Lemma 20.46.4. 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.**
We will use without further mention that formation of internal hom commutes with restriction to opens (Lemma 20.38.3). We may check $K^\vee $ is perfect locally on $X$. 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 20.38.9. It suffices to prove this map is an isomorphism locally. By Lemma 20.38.8 to see the final statement it suffices to check that the map (20.38.8.1)

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

is an isomorphism. This is local on $X$ as well. Hence it suffices to prove 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 20.42.9 that $K^\vee $ is represented by the complex whose terms are $(\mathcal{E}^{-n})^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-n}, \mathcal{O}_ X)$ 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. 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 (20.38.8.1) is an isomorphism, represent $M$ by a complex $\mathcal{F}^\bullet $. By Lemma 20.42.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

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

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 (20.38.8.1) 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