Lemma 48.20.8. Let $(S, \omega _ S^\bullet )$ be as in Situation 48.20.1. Let $f : X \to Y$ be a proper morphism of schemes of finite type over $S$. Let $\omega _ X^\bullet$ and $\omega _ Y^\bullet$ be dualizing complexes normalized relative to $\omega _ S^\bullet$. Let $a$ be the right adjoint of Lemma 48.3.1 for $f$. Then there is a canonical isomorphism $a(\omega _ Y^\bullet ) = \omega _ X^\bullet$.

Proof. Let $p : X \to S$ and $q : Y \to S$ be the structure morphisms. If $X$ and $Y$ are separated over $S$, then this follows from the fact that $\omega _ X^\bullet = p^!\omega _ S^\bullet$, $\omega _ Y^\bullet = q^!\omega _ S^\bullet$, $f^! = a$, and $f^! \circ q^! = p^!$ (Lemma 48.16.3). In the general case we first use Lemma 48.20.6 to reduce to the case $Y = S$. In this case $X$ and $Y$ are separated over $S$ and we've just seen the result. $\square$

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