The Stacks project

Proof. After replacing $X$ by the members of an open covering we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to E$ such that $H^ i(\alpha )$ is an isomorphism for $i \geq a$. We may and do replace $\mathcal{E}^\bullet $ by $\sigma _{\geq a - 1}\mathcal{E}^\bullet $. Choose a distinguished triangle

From the vanishing of cohomology sheaves of $E$ and $\mathcal{E}^\bullet $ and the assumption on $\alpha $ we obtain $C \cong \mathcal{K}[a - 2]$ with $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Applying $- \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}$ the assumption that $E$ has tor amplitude in $[a, b]$ implies $\mathcal{K} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{E}^{a - 1} \otimes _{\mathcal{O}_ X} \mathcal{F}$ has image $\mathop{\mathrm{Ker}}(\mathcal{E}^{a - 1} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{E}^ a \otimes _{\mathcal{O}_ X} \mathcal{F})$. It follows that $\text{Tor}_1^{\mathcal{O}_ X}(\mathcal{E}', \mathcal{F}) = 0$ where $\mathcal{E}' = \mathop{\mathrm{Coker}}(\mathcal{E}^{a - 1} \to \mathcal{E}^ a)$. Hence $\mathcal{E}'$ is flat (Lemma 20.26.16). Thus $\mathcal{E}'$ is locally a direct summand of a finite free module by Modules, Lemma 17.18.3. Thus locally the complex

is quasi-isomorphic to $E$ and $E$ is perfect. $\square$