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

$Rf_*E \to Ra_*(E|_ U) \oplus Rb_*(E|_ V) \to Rc_*(E|_{U \cap V}) \to Rf_*E[1]$

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

$0 \to f_*\mathcal{I}^\bullet \to a_*\mathcal{I}^\bullet |_ U \oplus b_*\mathcal{I}^\bullet |_ V \to c_*\mathcal{I}^\bullet |_{U \cap V} \to 0.$

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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