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It is well known that derived categories (I'm particularly thinking of constructible derived categories and derived categories of D-modules) don't form a stack. In particular given morphisms in the constructible derived category are essentially cohomology it would be absurd if they were stacky.

However as far as I understand it the obstruction to this stacky-ness is the fact that morphisms between cones in a triangulated category aren't unique (though this may be technically more to do with descent categories). Since cones (as objects) are unique up to isomorphism and since every counterexample to stacky-ness for a derived category I have ever seen was from the morphisms it leads me to the following question: If I have a complex in one of the two derived categories mentioned at the beginning and I can find an open cover such that this complex vanishes in certain degrees when restricted to this open cover, does it follow that the complex vanishes in those degrees globally?

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    $\begingroup$ As a side remark, if you take the derived $\infty$-categories, the formation $X\mapsto\mathcal D(X)$ does satisfy descent in different setups. See, for example, the introduction of Hansen–Scholze, Relative Perversity. $\endgroup$
    – Z. M
    Commented Nov 23, 2023 at 14:53

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What do you mean vanishes in those degrees? Do you mean the cohomology sheaf vanishes? This would just follow from the analogous vanishing statement for sheaves.

Or for the literal complex do you mean the $i$'th sheaf of cochains vanishes? Again this follows from the analogous statement for sheaves?

Or do you mean that the complex is quasi-isomorphic to a complex whose cochains vanish in $i$'th degree? This is not true. A nontrivial class in $\operatorname{Ext}^3 (\mathbb Q, \mathbb Q)$ produces a complex whose cohomology sheaves are $\mathbb Q$ in degrees $0$ and $2$ and $0$ in all other degrees. Over any open set where the class vanishes, the complex is quasi-isomorphic to a split complex with chains in exactly degrees $0$ and $2$, but globally it cannot be isomorphic to a complex with zero chains in degree $1$ as then it would split and force the cohomology class to be trivial.

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  • $\begingroup$ I meant the third version you answered so thank you for clearing that up. $\endgroup$
    – l-r-b
    Commented Nov 23, 2023 at 12:46

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