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 $H^0(L) = \mathcal{O}_ X$. $\square$

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