**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 (2)

Comment #1843 by Keenan Kidwell on

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