Lemma 36.13.3. Let $X$ be an affine scheme and let $U \subset X$ be a quasi-compact open subscheme. For any pseudo-coherent object $E$ of $D(\mathcal{O}_ U)$ there exists a bounded above complex of finite free $\mathcal{O}_ X$-modules whose restriction to $U$ is isomorphic to $E$.

**Proof.**
By Lemma 36.10.1 we see that $E$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ U)$. By Lemma 36.13.1 we may assume $E = E'|U$ for some object $E'$ of $D_\mathit{QCoh}(\mathcal{O}_ X)$. Write $X = \mathop{\mathrm{Spec}}(A)$. By Lemma 36.3.5 we can find a complex $M^\bullet $ of $A$-modules whose associated complex of $\mathcal{O}_ X$-modules is a representative of $E'$.

Choose $f_1, \ldots , f_ r \in A$ such that $U = D(f_1) \cup \ldots \cup D(f_ r)$. By Lemma 36.10.2 the complexes $M^\bullet _{f_ j}$ are pseudo-coherent complexes of $A_{f_ j}$-modules. Let $n$ be an integer. Assume we have a map of complexes $\alpha : F^\bullet \to M^\bullet $ where $F^\bullet $ is bounded above, $F^ i = 0$ for $i < n$, each $F^ i$ is a finite free $R$-module, such that

is an isomorphism for $i > n$ and surjective for $i = n$. Picture

Since each $M^\bullet _{f_ j}$ has vanishing cohomology in large degrees we can find such a map for $n \gg 0$. By induction on $n$ we are going to extend this to a map of complexes $F^\bullet \to M^\bullet $ such that $H^ i(\alpha _{f_ j})$ is an isomorphism for all $i$. The lemma will follow by taking $\widetilde{F^\bullet }$.

The induction step will be to extend the diagram above by adding $F^{n - 1}$. Let $C^\bullet $ be the cone on $\alpha $ (Derived Categories, Definition 13.9.1). The long exact sequence of cohomology shows that $H^ i(C^\bullet _{f_ j}) = 0$ for $i \geq n$. By More on Algebra, Lemma 15.64.2 we see that $C^\bullet _{f_ j}$ is $(n - 1)$-pseudo-coherent. By More on Algebra, Lemma 15.64.3 we see that $H^{-1}(C^\bullet _{f_ j})$ is a finite $A_{f_ j}$-module. Choose a finite free $A$-module $F^{n - 1}$ and an $A$-module $\beta : F^{n - 1} \to C^{-1}$ such that the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ is zero and such that $F^{n - 1}_{f_ j}$ surjects onto $H^{n - 1}(C^\bullet _{f_ j})$. (Some details omitted; hint: clear denominators.) Since $C^{n - 1} = M^{n - 1} \oplus F^ n$ we can write $\beta = (\alpha ^{n - 1}, -d^{n - 1})$. The vanishing of the composition $F^{n - 1} \to C^{n - 1} \to C^ n$ implies these maps fit into a morphism of complexes

Moreover, these maps define a morphism of distinguished triangles

Hence our choice of $\beta $ implies that the map of complexes $(F^{-1} \to \ldots ) \to M^\bullet $ induces an isomorphism on cohomology localized at $f_ j$ in degrees $\geq n$ and a surjection in degree $-1$. This finishes the proof of the lemma. $\square$

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