Lemma 48.20.3. In Situation 48.20.1 let $X$ be a scheme of finite type over $S$ and let $\mathcal{U}$, $\mathcal{V}$ be two finite open coverings of $X$ by schemes separated over $S$. If there exists a dualizing complex normalized relative to $\omega _ S^\bullet$ and $\mathcal{U}$, then there exists a dualizing complex normalized relative to $\omega _ S^\bullet$ and $\mathcal{V}$ and these complexes are canonically isomorphic.

Proof. It suffices to prove this when $\mathcal{U}$ is given by the opens $U_1, \ldots , U_ n$ and $\mathcal{V}$ by the opens $U_1, \ldots , U_{n + m}$. In fact, we may and do even assume $m = 1$. To go from a dualizing complex $(K, \alpha _ i)$ normalized relative to $\omega _ S^\bullet$ and $\mathcal{V}$ to a dualizing complex normalized relative to $\omega _ S^\bullet$ and $\mathcal{U}$ is achieved by forgetting about $\alpha _ i$ for $i = n + 1$. Conversely, let $(K, \alpha _ i)$ be a dualizing complex normalized relative to $\omega _ S^\bullet$ and $\mathcal{U}$. To finish the proof we need to construct a map $\alpha _{n + 1} : K|_{U_{n + 1}} \to \omega _{n + 1}^\bullet$ satisfying the desired conditions. To do this we observe that $U_{n + 1} = \bigcup U_ i \cap U_{n + 1}$ is an open covering. It is clear that $(K|_{U_{n + 1}}, \alpha _ i|_{U_ i \cap U_{n + 1}})$ is a dualizing complex normalized relative to $\omega _ S^\bullet$ and the covering $U_{n + 1} = \bigcup U_ i \cap U_{n + 1}$. On the other hand, by condition (3) the pair $(\omega _{n + 1}^\bullet |_{U_{n + 1}}, \varphi _{n + 1i})$ is another dualizing complex normalized relative to $\omega _ S^\bullet$ and the covering $U_{n + 1} = \bigcup U_ i \cap U_{n + 1}$. By Lemma 48.20.2 we obtain a unique isomorphism

$\alpha _{n + 1} : K|_{U_{n + 1}} \longrightarrow \omega _{n + 1}^\bullet$

compatible with the given local isomorphisms. It is a pleasant exercise to show that this means it satisfies the required property. $\square$

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