Lemma 21.48.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed space. The category of complexes of $\mathcal{O}$-modules with tensor product defined by $\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet = \text{Tot}(\mathcal{F}^\bullet \otimes _\mathcal {O} \mathcal{G}^\bullet )$ is a symmetric monoidal category.

## 21.48 Duals

In this section we characterize the dualizable objects of the category of complexes and of the derived category. In particular, we will see that an object of $D(\mathcal{O})$ has a dual if and only if it is perfect (this follows from Example 21.48.6 and Lemma 21.48.7).

**Proof.**
Omitted. Hints: as unit $\mathbf{1}$ we take the complex having $\mathcal{O}$ in degree $0$ and zero in other degrees with obvious isomorphisms $\text{Tot}(\mathbf{1} \otimes _\mathcal {O} \mathcal{G}^\bullet ) = \mathcal{G}^\bullet $ and $\text{Tot}(\mathcal{F}^\bullet \otimes _\mathcal {O} \mathbf{1}) = \mathcal{F}^\bullet $. to prove the lemma you have to check the commutativity of various diagrams, see Categories, Definitions 4.43.1 and 4.43.9. The verifications are straightforward in each case.
$\square$

Example 21.48.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}$-modules such that for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ there exists a covering $\{ U_ i \to U\} $ such that $\mathcal{F}^\bullet |_{U_ i}$ is strictly perfect. Consider the complex

as in Section 21.34. Let

be $\eta = \sum \eta _ n$ and $\epsilon = \sum \epsilon _ n$ where $\eta _ n : \mathcal{O} \to \mathcal{F}^ n \otimes _\mathcal {O} \mathcal{G}^{-n}$ and $\epsilon _ n : \mathcal{G}^{-n} \otimes _\mathcal {O} \mathcal{F}^ n \to \mathcal{O}$ are as in Modules on Sites, Example 18.29.1. Then $\mathcal{G}^\bullet , \eta , \epsilon $ is a left dual for $\mathcal{F}^\bullet $ as in Categories, Definition 4.43.5. We omit the verification that $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_{\mathcal{F}^\bullet }$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{\mathcal{G}^\bullet }$. Please compare with More on Algebra, Lemma 15.72.2.

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.
$\square$

Lemma 21.48.4. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K$ be a perfect object of $D(\mathcal{O})$. Then $K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O})$ is a perfect object too and $(K^\vee )^\vee \cong K$. There are functorial isomorphisms

and

for $M$ in $D(\mathcal{O})$.

**Proof.**
We will us without further mention that formation of internal hom commutes with restriction (Lemma 21.35.3). Let $U$ be an arbitrary object of $\mathcal{C}$. To check that $K^\vee $ is perfect, it suffices to show that there exists a covering $\{ U_ i \to U\} $ such that $K^\vee |_{U_ i}$ is perfect for all $i$. There is a canonical map

see Lemma 21.35.5. It suffices to prove there is a covering $\{ U_ i \to U\} $ such that the restriction of this map to $\mathcal{C}/U_ i$ is an isomorphism for all $i$. By Lemma 21.35.9 to see the final statement it suffices to check that the map (21.35.9.1)

is an isomorphism. This is a local question as well (in the sense above). Hence it suffices to prove the lemma when $K$ is represented by a strictly perfect complex.

Assume $K$ is represented by the strictly perfect complex $\mathcal{E}^\bullet $. Then it follows from Lemma 21.44.9 that $K^\vee $ is represented by the complex whose terms are $(\mathcal{E}^ n)^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^ n, \mathcal{O})$ in degree $-n$. Since $\mathcal{E}^ n$ is a direct summand of a finite free $\mathcal{O}$-module, so is $(\mathcal{E}^ n)^\vee $. Hence $K^\vee $ is represented by a strictly perfect complex too and we see that $K^\vee $ is perfect. The map $K \to (K^\vee )^\vee $ is an isomorphism as it is given up to sign by the evaluation maps $\mathcal{E}^ n \to ((\mathcal{E}^ n)^\vee )^\vee $ which are isomorphisms. To see that (21.35.9.1) is an isomorphism, represent $M$ by a K-flat complex $\mathcal{F}^\bullet $. By Lemma 21.44.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

On the other hand, the object $M \otimes ^\mathbf {L}_\mathcal {O} K^\vee $ is represented by the complex with terms

Thus the assertion that (21.35.9.1) is an isomorphism reduces to the assertion that the canonical map

is an isomorphism when $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}$-module and $\mathcal{F}$ is any $\mathcal{O}$-module. This follows immediately from the corresponding statement when $\mathcal{E}$ is finite free. $\square$

Lemma 21.48.5. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. The derived category $D(\mathcal{O})$ is a symmetric monoidal category with tensor product given by derived tensor product with usual associativity and commutativity constraints (for sign rules, see More on Algebra, Section 15.72).

**Proof.**
Omitted. Compare with Lemma 21.48.1.
$\square$

Example 21.48.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $K$ be a perfect object of $D(\mathcal{O})$. Set $K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O})$ as in Lemma 21.48.4. Then the map

is an isomorphism (by the lemma). Denote

the map sending $1$ to the section corresponding to $\text{id}_ K$ under the isomorphism above. Denote

the evaluation map (to construct it you can use Lemma 21.35.6 for example). Then $K^\vee , \eta , \epsilon $ is a left dual for $K$ as in Categories, Definition 4.43.5. We omit the verification that $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_ K$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{K^\vee }$.

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. $\square$

Lemma 21.48.8. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $(K_ n)_{n \in \mathbf{N}}$ be a system of perfect objects of $D(\mathcal{O})$. Let $K = \text{hocolim} K_ n$ be the derived colimit (Derived Categories, Definition 13.33.1). Then for any object $E$ of $D(\mathcal{O})$ we have

where $(K_ n^\vee )$ is the inverse system of dual perfect complexes.

**Proof.**
By Lemma 21.48.4 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_\mathcal {O} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (21.35.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

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