The Stacks project

Proof. Let $x \in X$. It suffices to find an open neighbourhood $U$ of $x$ such that $M$ restricts to a perfect complex over $U$. Hence during the proof we can (finitely often) replace $X$ by an open neighbourhood of $x$. Let $N, \eta , \epsilon $ be a left dual.

We are going to use the following argument several times. Choose any complex $\mathcal{M}^\bullet $ of $\mathcal{O}_ X$-modules representing $M$. Choose a K-flat complex $\mathcal{N}^\bullet $ representing $N$ whose terms are flat $\mathcal{O}_ X$-modules, see Lemma 20.26.12. Consider the map

After shrinking $X$ we can find an integer $N$ and for $i = 1, \ldots , N$ integers $n_ i \in \mathbf{Z}$ and sections $f_ i$ and $g_ i$ of $\mathcal{M}^{n_ i}$ and $\mathcal{N}^{-n_ i}$ such that

Let $\mathcal{K}^\bullet \subset \mathcal{M}^\bullet $ be any subcomplex of $\mathcal{O}_ X$-modules containing the sections $f_ i$ for $i = 1, \ldots , N$. Since $\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{N}^\bullet ) \subset \text{Tot}(\mathcal{M}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{N}^\bullet )$ by flatness of the modules $\mathcal{N}^ n$, we see that $\eta $ factors through

Denoting $K$ the object of $D(\mathcal{O}_ X)$ represented by $\mathcal{K}^\bullet $ we find a commutative diagram

Since the composition of the upper row is the identity on $M$ we conclude that $M$ is a direct summand of $K$ in $D(\mathcal{O}_ X)$.

As a first use of the argument above, we can choose the subcomplex $\mathcal{K}^\bullet = \sigma _{\geq a} \tau _{\leq b}\mathcal{M}^\bullet $ with $a < n_ i < b$ for $i = 1, \ldots , N$. Thus $M$ is a direct summand in $D(\mathcal{O}_ X)$ of a bounded complex and we conclude we may assume $M$ is in $D^ b(\mathcal{O}_ X)$. (Recall that the process above involves shrinking $X$.)

Since $M$ is in $D^ b(\mathcal{O}_ X)$ we may choose $\mathcal{M}^\bullet $ to be a bounded above complex of flat modules (by Modules, Lemma 17.17.6 and Derived Categories, Lemma 13.15.4). Then we can choose $\mathcal{K}^\bullet = \sigma _{\geq a}\mathcal{M}^\bullet $ with $a < n_ i$ for $i = 1, \ldots , N$ in the argument above. Thus we find that we may assume $M$ is a direct summand in $D(\mathcal{O}_ X)$ of a bounded complex of flat modules. In particular, $M$ has finite tor amplitude.

Say $M$ has tor amplitude in $[a, b]$. Assuming $M$ is $m$-pseudo-coherent we are going to show that (after shrinking $X$) we may assume $M$ is $(m - 1)$-pseudo-coherent. This will finish the proof by Lemma 20.49.4 and the fact that $M$ is $(b + 1)$-pseudo-coherent in any case. After shrinking $X$ we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to M$ in $D(\mathcal{O}_ X)$ such that $H^ i(\alpha )$ is an isomorphism for $i > m$ and surjective for $i = m$. We may and do assume that $\mathcal{E}^ i = 0$ for $i < m$. Choose a distinguished triangle

Observe that $H^ i(L) = 0$ for $i \geq m$. Thus we may represent $L$ by a complex $\mathcal{L}^\bullet $ with $\mathcal{L}^ i = 0$ for $i \geq m$. The map $L \to \mathcal{E}^\bullet [1]$ is given by a map of complexes $\mathcal{L}^\bullet \to \mathcal{E}^\bullet [1]$ which is zero in all degrees except in degree $m - 1$ where we obtain a map $\mathcal{L}^{m - 1} \to \mathcal{E}^ m$, see Derived Categories, Lemma 13.27.3. Then $M$ is represented by the complex

Apply the discussion in the second paragraph to this complex to get sections $f_ i$ of $\mathcal{M}^{n_ i}$ for $i = 1, \ldots , N$. For $n < m$ let $\mathcal{K}^ n \subset \mathcal{L}^ n$ be the $\mathcal{O}_ X$-submodule generated by the sections $f_ i$ for $n_ i = n$ and $d(f_ i)$ for $n_ i = n - 1$. For $n \geq m$ set $\mathcal{K}^ n = \mathcal{E}^ n$. Clearly, we have a morphism of distinguished triangles

where all the morphisms are as indicated above. Denote $K$ the object of $D(\mathcal{O}_ X)$ corresponding to the complex $\mathcal{K}^\bullet $. By the arguments in the second paragraph of the proof we obtain a morphism $s : M \to K$ in $D(\mathcal{O}_ X)$ such that the composition $M \to K \to M$ is the identity on $M$. We don't know that the diagram

commutes, but we do know it commutes after composing with the map $K \to M$. By Lemma 20.46.8 after shrinking $X$ we may assume that $s \circ i$ is given by a map of complexes $\sigma : \mathcal{E}^\bullet \to \mathcal{K}^\bullet $. By the same lemma we may assume the composition of $\sigma $ with the inclusion $\mathcal{K}^\bullet \subset \mathcal{M}^\bullet $ is homotopic to zero by some homotopy $\{ h^ i : \mathcal{E}^ i \to \mathcal{M}^{i - 1}\} $. Thus, after replacing $\mathcal{K}^{m - 1}$ by $\mathcal{K}^{m - 1} + \mathop{\mathrm{Im}}(h^ m)$ (note that after doing this it is still the case that $\mathcal{K}^{m - 1}$ is generated by finitely many global sections), we see that $\sigma $ itself is homotopic to zero! This means that we have a commutative solid diagram

By the axioms of triangulated categories we obtain a dotted arrow fitting into the diagram. Looking at cohomology sheaves in degree $m - 1$ we see that we obtain

Since the vertical compositions are the identity in both the left and right column, we conclude the vertical composition $H^{m - 1}(\mathcal{L}^\bullet ) \to H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \to H^{m - 1}(\mathcal{L}^\bullet )$ in the middle is surjective! In particular $H^{m - 1}(\sigma _{\leq m - 1}\mathcal{K}^\bullet ) \to H^{m - 1}(\mathcal{L}^\bullet )$ is surjective. Using the induced map of long exact sequences of cohomology sheaves from the morphism of triangles above, a diagram chase shows this implies $H^ i(K) \to H^ i(M)$ is an isomorphism for $i \geq m$ and surjective for $i = m - 1$. By construction we can choose an $r \geq 0$ and a surjection $\mathcal{O}_ X^{\oplus r} \to \mathcal{K}^{m - 1}$. Then the composition

induces an isomorphism on cohomology sheaves in degrees $\geq m$ and a surjection in degree $m - 1$ and the proof is complete. $\square$