Remark 48.12.6. Let $f : X \to Y$, $\omega ^\bullet _{X/Y}$, and $\text{Tr}_{f, \mathcal{O}_ Y}$ be as in Remark 48.12.5. Let $K$ and $M$ be in $D_\mathit{QCoh}(\mathcal{O}_ X)$ with $M$ pseudo-coherent (for example perfect). Suppose given a map $K \otimes _{\mathcal{O}_ X}^\mathbf {L} M \to \omega ^\bullet _{X/Y}$ which corresponds to an isomorphism $K \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(M, \omega ^\bullet _{X/Y})$ via Cohomology, Equation (20.39.0.1). Then the relative cup product (Cohomology, Remark 20.28.7)

$Rf_*K \otimes _{\mathcal{O}_ Y}^\mathbf {L} Rf_*M \to Rf_*(K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \to Rf_*\omega ^\bullet _{X/Y} \xrightarrow {\text{Tr}_{f, \mathcal{O}_ Y}} \mathcal{O}_ Y$

determines an isomorphism $Rf_*K \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*M, \mathcal{O}_ Y)$. Namely, since $\omega ^\bullet _{X/Y} = a(\mathcal{O}_ Y)$ the canonical map (48.3.5.1)

$Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(M, \omega ^\bullet _{X/Y}) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*M, \mathcal{O}_ Y)$

is an isomorphism by Lemma 48.3.6 and Remark 48.3.8 and the fact that $M$ and $Rf_*M$ are pseudo-coherent, see Derived Categories of Schemes, Lemma 36.30.5. To see that the relative cup product induces this isomorphism use the commutativity of the diagram in Cohomology, Remark 20.39.12.

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