0}(X)\to\mathcal D^{\geq0}(\Lambda)$ be the derived functor of global sections of the pullback of $\mathcal F$ (defined via HA.1.3.3.2 or equivalently as discussed here as a (homotopy) Kan extension). Then $R\Gamma(-,\mathcal F)$ takes finite disjoint unions to finite products and satisfies the $\infty$-categorical sheaf axiom that you state; i.e. for every covering map $f:U\to X$ in the topology $\tau$ we have $$R\Gamma(X_\tau,\mathcal F)=\lim\Big(R\Gamma(U,\mathcal F)\rightrightarrows R\Gamma(U\times_XU,\mathcal F)\mathrel{\substack{\textstyle\rightarrow\\[-0.6ex]\textstyle\rightarrow \\[-0.6ex]\textstyle\rightarrow}}\cdots\Big).$$ To see this last point, since $R\Gamma(X_\tau,-)$ is a right adjoint, it will suffice to see that $$\mathcal F=\lim\Big(Rf_*f^*\mathcal F\rightrightarrows Rf_{1*}f_1^*\mathcal F\mathrel{\substack{\textstyle\rightarrow\\[-0.6ex]\textstyle\rightarrow \\[-0.6ex]\textstyle\rightarrow}}\cdots\Big),\tag{$\ast$}$$ in $\mathcal D^{\geq0}(X)$, where $f_0=f:U\to X$ and $f_n:U_n=\underbrace{U\times_X\ldots\times_XU}_{n+1\text{ times}}\to X$. Replacing $\mathcal F$ by a bounded below complex of injectives in $\operatorname{Mod}_\Lambda$, we may drop the $R$, since the maps $f_n^*$ preserve injectives as the $f_n$ belong to our topology, and our task is to compute the homotopy limit of the diagram appearing in $(\ast)$. Such homotopy limits are called totalizations, and much has been written about computing totalizations of diagrams of chain complexes (nLab, Reference for homotopy (co)limits of (co)chain complexes via totalization of double complexes, Reference for homotopy colimit = total complex, §15 of this article by McClure & Smith). Unfortunately none of these accounts are really satisfactory, but the gist of it seems to be that given a cosimplicial map $\Delta\to\operatorname{Ch}$, we form the normalized Moore cochains in $\operatorname{Ch}$; i.e. a double complex $K^{ij}$, and then take the total complex. (I restrict to non-negative chain complexes because I don’t know if this recipe holds in the unbounded case.)

If we think about the complexes appearing in $(\ast)$ in the horizontal direction and the cosimplicial maps in the vertical direction, and we think of $\mathcal F$ as a double complex concentrated in row 0, then we have a map of double complexes $F\to K^{ij}$ that we wish to see induces a quasi-isomorphism after totalizing. Now it’s helpful to know that normalized cochains are quasi-isomorphic to un-normalized ones (019H), so by the spectral sequence associated to the filtration by columns it will suffice to show that the augmented relative Čech complex $$\mathcal F\to f_*f^*\mathcal F\xrightarrow{d^0-d^1}f_{1*}f_1^*\mathcal F\xrightarrow{d^0-d^1+d^2}\cdots$$ is acyclic when $\mathcal F\in\operatorname{Mod}_\Lambda$ and nothing is derived. As $f:U\to X$ is a covering map, it will suffice to check this after pulling back to $U$, but now we’re in the situation of forming the relative Čech complex associated to a map which admits a section: the resulting complex is homotopic to zero (06X6).

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