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

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

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 21.35.3). 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

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

see Lemma 21.35.5. 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)

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

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 21.44.9 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

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

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 (21.35.9.1) 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. $\square$

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