Lemma 36.9.6. In Situation 36.9.1. Let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Assume that $H^ i(E)|_ U = 0$ for $i = - r + 1, \ldots , 0$. Then given $s \in H^0(X, E)$ there exists an $e \geq 0$ and a morphism $K_ e \to E$ such that $s$ is in the image of $H^0(X, K_ e) \to H^0(X, E)$.
Proof. Since $U$ is covered by $r$ affine opens we have $H^ j(U, \mathcal{F}) = 0$ for $j \geq r$ and any quasi-coherent module (Cohomology of Schemes, Lemma 30.4.2). By Lemma 36.3.4 we see that $H^0(U, E)$ is equal to $H^0(U, \tau _{\geq -r + 1}E)$. There is a spectral sequence
\[ H^ j(U, H^ i(\tau _{\geq -r + 1}E)) \Rightarrow H^{i + j}(U, \tau _{\geq -N}E) \]
see Derived Categories, Lemma 13.21.3. Hence $H^0(U, E) = 0$ by our assumed vanishing of cohomology sheaves of $E$. We conclude that $s|_ U = 0$. Think of $s$ as a morphism $\mathcal{O}_ X \to E$ in $D(\mathcal{O}_ X)$. By Proposition 36.9.5 the composition $I_ e \to \mathcal{O}_ X \to E$ is zero for some $e$. By the distinguished triangle $I_ e \to \mathcal{O}_ X \to K_ e \to I_ e[1]$ we obtain a morphism $K_ e \to E$ such that $s$ is the composition $\mathcal{O}_ X \to K_ e \to E$. $\square$
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