Lemma 48.20.4. 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 $S$. Then there exists a dualizing complex normalized relative to $\omega _ S^\bullet$ and $\mathcal{U}$.

Proof. Say $\mathcal{U} : X = \bigcup _{i = 1, \ldots , n} U_ i$. We prove the lemma by induction on $n$. The base case $n = 1$ is immediate. Assume $n > 1$. Set $X' = U_1 \cup \ldots \cup U_{n - 1}$ and let $(K', \{ \alpha '_ i\} _{i = 1, \ldots , n - 1})$ be a dualizing complex normalized relative to $\omega _ S^\bullet$ and $\mathcal{U}' : X' = \bigcup _{i = 1, \ldots , n - 1} U_ i$. It is clear that $(K'|_{X' \cap U_ n}, \alpha '_ i|_{U_ i \cap U_ n})$ is a dualizing complex normalized relative to $\omega _ S^\bullet$ and the covering $X' \cap U_ n = \bigcup _{i = 1, \ldots , n - 1} U_ i \cap U_ n$. On the other hand, by condition (3) the pair $(\omega _ n^\bullet |_{X' \cap U_ n}, \varphi _{ni})$ is another dualizing complex normalized relative to $\omega _ S^\bullet$ and the covering $X' \cap U_ n = \bigcup _{i = 1, \ldots , n - 1} U_ i \cap U_ n$. By Lemma 48.20.2 we obtain a unique isomorphism

$\epsilon : K'|_{X' \cap U_ n} \longrightarrow \omega _ i^\bullet |_{X' \cap U_ n}$

compatible with the given local isomorphisms. By Cohomology, Lemma 20.42.1 we obtain $K \in D(\mathcal{O}_ X)$ together with isomorphisms $\beta : K|_{X'} \to K'$ and $\gamma : K|_{U_ n} \to \omega _ n^\bullet$ such that $\epsilon = \gamma |_{X'\cap U_ n} \circ \beta |_{X' \cap U_ n}^{-1}$. Then we define

$\alpha _ i = \alpha '_ i \circ \beta |_{U_ i}, i = 1, \ldots , n - 1, \text{ and } \alpha _ n = \gamma$

We still need to verify that $\varphi _{ij}$ is given by $\alpha _ j|_{U_ i \cap U_ j} \circ \alpha _ i^{-1}|_{U_ i \cap U_ j}$. For $i, j \leq n - 1$ this follows from the corresponding condition for $\alpha _ i'$. For $i = j = n$ it is clear as well. If $i < j = n$, then we get

$\alpha _ n|_{U_ i \cap U_ n} \circ \alpha _ i^{-1}|_{U_ i \cap U_ n} = \gamma |_{U_ i \cap U_ n} \circ \beta ^{-1}|_{U_ i \cap U_ n} \circ (\alpha '_ i)^{-1}|_{U_ i \cap U_ n} = \epsilon |_{U_ i \cap U_ n} \circ (\alpha '_ i)^{-1}|_{U_ i \cap U_ n}$

This is equal to $\alpha _{in}$ exactly because $\epsilon$ is the unique map compatible with the maps $\alpha _ i'$ and $\alpha _{ni}$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).