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Let $X$ be a scheme and $\mathcal S$ a site which is a full subcategory of the category $Aff/X$ of affine schemes with a map to $X$. If I understand correctly, the category $QCoh^\mathcal S(X)$ of $\mathcal S$-quasicoherent sheaves is the global sections of the $\mathcal S$-stackification of the functor $Mod: \mathcal S^{op} \to Cat$, $(Spec A \to X) \mapsto Mod_A$, where $Mod_A$ is the category of $A$-modules.

The interesting phenomenon (cf. the stacks project, which is probably using slightly different definitions) is that for reasonable $\mathcal S$ (the above link gives precise conditions), $QCoh^\mathcal S(X)$ is actually independent of $\mathcal S$. I'm looking for a high-concept explanation of this fact.

The best I can figure is the following. For reasonable topologies, the functor $Mod: \mathcal S \to Cat$ is already a stack, so its global sections can be computed using an $\mathcal S$-cover of the terminal object (which I've technically left out of the category $\mathcal S$, but that's okay). Since $X$ is a scheme, it comes with a Zariski cover, which is also a $\mathcal S$-cover for reasonable topologies. Thus $QCoh^\mathcal S(X)$ is simply computed in the same way for reasonable topologies $\mathcal S$, so of course it agrees.

This suggests that a statement of this meta-principle (an alternative to the one found in the stacks project) would say that

Claim: Let $X$ be a scheme $\mathcal S$ be a full subcategory of $Aff/X$, equipped with a Grothendieck topology. Then $QCoh^\mathcal S(X) = QCoh^{Zariski}(X)$ if

  1. $Mod : \mathcal S^{op} \to Cat$ is a stack, and

  2. Every Zariski cover is an $\mathcal S$-cover.

Questions:

  1. Is the above claim correct?

  2. If not (or if so!) is there some other high-concept way to see that $QCoh^\mathcal S(X)$ is independent of $\mathcal S$ for reasonable $\mathcal S$?

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  • $\begingroup$ Isn't this fpqc descent for quasicoherent sheaves: stacks.math.columbia.edu/tag/023R ? $\endgroup$ – Harry Gindi Dec 12 '18 at 20:06
  • $\begingroup$ @HarryGindi yes $\endgroup$ – Tim Campion Dec 12 '18 at 20:07
  • $\begingroup$ Condition 2 does not really make sense as $\mathcal S$ is defined as a topology on $Aff/X$, which presumably means the topology defines coverings within $Aff/X$. In this generality one can define global sections as choices of an object in $\mathcal F(U)$ for each open set $U$, compatible with maps, and then 1 is obviously sufficient because the maps are the same in each topology. $\endgroup$ – Will Sawin Dec 12 '18 at 20:12
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The basic reason is that a sheaf of modules is quasi-coherent if and only if it is a Cartesian presheaf. Therefore a certain covering suffices for the condition to hold. The condition on coarse covers determine the behavior on fine covers, therefore being quasi-coherent in Zariski topology (or in étale topology on an algebraic space, DM-stack) already makes the sheaf quasi-coherent in any finer topology. The theory is specially simple when you have a base of "affine" schemes that cover the base space.

The Cartesian condition is just a condition on presheaves, so, in a sense, topology is not important. Risking to commit self-promotion, this is treated in the context of geometric stacks in the paper "A functorial formalism for quasi-coherent sheaves on a geometric stack", https://arxiv.org/abs/1304.2520 (see also Expo. Math., 33 (2015) pp. 452-501). The case of affine schemes is explained in section 2 and it is generalized to stacks with affine diagonal in section 3. The case of a semi-separated scheme follows adapting (and somehow simplifying) the arguments there. In particular, one does not need to use the small flat site in this case

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    $\begingroup$ Sorry, what is a Cartesian sheaf? Is there a site called the "cartesian site"? $\endgroup$ – Tim Campion Dec 13 '18 at 5:40
  • $\begingroup$ @TimCampion I've made an edit. Hope this is more explicit now. $\endgroup$ – Leo Alonso Dec 13 '18 at 9:25
  • $\begingroup$ @Tim Campiom: see definition 3.1 on page 44 of Vistoli's notes (homepage.sns.it/vistoli/descent.pdf). I think a presheaf $\mathcal{F}$ on a category $\mathcal{C}$ is cartesian when it is cartesian when you think of it as a fibered (in sets) category $\mathcal{F}\to\mathcal{C}$. Compare with Définition (12.3) in Champ algébriques by Laumon--Moret-Bailly $\endgroup$ – Qfwfq Dec 14 '18 at 16:47

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