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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)