Lemma 48.20.6. Let $(S, \omega _ S^\bullet )$ be as in Situation 48.20.1. Let $f : X \to Y$ be a morphism of finite type schemes over $S$. Let $\omega _ X^\bullet $ and $\omega _ Y^\bullet $ be dualizing complexes normalized relative to $\omega _ S^\bullet $. Then $\omega _ X^\bullet $ is a dualizing complex normalized relative to $\omega _ Y^\bullet $.
Proof. This is just a matter of bookkeeping. Choose a finite affine open covering $\mathcal{V} : Y = \bigcup V_ j$. For each $j$ choose a finite affine open covering $f^{-1}(V_ j) = U_{ji}$. Set $\mathcal{U} : X = \bigcup U_{ji}$. The schemes $V_ j$ and $U_{ji}$ are separated over $S$, hence we have the upper shriek functors for $q_ j : V_ j \to S$, $p_{ji} : U_{ji} \to S$ and $f_{ji} : U_{ji} \to V_ j$ and $f_{ji}' : U_{ji} \to Y$. Let $(L, \beta _ j)$ be a dualizing complex normalized relative to $\omega _ S^\bullet $ and $\mathcal{V}$. Let $(K, \gamma _{ji})$ be a dualizing complex normalized relative to $\omega _ S^\bullet $ and $\mathcal{U}$. (In other words, $L = \omega _ Y^\bullet $ and $K = \omega _ X^\bullet $.) We can define
\[ \alpha _{ji} : K|_{U_{ji}} \xrightarrow {\gamma _{ji}} p_{ji}^!\omega _ S^\bullet = f_{ji}^!q_ j^!\omega _ S^\bullet \xrightarrow {f_{ji}^!\beta _ j^{-1}} f_{ji}^!(L|_{V_ j}) = (f_{ji}')^!(L) \]
To finish the proof we have to show that $\alpha _{ji}|_{U_{ji} \cap U_{j'i'}} \circ \alpha _{j'i'}^{-1}|_{U_{ji} \cap U_{j'i'}}$ is the canonical isomorphism $(f_{ji}')^!(L)|_{U_{ji} \cap U_{j'i'}} \to (f_{j'i'}')^!(L)|_{U_{ji} \cap U_{j'i'}}$. This is formal and we omit the details. $\square$
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