Lemma 75.10.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $(U \subset X, V \to X)$ be an elementary distinguished square. Denote $a = f|_ U : U \to Y$, $b = f|_ V : V \to Y$, and $c = f|_{U \times _ X V} : U \times _ X V \to Y$ the restrictions. For every object $E$ of $D(\mathcal{O}_ X)$ there exists a distinguished triangle
\[ Rf_*E \to Ra_*(E|_ U) \oplus Rb_*(E|_ V) \to Rc_*(E|_{U \times _ X V}) \to Rf_*E[1] \]
in $D(\mathcal{O}_ Y)$. 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 Cohomology on Sites, Lemma 21.20.1. 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
\[ 0 \to f_*\mathcal{I}^\bullet \to a_*\mathcal{I}^\bullet |_ U \oplus b_*\mathcal{I}^\bullet |_ V \to c_*\mathcal{I}^\bullet |_{U \times _ X V} \to 0. \]
To see this is a short exact sequence of complexes we argue as follows. Pick an injective object $\mathcal{I}$ of $\textit{Mod}(\mathcal{O}_ X)$. Apply $f_*$ to the short exact sequence
\[ 0 \to \mathcal{I} \to j_{U, *}\mathcal{I}|_ U \oplus j_{V, *}\mathcal{I}|_ V \to j_{U \times _ X V, *}\mathcal{I}|_{U \times _ X V} \to 0 \]
of Lemma 75.10.2 and use that $R^1f_*\mathcal{I} = 0$ to get a short exact sequence
\[ 0 \to f_*\mathcal{I} \to f_*j_{U, *}\mathcal{I}|_ U \oplus f_*j_{V, *}\mathcal{I}|_ V \to f_*j_{U \times _ X V, *}\mathcal{I}|_{U \times _ X V} \to 0 \]
The proof is finished by observing that $a_* = f_*j_{U, *}$ and similarly for $b_*$ and $c_*$.
$\square$
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