
Lemma 67.10.5. Let $S$ be a scheme. Let $(U \subset X, V \to X)$ be an elementary distinguished square of algebraic spaces over $S$. For an object $E$ of $D(\mathcal{O}_ X)$ we have a distinguished triangle

$R\Gamma (X, E) \to R\Gamma (U, E) \oplus R\Gamma (V, E) \to R\Gamma (U \times _ X V, E) \to R\Gamma (X, E)[1]$

and in particular a long exact cohomology sequence

$\ldots \to H^ n(X, E) \to H^ n(U, E) \oplus H^ n(V, E) \to H^ n(U \times _ X V, E) \to H^{n + 1}(X, E) \to \ldots$

The construction of the distinguished triangle and the long exact sequence is functorial in $E$.

Proof. Choose a K-injective complex $\mathcal{I}^\bullet$ representing $E$ whose terms $\mathcal{I}^ n$ are injective objects of $\textit{Mod}(\mathcal{O}_ X)$, see Injectives, Theorem 19.12.6. In the proof of Lemma 67.10.2 we found a short exact sequence of complexes

$0 \to \mathcal{I}^\bullet \to j_{U, *}\mathcal{I}^\bullet |_ U \oplus j_{V, *}\mathcal{I}^\bullet |_ V \to j_{U \times _ X V, *}\mathcal{I}^\bullet |_{U \times _ X V} \to 0$

Since $H^1(X, \mathcal{I}^ n) = 0$, we see that taking global sections gives an exact sequence of complexes

$0 \to \Gamma (X, \mathcal{I}^\bullet ) \to \Gamma (U, \mathcal{I}^\bullet ) \oplus \Gamma (V, \mathcal{I}^\bullet ) \to \Gamma (U \times _ X V, \mathcal{I}^\bullet ) \to 0$

Since these complexes represent $R\Gamma (X, E)$, $R\Gamma (U, E)$, $R\Gamma (V, E)$, and $R\Gamma (U \times _ X V, E)$ we get a distinguished triangle by Derived Categories, Section 13.12 and especially Lemma 13.12.1. $\square$

Comment #2792 by on

There shouldn't be a 0 in the long exact cohomology sequence, rather an $n$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).