Lemma 20.9.2 (Mayer-Vietoris). Let $X$ be a ringed space. Suppose that $X = U \cup V$ is a union of two open subsets. For every $\mathcal{O}_ X$-module $\mathcal{F}$ there exists a long exact cohomology sequence

$0 \to H^0(X, \mathcal{F}) \to H^0(U, \mathcal{F}) \oplus H^0(V, \mathcal{F}) \to H^0(U \cap V, \mathcal{F}) \to H^1(X, \mathcal{F}) \to \ldots$

This long exact sequence is functorial in $\mathcal{F}$.

Proof. The sheaf condition says that the kernel of $(1, -1) : \mathcal{F}(U) \oplus \mathcal{F}(V) \to \mathcal{F}(U \cap V)$ is equal to the image of $\mathcal{F}(X)$ by the first map for any abelian sheaf $\mathcal{F}$. Lemma 20.9.1 above implies that the map $(1, -1) : \mathcal{I}(U) \oplus \mathcal{I}(V) \to \mathcal{I}(U \cap V)$ is surjective whenever $\mathcal{I}$ is an injective $\mathcal{O}_ X$-module. Hence if $\mathcal{F} \to \mathcal{I}^\bullet$ is an injective resolution of $\mathcal{F}$, then we get a short exact sequence of complexes

$0 \to \mathcal{I}^\bullet (X) \to \mathcal{I}^\bullet (U) \oplus \mathcal{I}^\bullet (V) \to \mathcal{I}^\bullet (U \cap V) \to 0.$

Taking cohomology gives the result (use Homology, Lemma 12.12.12). We omit the proof of the functoriality of the sequence. $\square$

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