Lemma 48.28.10. Let $f : Y \to X$ and $X \to S$ be morphisms of schemes which are flat and of finite presentation. Let $(K, \xi )$ and $(M, \eta )$ be a relative dualizing complex for $X \to S$ and $Y \to X$. Set $E = M \otimes _{\mathcal{O}_ Y}^\mathbf {L} Lf^*K$. Then $(E, \zeta )$ is a relative dualizing complex for $Y \to S$ for a suitable $\zeta$.

Proof. Using Lemma 48.28.2 and the algebraic version of this lemma (Dualizing Complexes, Lemma 47.27.6) we see that $E$ is affine locally the first component of a relative dualizing complex. In particular we see that $E$ is $S$-perfect since this may be checked affine locally, see Derived Categories of Schemes, Lemma 36.35.3.

Let us first prove the existence of $\zeta$ in case the morphisms $X \to S$ and $Y \to X$ are separated so that $\Delta _{X/S}$, $\Delta _{Y/X}$, and $\Delta _{Y/S}$ are closed immersions. Consider the following diagram

$\xymatrix{ & & Y \ar@{=}[r] & Y \ar[d]^ f \\ Y \ar[r]_{\Delta _{Y/X}} & Y \times _ X Y \ar[d]_ m \ar[r]_\delta \ar[ru]_ q & Y \times _ S Y \ar[d]^{f \times f} \ar[ru]_ p & X\\ & X \ar[r]^{\Delta _{X/S}} & X \times _ S X \ar[ru]_ r }$

where $p$, $q$, $r$ are the first projections. By Lemma 48.9.4 we have

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{Y \times _ S Y}}( \Delta _{Y/S, *}\mathcal{O}_ Y, Lp^*E) = R\delta _*\left(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{Y \times _ X Y}}( \Delta _{Y/X, *}\mathcal{O}_ Y, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Y \times _ X Y}, Lp^*E))\right)$

By Lemma 48.10.3 we have

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Y \times _ X Y}, Lp^*E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Y \times _ X Y}, L(f \times f)^*Lr^*K) \otimes _{\mathcal{O}_{Y \times _ S Y}}^\mathbf {L} Lq^*M$

By Lemma 48.10.2 we have

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Y \times _ X Y}, L(f \times f)^*Lr^*K) = Lm^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, Lr^*K)$

The last expression is isomorphic (via $\xi$) to $Lm^*\mathcal{O}_ X = \mathcal{O}_{Y \times _ X Y}$. Hence the expression preceding is isomorphic to $Lq^*M$. Hence

$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{Y \times _ S Y}}( \Delta _{Y/S, *}\mathcal{O}_ Y, Lp^*E) = R\delta _*\left(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{Y \times _ X Y}}( \Delta _{Y/X, *}\mathcal{O}_ Y, Lq^*M)\right)$

The material inside the parentheses is isomorphic to $\Delta _{Y/X, *}*\mathcal{O}_ X$ via $\eta$. This finishes the proof in the separated case.

In the general case we choose an open $W \subset X \times _ S X$ such that $\Delta _{X/S}$ factors through a closed immersion $\Delta : X \to W$ and we choose an open $V \subset Y \times _ X Y$ such that $\Delta _{Y/X}$ factors through a closed immersion $\Delta ' : Y \to V$. Finally, choose an open $W' \subset Y \times _ S Y$ whose intersection with $Y \times _ X Y$ gives $V$ and which maps into $W$. Then we consider the diagram

$\xymatrix{ & & Y \ar@{=}[r] & Y \ar[d]^ f \\ Y \ar[r]_{\Delta '} & V \ar[d]_ m \ar[r]_\delta \ar[ru]_ q & W' \ar[d]^{f \times f} \ar[ru]_ p & X\\ & X \ar[r]^\Delta & W \ar[ru]_ r }$

and we use exactly the same argument as before. $\square$

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