The Stacks project

Lemma 75.10.1. Let $S$ be a scheme. Let $(U \subset X, V \to X)$ be an elementary distinguished square of algebraic spaces over $S$.

  1. For a sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$ we have a short exact sequence

    \[ 0 \to j_{U \times _ X V!}\mathcal{F}|_{U \times _ X V} \to j_{U!}\mathcal{F}|_ U \oplus j_{V!}\mathcal{F}|_ V \to \mathcal{F} \to 0 \]

  2. For an object $E$ of $D(\mathcal{O}_ X)$ we have a distinguished triangle

    \[ j_{U \times _ X V!}E|_{U \times _ X V} \to j_{U!}E|_ U \oplus j_{V!}E|_ V \to E \to j_{U \times _ X V!}E|_{U \times _ X V}[1] \]

    in $D(\mathcal{O}_ X)$.

Proof. To show the sequence of (1) is exact we may check on stalks at geometric points by Properties of Spaces, Theorem 66.19.12. Let $\overline{x}$ be a geometric point of $X$. By Equations (75.10.0.1) and (75.10.0.2) taking stalks at $\overline{x}$ we obtain the sequence

\[ 0 \to \bigoplus \nolimits _{(\overline{u}, \overline{v})} \mathcal{F}_{\overline{x}} \to \bigoplus \nolimits _{\overline{u}} \mathcal{F}_{\overline{x}} \oplus \bigoplus \nolimits _{\overline{v}} \mathcal{F}_{\overline{x}} \to \mathcal{F}_{\overline{x}} \to 0 \]

This sequence is exact because for every $\overline{x}$ there either is exactly one $\overline{u}$ mapping to $\overline{x}$, or there is no $\overline{u}$ and exactly one $\overline{v}$ mapping to $\overline{x}$.

Proof of (2). We have seen in Cohomology on Sites, Section 21.20 that the restriction functors and the extension by zero functors on derived categories are computed by just applying the functor to any complex. Let $\mathcal{E}^\bullet $ be a complex of $\mathcal{O}_ X$-modules representing $E$. 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 of $\mathcal{O}_ X$-modules

\[ 0 \to j_{U \times _ X V!}\mathcal{E}^\bullet |_{U \times _ X V} \to j_{U!}\mathcal{E}^\bullet |_ U \oplus j_{V!}\mathcal{E}^\bullet |_ V \to \mathcal{E}^\bullet \to 0 \]

which is short exact by (1). $\square$

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