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. $\square$

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08DQ. Beware of the difference between the letter 'O' and the digit '0'.