Lemma 20.47.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ be a perfect object of $D(\mathcal{O}_ X)$. Then $K^\vee = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \mathcal{O}_ X)$ is a perfect object too and $(K^\vee )^\vee \cong K$. There are functorial isomorphisms

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

and

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

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

**Proof.**
By Lemma 20.39.9 there is a canonical map

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

which is an isomorphism by Lemma 20.47.4. To check the other statements we will use without further mention that formation of internal hom commutes with restriction to opens (Lemma 20.39.3). We may check $K^\vee $ is perfect locally on $X$. By Lemma 20.39.8 to see the final statement it suffices to check that the map (20.39.8.1)

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

is an isomorphism. This is local on $X$ as well. Hence it suffices to prove these two statements $K$ is represented by a strictly perfect complex.

Assume $K$ is represented by the strictly perfect complex $\mathcal{E}^\bullet $. Then it follows from Lemma 20.43.9 that $K^\vee $ is represented by the complex whose terms are $(\mathcal{E}^{-n})^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-n}, \mathcal{O}_ X)$ in degree $n$. Since $\mathcal{E}^{-n}$ is a direct summand of a finite free $\mathcal{O}_ X$-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. To see that (20.39.8.1) is an isomorphism, represent $M$ by a complex $\mathcal{F}^\bullet $. By Lemma 20.43.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

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

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

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

Thus the assertion that (20.39.8.1) is an isomorphism reduces to the assertion that the canonical map

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

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

## Comments (1)

Comment #161 by Pieter Belmans on