Lemma 48.20.2. In Situation 48.20.1 let X be a scheme of finite type over S and let \mathcal{U} be a finite open covering of X by schemes separated over S. If there exists a dualizing complex normalized relative to \omega _ S^\bullet and \mathcal{U}, then it is unique up to unique isomorphism.
Proof. If (K, \alpha _ i) and (K', \alpha _ i') are two, then we consider L = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(K, K'). By Lemma 48.2.6 and its proof, this is an invertible object of D(\mathcal{O}_ X). Using \alpha _ i and \alpha '_ i we obtain an isomorphism
\alpha _ i^ t \otimes \alpha '_ i : L|_{U_ i} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\omega _ i^\bullet , \omega _ i^\bullet ) = \mathcal{O}_{U_ i}[0]
This already implies that L = H^0(L)[0] in D(\mathcal{O}_ X). Moreover, H^0(L) is an invertible sheaf with given trivializations on the opens U_ i of X. Finally, the condition that \alpha _ j|_{U_ i \cap U_ j} \circ \alpha _ i^{-1}|_{U_ i \cap U_ j} and \alpha '_ j|_{U_ i \cap U_ j} \circ (\alpha '_ i)^{-1}|_{U_ i \cap U_ j} both give \varphi _{ij} implies that the transition maps are 1 and we get an isomorphism H^0(L) = \mathcal{O}_ X. \square
Comments (0)