Lemma 48.28.9. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. If $f$ is flat, then $f^!\mathcal{O}_ Y$ is (the first component of) a relative dualizing complex for $X$ over $Y$ in the sense of Definition 48.28.1.

Proof. By Lemma 48.17.10 we have that $f^!\mathcal{O}_ Y$ is $Y$-perfect. As $f$ is separated the diagonal $\Delta : X \to X \times _ Y X$ is a closed immersion and $\Delta _*\Delta ^!(-) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X \times _ Y X}}(\mathcal{O}_ X, -)$, see Lemmas 48.9.7 and 48.9.3. Hence to finish the proof it suffices to show $\Delta ^!(L\text{pr}_1^*f^!(\mathcal{O}_ Y)) \cong \mathcal{O}_ X$ where $\text{pr}_1 : X \times _ Y X \to X$ is the first projection. We have

$\mathcal{O}_ X = \Delta ^! \text{pr}_1^!\mathcal{O}_ X = \Delta ^! \text{pr}_1^! L\text{pr}_2^*\mathcal{O}_ Y = \Delta ^!(L\text{pr}_1^* f^!\mathcal{O}_ Y)$

where $\text{pr}_2 : X \times _ Y X \to X$ is the second projection and where we have used the base change isomorphism $\text{pr}_1^! \circ L\text{pr}_2^* = L\text{pr}_1^* \circ f^!$ of Lemma 48.18.1. $\square$

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