## 21.46 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 21.46.6 and Lemma 21.46.7).

Lemma 21.46.1. 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.

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 4.42.1 and 4.42.9. The verifications are straightforward in each case. $\square$

Example 21.46.2. 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

$\mathcal{G}^\bullet = \mathop{\mathcal{H}\! \mathit{om}}\nolimits ^\bullet (\mathcal{F}^\bullet , \mathcal{O})$

as in Section 21.33. Let

$\eta : \mathcal{O} \to \text{Tot}(\mathcal{F}^\bullet \otimes _\mathcal {O} \mathcal{G}^\bullet ) \quad \text{and}\quad \epsilon : \text{Tot}(\mathcal{G}^\bullet \otimes _\mathcal {O} \mathcal{F}^\bullet ) \to \mathcal{O}$

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 4.42.5. 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 15.71.3.

Lemma 21.46.3. 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 4.42.5) 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 21.46.2.

Proof. By uniqueness of left duals (Categories, Remark 4.42.7) 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 4.42.5 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 21.46.4. 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

$M \otimes ^\mathbf {L}_\mathcal {O} K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, M)$

and

$H^0(\mathcal{C}, M \otimes ^\mathbf {L}_\mathcal {O} K^\vee ) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O})}(K, M)$

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

Proof. We will us without further mention that formation of internal hom commutes with restriction (Lemma 21.34.3). 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

$K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, \mathcal{O}_ X) \otimes _{\mathcal{O}_ X}^\mathbf {L} K \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O}_ X), \mathcal{O}_ X) = (K^\vee )^\vee$

see Lemma 21.34.5. 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 21.34.9 to see the final statement it suffices to check that the map (21.34.9.1)

$M \otimes ^\mathbf {L}_\mathcal {O} K^\vee \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$

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 21.42.9 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 (21.34.9.1) is an isomorphism, represent $M$ by a K-flat complex $\mathcal{F}^\bullet$. By Lemma 21.42.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

$\bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{F}^ p)$

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

$\bigoplus \nolimits _{n = p + q} \mathcal{F}^ p \otimes _\mathcal {O} (\mathcal{E}^{-q})^\vee$

Thus the assertion that (21.34.9.1) is an isomorphism reduces to the assertion that the canonical map

$\mathcal{F} \otimes _\mathcal {O} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}, \mathcal{O}) \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}, \mathcal{F})$

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 21.46.5. 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 15.71).

Proof. Omitted. Compare with Lemma 21.46.1. $\square$

Example 21.46.6. 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 21.46.4. Then the map

$K \otimes _\mathcal {O}^\mathbf {L} K^\vee \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K)$

is an isomorphism (by the lemma). Denote

$\eta : \mathcal{O} \longrightarrow K \otimes _\mathcal {O}^\mathbf {L} K^\vee$

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

$\epsilon : K^\vee \otimes _\mathcal {O}^\mathbf {L} K \longrightarrow \mathcal{O}$

the evaluation map (to construct it you can use Lemma 21.34.6 for example). Then $K^\vee , \eta , \epsilon$ is a left dual for $K$ as in Categories, Definition 4.42.5. 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 21.46.7. 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 4.42.5) then $M$ is perfect and the left dual is as constructed in Example 21.46.6.

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

$\eta : \mathcal{O} \to \text{Tot}(\mathcal{M}^\bullet \otimes _\mathcal {O} \mathcal{N}^\bullet )$

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

$\eta (1) = \sum \nolimits _ i f_ i \otimes g_ i$

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

$\tilde\eta : \mathcal{O} \to \text{Tot}(\mathcal{K}^\bullet \otimes _\mathcal {O} \mathcal{N}^\bullet )$

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

$\xymatrix{ M \ar[rr]_-{\eta \otimes 1} \ar[rrd]_{\tilde\eta \otimes 1} & & M \otimes ^\mathbf {L} N \otimes ^\mathbf {L} M \ar[r]_-{1 \otimes \epsilon } & M \\ & & K \otimes ^\mathbf {L} N \otimes ^\mathbf {L} M \ar[u] \ar[r]^-{1 \otimes \epsilon } & K \ar[u] }$

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 21.45.3 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

$\mathcal{E}^\bullet \to M \to L \to \mathcal{E}^\bullet [1]$

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

$\mathcal{M}^\bullet : \ldots \to \mathcal{L}^{m - 2} \to \mathcal{L}^{m - 1} \to \mathcal{E}^ m \to \mathcal{E}^{m + 1} \to \ldots$

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

$\xymatrix{ \mathcal{E}^\bullet \ar[r] & \mathcal{M}^\bullet \ar[r] & \mathcal{L}^\bullet \ar[r] & \mathcal{E}^\bullet [1] \\ \mathcal{E}^\bullet \ar[r] \ar[u] & \mathcal{K}^\bullet \ar[r] \ar[u] & \sigma _{\leq m - 1}\mathcal{K}^\bullet \ar[r] \ar[u] & \mathcal{E}^\bullet [1] \ar[u] }$

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

$\xymatrix{ \mathcal{E}^\bullet \ar[r] & \mathcal{K}^\bullet \ar@{=}[r] & K \\ \mathcal{E}^\bullet \ar[u]^{\text{id}} \ar[r]^ i & \mathcal{M}^\bullet \ar@{=}[r] & M \ar[u]_ s }$

commutes, but we do know it commutes after composing with the map $K \to M$. By Lemma 21.42.8 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

$\xymatrix{ \mathcal{E}^\bullet \ar[r] & M \ar[r] & \mathcal{L}^\bullet \ar[r] & \mathcal{E}^\bullet [1] \\ \mathcal{E}^\bullet \ar[r] \ar[u] & K \ar[r] \ar[u] & \sigma _{\leq m - 1}\mathcal{K}^\bullet \ar[r] \ar[u] & \mathcal{E}^\bullet [1] \ar[u] \\ \mathcal{E}^\bullet \ar[r] \ar[u] & M \ar[r] \ar[u]^ s & \mathcal{L}^\bullet \ar[r] \ar@{..>}[u] & \mathcal{E}^\bullet [1] \ar[u] }$

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

$\xymatrix{ H^{m - 1}(M) \ar[r] & H^{m - 1}(\mathcal{L}^\bullet ) \ar[r] & H^ m(\mathcal{E}^\bullet ) \\ H^{m - 1}(K) \ar[r] \ar[u] & H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \ar[r] \ar[u] & H^ m(\mathcal{E}^\bullet ) \ar[u] \\ H^{m - 1}(M) \ar[r] \ar[u] & H^{m - 1}(\mathcal{L}^\bullet ) \ar[r] \ar[u] & H^ m(\mathcal{E}^\bullet ) \ar[u] }$

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

$(\mathcal{O}^{\oplus r} \to \mathcal{E}^ m \to \mathcal{E}^{m + 1} \to \ldots ) \longrightarrow K \longrightarrow M$

induces an isomorphism on cohomology sheaves in degrees $\geq m$ and a surjection in degree $m - 1$ and the proof is complete. $\square$

Lemma 21.46.8. 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

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, E) = R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_\mathcal {O} K_ n^\vee$

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

Proof. By Lemma 21.46.4 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

$R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) \to \prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$

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 (21.34.0.1) and the fact that derived tensor product commutes with direct sums. $\square$

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).