Lemma 48.17.8. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Let $K$ be a dualizing complex on $Y$. Set $D_ Y(M) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(M, K)$ for $M \in D_{\textit{Coh}}(\mathcal{O}_ Y)$ and $D_ X(E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, f^!K)$ for $E \in D_{\textit{Coh}}(\mathcal{O}_ X)$. Then there is a canonical isomorphism

$f^!M \longrightarrow D_ X(Lf^*D_ Y(M))$

for $M \in D_{\textit{Coh}}^+(\mathcal{O}_ Y)$.

Proof. Choose compactification $j : X \subset \overline{X}$ of $X$ over $Y$ (More on Flatness, Theorem 38.33.8 and Lemma 38.32.2). Let $a$ be the right adjoint of Lemma 48.3.1 for $\overline{X} \to Y$. Set $D_{\overline{X}}(E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{\overline{X}}}(E, a(K))$ for $E \in D_{\textit{Coh}}(\mathcal{O}_{\overline{X}})$. Since formation of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits$ commutes with restriction to opens and since $f^! = j^* \circ a$ we see that it suffices to prove that there is a canonical isomorphism

$a(M) \longrightarrow D_{\overline{X}}(L\overline{f}^*D_ Y(M))$

for $M \in D_{\textit{Coh}}(\mathcal{O}_ Y)$. For $F \in D_\mathit{QCoh}(\mathcal{O}_ X)$ we have

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{\overline{X}}( F, D_{\overline{X}}(L\overline{f}^*D_ Y(M))) & = \mathop{\mathrm{Hom}}\nolimits _{\overline{X}}( F \otimes _{\mathcal{O}_ X}^\mathbf {L} L\overline{f}^*D_ Y(M), a(K)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y( R\overline{f}_*(F \otimes _{\mathcal{O}_ X}^\mathbf {L} L\overline{f}^*D_ Y(M)), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y( R\overline{f}_*(F) \otimes _{\mathcal{O}_ Y}^\mathbf {L} D_ Y(M), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y( R\overline{f}_*(F), D_ Y(D_ Y(M))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(R\overline{f}_*(F), M) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\overline{X}}(F, a(M)) \end{align*}

The first equality by Cohomology, Lemma 20.42.2. The second by definition of $a$. The third by Derived Categories of Schemes, Lemma 36.22.1. The fourth equality by Cohomology, Lemma 20.42.2 and the definition of $D_ Y$. The fifth equality by Lemma 48.2.5. The final equality by definition of $a$. Hence we see that $a(M) = D_{\overline{X}}(L\overline{f}^*D_ Y(M))$ by Yoneda's lemma. $\square$

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).