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Aug 30, 2022 at 23:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jul 31, 2022 at 21:07 answer added Tomo timeline score: 4
Apr 11, 2019 at 13:45 comment added Ben Wieland ...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.)
Apr 11, 2019 at 13:42 comment added Ben Wieland 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...
Apr 10, 2019 at 20:14 comment added user662742 @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?
Apr 10, 2019 at 19:50 comment added Yonatan Harpaz 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.
Apr 10, 2019 at 19:45 review First posts
Apr 10, 2019 at 20:12
Apr 10, 2019 at 19:41 history asked user662742 CC BY-SA 4.0