The Stacks project

Proof. Let $\mathcal{C}' \subset \mathcal{C}$ be the full subcategory of objects $U$ such that $\mathcal{I}|_ U$ is generated by finitely many sections. Then $\mathcal{C}' \to \mathcal{C}$ is a special cocontinuous functor (Sites, Definition 7.29.2). Hence it suffices to work with $\mathcal{C}'$, see Sites, Lemma 7.29.1. In other words we may assume that for every object $U$ of $\mathcal{C}$ there exists a finitely generated ideal $I \subset \mathcal{I}(U)$ such that $\mathcal{I}|_ U = \mathop{\mathrm{Im}}(I \otimes \mathcal{O}_ U \to \mathcal{O}_ U)$. We will say that $I$ generates $\mathcal{I}|_ U$. Warning: We do not know that $\mathcal{I}(U)$ is a finitely generated ideal in $\mathcal{O}(U)$.

Let $U$ be an object and $I \subset \mathcal{O}(U)$ a finitely generated ideal which generates $\mathcal{I}|_ U$. On the category $\mathcal{C}/U$ consider the complex of presheaves

with $K(-, -)$ as in Lemma 52.6.10. We claim that the sheafification of this is independent of the choice of $I$. Indeed, if $I' \subset \mathcal{O}(U)$ is a finitely generated ideal which also generates $\mathcal{I}|_ U$, then there exists a covering $\{ U_ j \to U\}$ such that $I\mathcal{O}(U_ j) = I'\mathcal{O}(U_ j)$. (Hint: this works because both $I$ and $I'$ are finitely generated and generate $\mathcal{I}|_ U$.) Hence $K_{U, I}^\bullet$ and $K_{U, I'}^\bullet$ are the same for any object lying over one of the $U_ j$. The statement on sheafifications follows. Denote $K_ U^\bullet$ the common value.

The independence of choice of $I$ also shows that $K_ U^\bullet |_{\mathcal{C}/U'} = K_{U'}^\bullet$ whenever we are given a morphism $U' \to U$ and hence a localization morphism $\mathcal{C}/U' \to \mathcal{C}/U$. Thus the complexes $K_ U^\bullet$ glue to give a single well defined complex $K^\bullet$ of $\mathcal{O}$-modules. The existence of the map $K^\bullet \to \mathcal{O}$ and the quasi-isomorphism of the lemma follow immediately from the corresponding properties of the complexes $K(-, -)$ in Lemma 52.6.10. $\square$