Lemma 21.48.7. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $M$ be an object of $D(\mathcal{O})$. If $M$ has a left dual in the monoidal category $D(\mathcal{O})$ (Categories, Definition 4.43.5) then $M$ is perfect and the left dual is as constructed in Example 21.48.6.

**Proof.**
Let $N, \eta , \epsilon $ be a left dual. Observe that for any object $U$ of $\mathcal{C}$ the restriction $N|_ U, \eta |_ U, \epsilon |_ U$ is a left dual for $M|_ U$.

Let $U$ be an object of $\mathcal{C}$. It suffices to find a covering $\{ U_ i \to U\} _{i \in I}$ fo $\mathcal{C}$ such that $M|_{U_ i}$ is a perfect object of $D(\mathcal{O}_{U_ i})$. Hence we may replace $\mathcal{C}, \mathcal{O}, M, N, \eta , \epsilon $ by $\mathcal{C}/U, \mathcal{O}_ U, M|_ U, N|_ U, \eta |_ U, \epsilon |_ U$ and assume $\mathcal{C}$ has a final object $X$. Moreover, during the proof we can (finitely often) replace $X$ by the members of a covering $\{ U_ i \to X\} $ of $X$.

We are going to use the following argument several times. Choose any complex $\mathcal{M}^\bullet $ of $\mathcal{O}$-modules representing $M$. Choose a K-flat complex $\mathcal{N}^\bullet $ representing $N$ whose terms are flat $\mathcal{O}$-modules, see Lemma 21.17.11. Consider the map

After replacing $X$ by the members of a covering, we can find an integer $N$ and for $i = 1, \ldots , N$ integers $n_ i \in \mathbf{Z}$ and sections $f_ i$ and $g_ i$ of $\mathcal{M}^{n_ i}$ and $\mathcal{N}^{-n_ i}$ such that

Let $\mathcal{K}^\bullet \subset \mathcal{M}^\bullet $ be any subcomplex of $\mathcal{O}$-modules containing the sections $f_ i$ for $i = 1, \ldots , N$. Since $\text{Tot}(\mathcal{K}^\bullet \otimes _\mathcal {O} \mathcal{N}^\bullet ) \subset \text{Tot}(\mathcal{M}^\bullet \otimes _\mathcal {O} \mathcal{N}^\bullet )$ by flatness of the modules $\mathcal{N}^ n$, we see that $\eta $ factors through

Denoting $K$ the object of $D(\mathcal{O})$ represented by $\mathcal{K}^\bullet $ we find a commutative diagram

Since the composition of the upper row is the identity on $M$ we conclude that $M$ is a direct summand of $K$ in $D(\mathcal{O})$.

As a first use of the argument above, we can choose the subcomplex $\mathcal{K}^\bullet = \sigma _{\geq a} \tau _{\leq b}\mathcal{M}^\bullet $ with $a < n_ i < b$ for $i = 1, \ldots , N$. Thus $M$ is a direct summand in $D(\mathcal{O})$ of a bounded complex and we conclude we may assume $M$ is in $D^ b(\mathcal{O})$. (Recall that the process above involves replacing $X$ by the members of a covering.)

Since $M$ is in $D^ b(\mathcal{O})$ we may choose $\mathcal{M}^\bullet $ to be a bounded above complex of flat modules (by Modules, Lemma 17.17.6 and Derived Categories, Lemma 13.15.4). Then we can choose $\mathcal{K}^\bullet = \sigma _{\geq a}\mathcal{M}^\bullet $ with $a < n_ i$ for $i = 1, \ldots , N$ in the argument above. Thus we find that we may assume $M$ is a direct summand in $D(\mathcal{O})$ of a bounded complex of flat modules. In particular, we find $M$ has finite tor amplitude.

Say $M$ has tor amplitude in $[a, b]$. Assuming $M$ is $m$-pseudo-coherent we are going to show that (after replacing $X$ by the members of a covering) we may assume $M$ is $(m - 1)$-pseudo-coherent. This will finish the proof by Lemma 21.47.3 and the fact that $M$ is $(b + 1)$-pseudo-coherent in any case. After replacing $X$ by the members of a covering we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to M$ in $D(\mathcal{O})$ such that $H^ i(\alpha )$ is an isomorphism for $i > m$ and surjective for $i = m$. We may and do assume that $\mathcal{E}^ i = 0$ for $i < m$. Choose a distinguished triangle

Observe that $H^ i(L) = 0$ for $i \geq m$. Thus we may represent $L$ by a complex $\mathcal{L}^\bullet $ with $\mathcal{L}^ i = 0$ for $i \geq m$. The map $L \to \mathcal{E}^\bullet [1]$ is given by a map of complexes $\mathcal{L}^\bullet \to \mathcal{E}^\bullet [1]$ which is zero in all degrees except in degree $m - 1$ where we obtain a map $\mathcal{L}^{m - 1} \to \mathcal{E}^ m$, see Derived Categories, Lemma 13.27.3. Then $M$ is represented by the complex

Apply the discussion in the second paragraph to this complex to get sections $f_ i$ of $\mathcal{M}^{n_ i}$ for $i = 1, \ldots , N$. For $n < m$ let $\mathcal{K}^ n \subset \mathcal{L}^ n$ be the $\mathcal{O}$-submodule generated by the sections $f_ i$ for $n_ i = n$ and $d(f_ i)$ for $n_ i = n - 1$. For $n \geq m$ set $\mathcal{K}^ n = \mathcal{E}^ n$. Clearly, we have a morphism of distinguished triangles

where all the morphisms are as indicated above. Denote $K$ the object of $D(\mathcal{O})$ corresponding to the complex $\mathcal{K}^\bullet $. By the arguments in the second paragraph of the proof we obtain a morphism $s : M \to K$ in $D(\mathcal{O})$ such that the composition $M \to K \to M$ is the identity on $M$. We don't know that the diagram

commutes, but we do know it commutes after composing with the map $K \to M$. By Lemma 21.44.8 after replacing $X$ by the members of a covering, we may assume that $s \circ i$ is given by a map of complexes $\sigma : \mathcal{E}^\bullet \to \mathcal{K}^\bullet $. By the same lemma we may assume the composition of $\sigma $ with the inclusion $\mathcal{K}^\bullet \subset \mathcal{M}^\bullet $ is homotopic to zero by some homotopy $\{ h^ i : \mathcal{E}^ i \to \mathcal{M}^{i - 1}\} $. Thus, after replacing $\mathcal{K}^{m - 1}$ by $\mathcal{K}^{m - 1} + \mathop{\mathrm{Im}}(h^ m)$ (note that after doing this it is still the case that $\mathcal{K}^{m - 1}$ is generated by finitely many global sections), we see that $\sigma $ itself is homotopic to zero! This means that we have a commutative solid diagram

By the axioms of triangulated categories we obtain a dotted arrow fitting into the diagram. Looking at cohomology sheaves in degree $m - 1$ we see that we obtain

Since the vertical compositions are the identity in both the left and right column, we conclude the vertical composition $H^{m - 1}(\mathcal{L}^\bullet ) \to H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \to H^{m - 1}(\mathcal{L}^\bullet )$ in the middle is surjective! In particular $H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \to H^{m - 1}(\mathcal{L}^\bullet )$ is surjective. Using the induced map of long exact sequences of cohomology sheaves from the morphism of triangles above, a diagram chase shows this implies $H^ i(K) \to H^ i(M)$ is an isomorphism for $i \geq m$ and surjective for $i = m - 1$. By construction we can choose an $r \geq 0$ and a surjection $\mathcal{O}^{\oplus r} \to \mathcal{K}^{m - 1}$. Then the composition

induces an isomorphism on cohomology sheaves in degrees $\geq m$ and a surjection in degree $m - 1$ 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)