The Stacks project

Proof. Let $\mathcal{B}$ be the collection of affine opens of $X$ which map into an affine open of $S$. For each $U$ we have a relative dualizing complex $(K_ U, \xi _ U)$ for $U$ over $S$. Namely, choose an affine open $V \subset S$ such that $U \to X \to S$ factors through $V$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(R)$. By Dualizing Complexes, Lemma 47.27.4 there exists a relative dualizing complex $K_ A \in D(A)$ for $R \to A$. Arguing backwards through the proof of Lemma 48.28.2 this determines an $V$-perfect object $K_ U \in D(\mathcal{O}_ U)$ and a map

in $D(\mathcal{O}_{U \times _ V U})$. Since being $V$-perfect is the same as being $S$-perfect and since $U \times _ V U = U \times _ S U$ we find that $(K_ U, \xi _ U)$ is as desired.

If $U' \subset U \subset X$ with $U', U \in \mathcal{B}$, then we have a unique isomorphism $\rho _{U'}^ U : K_ U|_{U'} \to K_{U'}$ in $D(\mathcal{O}_{U'})$ sending $\xi _ U|_{U' \times _ S U'}$ to $\xi _{U'}$ by Lemma 48.28.4 (note that trivially the restriction of a relative dualizing complex to an open is a relative dualizing complex). The uniqueness guarantees that $\rho ^ U_{U''} = \rho ^ V_{U''} \circ \rho ^ U_{U'}|_{U''}$ for $U'' \subset U' \subset U$ in $\mathcal{B}$. Observe that $\text{Ext}^ i(K_ U, K_ U) = 0$ for $i < 0$ for $U \in \mathcal{B}$ by Lemma 48.28.3 applied to $U/S$ and $K_ U$. Thus the BBD glueing lemma (Cohomology, Theorem 20.45.8) tells us there is a unique solution, namely, an object $K \in D(\mathcal{O}_ X)$ and isomorphisms $\rho _ U : K|_ U \to K_ U$ such that we have $\rho ^ U_{U'} \circ \rho _ U|_{U'} = \rho _{U'}$ for all $U' \subset U$, $U, U' \in \mathcal{B}$.

To finish the proof we have to construct the map

in $D(\mathcal{O}_ W)$ inducing an isomorphism from $\Delta _*\mathcal{O}_ X$ to $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W)$. Since we may change $W$, we choose $W = \bigcup _{U \in \mathcal{B}} U \times _ S U$. We can use $\rho _ U$ to get isomorphisms

As $W$ is covered by the opens $U \times _ S U$ we conclude that the cohomology sheaves of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W)$ are zero except in degree $0$. Moreover, we obtain isomorphisms

Let $\tau _ U$ in the LHS be an element mapping to $\xi _ U$ under this map. The compatibilities between $\rho ^ U_{U'}$, $\xi _ U$, $\xi _{U'}$, $\rho _ U$, and $\rho _{U'}$ for $U' \subset U \subset X$ open $U', U \in \mathcal{B}$ imply that $\tau _ U|_{U' \times _ S U'} = \tau _{U'}$. Thus we get a global section $\tau$ of the $0$th cohomology sheaf $H^0(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W))$. Since the other cohomology sheaves of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ W}(\Delta _*\mathcal{O}_ X, L\text{pr}_1^*K|_ W)$ are zero, this global section $\tau$ determines a morphism $\xi$ as desired. Since the restriction of $\xi$ to $U \times _ S U$ gives $\xi _ U$, we see that it satisfies the final condition of Definition 48.28.1. $\square$