The Stacks project

Lemma 21.48.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}$-modules. If $\mathcal{F}^\bullet $ has a left dual in the monoidal category of complexes of $\mathcal{O}$-modules (Categories, Definition 4.43.5) then for every object $U$ of $\mathcal{C}$ there exists a covering $\{ U_ i \to U\} $ such that $\mathcal{F}^\bullet |_{U_ i}$ is strictly perfect and the left dual is as constructed in Example 21.48.2.

Proof. By uniqueness of left duals (Categories, Remark 4.43.7) we get the final statement provided we show that $\mathcal{F}^\bullet $ is as stated. Let $\mathcal{G}^\bullet , \eta , \epsilon $ be a left dual. Write $\eta = \sum \eta _ n$ and $\epsilon = \sum \epsilon _ n$ where $\eta _ n : \mathcal{O} \to \mathcal{F}^ n \otimes _\mathcal {O} \mathcal{G}^{-n}$ and $\epsilon _ n : \mathcal{G}^{-n} \otimes _\mathcal {O} \mathcal{F}^ n \to \mathcal{O}$. Since $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_{\mathcal{F}^\bullet }$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{\mathcal{G}^\bullet }$ by Categories, Definition 4.43.5 we see immediately that we have $(1 \otimes \epsilon _ n) \circ (\eta _ n \otimes 1) = \text{id}_{\mathcal{F}^ n}$ and $(\epsilon _ n \otimes 1) \circ (1 \otimes \eta _ n) = \text{id}_{\mathcal{G}^{-n}}$. In other words, we see that $\mathcal{G}^{-n}$ is a left dual of $\mathcal{F}^ n$ and we see that Modules on Sites, Lemma 18.29.2 applies to each $\mathcal{F}^ n$. Let $U$ be an object of $\mathcal{C}$. There exists a covering $\{ U_ i \to U\} $ such that for every $i$ only a finite number of $\eta _ n|_{U_ i}$ are nonzero. Thus after replacing $U$ by $U_ i$ we may assume only a finite number of $\eta _ n|_ U$ are nonzero and by the lemma cited this implies only a finite number of $\mathcal{F}^ n|_ U$ are nonzero. Using the lemma again we can then find a covering $\{ U_ i \to U\} $ such that each $\mathcal{F}^ n|_{U_ i}$ is a direct summand of a finite free $\mathcal{O}$-module and the proof is complete. $\square$

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