The Stacks project

Proof. Let $a \leq b$ be such that $H^ i(K) = 0$ for $i \not\in [a, b]$. By induction on $b - a$ we will prove there exists a covering $\{ U_ i \to X\}$ such that $K|_{U_ i}$ can be represented by a complex $\underline{M^\bullet }_{U_ i}$ with $M^ p$ a finite type $\Lambda$-module and $M^ p = 0$ for $p \not\in [a, b]$. If $b = a$, then this is clear. In general, we may replace $X$ by the members of a covering and assume that $H^ b(K)$ is constant, say $H^ b(K) = \underline{M}$. By Modules on Sites, Lemma 18.42.5 the module $M$ is a finite $\Lambda$-module. Choose a surjection $\Lambda ^{\oplus r} \to M$ given by generators $x_1, \ldots , x_ r$ of $M$.

By a slight generalization of Lemma 21.7.3 (details omitted) there exists a covering $\{ U_ i \to X\}$ such that $x_ i \in H^0(X, H^ b(K))$ lifts to an element of $H^ b(U_ i, K)$. Thus, after replacing $X$ by the $U_ i$ we reach the situation where there is a map $\underline{\Lambda ^{\oplus r}}[-b] \to K$ inducing a surjection on cohomology sheaves in degree $b$. Choose a distinguished triangle

Now the cohomology sheaves of $L$ are nonzero only in the interval $[a, b - 1]$, agree with the cohomology sheaves of $K$ in the interval $[a, b - 2]$ and there is a short exact sequence

in degree $b - 1$. By Modules on Sites, Lemma 18.43.5 we see that $H^{b - 1}(L)$ is locally constant of finite type. By induction hypothesis we obtain an isomorphism $\underline{M^\bullet } \to L$ in $D(\mathcal{C}, \underline{\Lambda })$ with $M^ p$ a finite $\Lambda$-module and $M^ p = 0$ for $p \not\in [a, b - 1]$. The map $L \to \Lambda ^{\oplus r}[-b + 1]$ gives a map $\underline{M^{b - 1}} \to \underline{\Lambda ^{\oplus r}}$ which locally is constant (Modules on Sites, Lemma 18.43.3). Thus we may assume it is given by a map $M^{b - 1} \to \Lambda ^{\oplus r}$. The distinguished triangle shows that the composition $M^{b - 2} \to M^{b - 1} \to \Lambda ^{\oplus r}$ is zero and the axioms of triangulated categories produce an isomorphism

in $D(\mathcal{C}, \Lambda )$. $\square$