Lemma 48.3.6. Let $f : X \to Y$ be a morphism of quasi-compact and quasi-separated schemes. Let $a$ be the right adjoint to $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$. Let $L \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $K \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. Then the map (48.3.5.1)

\[ Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \]

becomes an isomorphism after applying the functor $DQ_ Y : D(\mathcal{O}_ Y) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ discussed in Derived Categories of Schemes, Section 36.21.

**Proof.**
The statement makes sense as $DQ_ Y$ exists by Derived Categories of Schemes, Lemma 36.21.1. Since $DQ_ Y$ is the right adjoint to the inclusion functor $D_\mathit{QCoh}(\mathcal{O}_ Y) \to D(\mathcal{O}_ Y)$ to prove the lemma we have to show that for any $M \in D_\mathit{QCoh}(\mathcal{O}_ Y)$ the map (48.3.5.1) induces an bijection

\[ \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ Y(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)) \]

To see this we use the following string of equalities

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Y(M, Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(Lf^*M \otimes _{\mathcal{O}_ X}^\mathbf {L} L, a(K)) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*(Lf^*M \otimes _{\mathcal{O}_ X}^\mathbf {L} L), K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*L, K) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)) \end{align*}

The first equality holds by Cohomology, Lemma 20.28.1. The second equality by Cohomology, Lemma 20.42.2. The third equality by construction of $a$. The fourth equality by Derived Categories of Schemes, Lemma 36.22.1 (this is the important step). The fifth by Cohomology, Lemma 20.42.2.
$\square$

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