Proof.
Let X = \bigcup U_ i be a finite affine open covering. Set U = \coprod U_ i and denote j : U \to X the morphism inducing the given open immersions U_ i \to X. Since U is an affine scheme and X has affine diagonal, the morphism j is affine, see Morphisms, Lemma 29.11.11. For every \mathcal{O}_ X-module \mathcal{F} there is a canonical map \mathcal{F} \to j_*j^*\mathcal{F}. This map is injective as can be seen by checking on stalks: if x \in U_ i, then we have a factorization
\mathcal{F}_ x \to (j_*j^*\mathcal{F})_ x \to (j^*\mathcal{F})_{x'} = \mathcal{F}_ x
where x' \in U is the point x viewed as a point of U_ i \subset U. Now if \mathcal{F} is quasi-coherent, then j^*\mathcal{F} is quasi-coherent on the affine scheme U hence has vanishing higher cohomology by Lemma 30.2.2. Then H^ p(X, j_*j^*\mathcal{F}) = 0 for p > 0 by Lemma 30.2.4 as j is affine. This proves (1). Finally, we see that the map H^ p(X, \mathcal{F}) \to H^ p(X, j_*j^*\mathcal{F}) is zero and part (2) follows from Homology, Lemma 12.12.4.
\square
Comments (4)
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