Remark 48.23.4. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Assume $f$ is a Cohen-Macaulay morphism of relative dimension $d$. Let $\omega _{X/Y} = H^{-d}(f^!\mathcal{O}_ Y)$ be the unique nonzero cohomology sheaf of $f^!\mathcal{O}_ Y$, see Lemma 48.21.7. Then there is a canonical isomorphism

$f^!K = Lf^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}[d]$

for $K \in D^+_\mathit{QCoh}(\mathcal{O}_ Y)$, see Lemma 48.17.9. In particular, if $S$ has a dualizing complex $\omega _ S^\bullet$, $\omega _ Y^\bullet = (Y \to S)^!\omega _ S^\bullet$, and $\omega _ X^\bullet = (X \to S)^!\omega _ S^\bullet$ then we have

$\omega _ X^\bullet = Lf^*\omega _ Y^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} \omega _{X/Y}[d]$

Thus if further $X$ and $Y$ are connected and Cohen-Macaulay and if $\omega _ Y$ and $\omega _ X$ denote the unique nonzero cohomology sheaves of $\omega _ Y^\bullet$ and $\omega _ X^\bullet$, then we have

$\omega _ X = f^*\omega _ Y \otimes _{\mathcal{O}_ X} \omega _{X/Y}.$

Similar results hold for $X$ and $Y$ arbitrary finite type schemes over $S$ (i.e., not necessarily separated over $S$) with dualizing complexes normalized with respect to $\omega _ S^\bullet$ as in Section 48.20.

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