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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