Lemma 20.33.5. Let $f : X \to Y$ be a morphism of ringed spaces. Suppose that $X = U \cup V$ is a union of two open subsets. Denote $a = f|_ U : U \to Y$, $b = f|_ V : V \to Y$, and $c = f|_{U \cap V} : U \cap V \to Y$. For every object $E$ of $D(\mathcal{O}_ X)$ there exists a distinguished triangle
This triangle is functorial in $E$.
Proof. Choose a K-injective complex $\mathcal{I}^\bullet $ representing $E$. We may assume $\mathcal{I}^ n$ is an injective object of $\textit{Mod}(\mathcal{O}_ X)$ for all $n$, see Injectives, Theorem 19.12.6. Then $Rf_*E$ is computed by $f_*\mathcal{I}^\bullet $. Similarly for $U$, $V$, and $U \cap V$ by Lemma 20.32.1. Hence the distinguished triangle of the lemma is the distinguished triangle associated (by Derived Categories, Section 13.12 and especially Lemma 13.12.1) to the short exact sequence of complexes
This is a short exact sequence of complexes by Lemma 20.8.3 and the fact that $R^1f_*\mathcal{I} = 0$ for an injective object $\mathcal{I}$ of $\textit{Mod}(\mathcal{O}_ X)$. The final statement follows from the functoriality of the construction in Injectives, Theorem 19.12.6. $\square$
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