The Stacks project

Proof. Let $\mathcal{F}^\bullet $ be any complex representing $E$ and let $X = \bigcup U_ i$ and $\alpha _ i : \mathcal{E}_ i^\bullet \to E|_{U_ i}$ be as in (1). We will show that $\mathcal{F}^\bullet $ is $m$-pseudo-coherent as a complex, which will prove (1) and (2) simultaneously. By Lemma 20.46.8 we can after refining the open covering $X = \bigcup U_ i$ represent the maps $\alpha _ i$ by maps of complexes $\alpha _ i : \mathcal{E}_ i^\bullet \to \mathcal{F}^\bullet |_{U_ i}$. By assumption $H^ j(\alpha _ i)$ are isomorphisms for $j > m$, and $H^ m(\alpha _ i)$ is surjective whence $\mathcal{F}^\bullet $ is $m$-pseudo-coherent. $\square$

Proof. Represent $E$ by a complex $\mathcal{E}^\bullet $ of $\mathcal{O}_ Y$-modules and choose an open covering $Y = \bigcup V_ i$ and $\alpha _ i : \mathcal{E}_ i^\bullet \to \mathcal{E}^\bullet |_{V_ i}$ as in Definition 20.47.1. Set $U_ i = f^{-1}(V_ i)$. By Lemma 20.47.2 it suffices to show that $Lf^*\mathcal{E}^\bullet |_{U_ i}$ is $m$-pseudo-coherent. Choose a distinguished triangle

The assumption on $\alpha _ i$ means exactly that the cohomology sheaves $H^ j(C)$ are zero for all $j \geq m$. Denote $f_ i : U_ i \to V_ i$ the restriction of $f$. Note that $Lf^*\mathcal{E}^\bullet |_{U_ i} = Lf_ i^*(\mathcal{E}|_{V_ i})$. Applying $Lf_ i^*$ we obtain the distinguished triangle

By the construction of $Lf_ i^*$ as a left derived functor we see that $H^ j(Lf_ i^*C) = 0$ for $j \geq m$ (by the dual of Derived Categories, Lemma 13.16.1). Hence $H^ j(Lf_ i^*\alpha _ i)$ is an isomorphism for $j > m$ and $H^ m(Lf^*\alpha _ i)$ is surjective. On the other hand, $Lf_ i^*\mathcal{E}_ i^\bullet = f_ i^*\mathcal{E}_ i^\bullet $. is strictly perfect by Lemma 20.46.4. Thus we conclude. $\square$

Proof. Proof of (1). Choose an open covering $X = \bigcup U_ i$ and maps $\alpha _ i : \mathcal{K}_ i^\bullet \to K|_{U_ i}$ in $D(\mathcal{O}_{U_ i})$ with $\mathcal{K}_ i^\bullet $ strictly perfect and $H^ j(\alpha _ i)$ isomorphisms for $j > m + 1$ and surjective for $j = m + 1$. We may replace $\mathcal{K}_ i^\bullet $ by $\sigma _{\geq m + 1}\mathcal{K}_ i^\bullet $ and hence we may assume that $\mathcal{K}_ i^ j = 0$ for $j < m + 1$. After refining the open covering we may choose maps $\beta _ i : \mathcal{L}_ i^\bullet \to L|_{U_ i}$ in $D(\mathcal{O}_{U_ i})$ with $\mathcal{L}_ i^\bullet $ strictly perfect such that $H^ j(\beta )$ is an isomorphism for $j > m$ and surjective for $j = m$. By Lemma 20.46.7 we can, after refining the covering, find maps of complexes $\gamma _ i : \mathcal{K}^\bullet \to \mathcal{L}^\bullet $ such that the diagrams

are commutative in $D(\mathcal{O}_{U_ i})$ (this requires representing the maps $\alpha _ i$, $\beta _ i$ and $K|_{U_ i} \to L|_{U_ i}$ by actual maps of complexes; some details omitted). The cone $C(\gamma _ i)^\bullet $ is strictly perfect (Lemma 20.46.2). The commutativity of the diagram implies that there exists a morphism of distinguished triangles

It follows from the induced map on long exact cohomology sequences and Homology, Lemmas 12.5.19 and 12.5.20 that $C(\gamma _ i)^\bullet \to M|_{U_ i}$ induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. Hence $M$ is $m$-pseudo-coherent by Lemma 20.47.2.

Assertions (2) and (3) follow from (1) by rotating the distinguished triangle. $\square$

Proof. Proof of (1). By replacing $X$ by the members of an open covering we may assume there exist strictly perfect complexes $\mathcal{K}^\bullet $ and $\mathcal{L}^\bullet $ and maps $\alpha : \mathcal{K}^\bullet \to K$ and $\beta : \mathcal{L}^\bullet \to L$ with $H^ i(\alpha )$ and isomorphism for $i > n$ and surjective for $i = n$ and with $H^ i(\beta )$ and isomorphism for $i > m$ and surjective for $i = m$. Then the map

induces isomorphisms on cohomology sheaves in degree $i$ for $i > t$ and a surjection for $i = t$. This follows from the spectral sequence of tors (details omitted).

Proof of (2). We may first replace $X$ by the members of an open covering to reduce to the case that $K$ and $L$ are bounded above. Then the statement follows immediately from case (1). $\square$

Proof. Assume that $K \oplus L$ is $m$-pseudo-coherent. After replacing $X$ by the members of an open covering we may assume $K \oplus L \in D^-(\mathcal{O}_ X)$, hence $L \in D^-(\mathcal{O}_ X)$. Note that there is a distinguished triangle

see Derived Categories, Lemma 13.4.10. By Lemma 20.47.4 we see that $L \oplus L[1]$ is $m$-pseudo-coherent. Hence also $L[1] \oplus L[2]$ is $m$-pseudo-coherent. By induction $L[n] \oplus L[n + 1]$ is $m$-pseudo-coherent. Since $L$ is bounded above we see that $L[n]$ is $m$-pseudo-coherent for large $n$. Hence working backwards, using the distinguished triangles

we conclude that $L[n - 1], L[n - 2], \ldots , L$ are $m$-pseudo-coherent as desired. $\square$

Proof. Omitted. Hint: use Lemma 20.47.4 and truncations as in the proof of More on Algebra, Lemma 15.64.9. $\square$

Proof. Omitted. Hint: use Lemma 20.47.4 and truncations as in the proof of More on Algebra, Lemma 15.64.10. $\square$

Proof. Proof of (1). We may work locally on $X$. Hence we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to K$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. It suffices to prove the result for $\mathcal{E}^\bullet $. Let $n$ be the largest integer such that $\mathcal{E}^ n \not= 0$. If $n = m$, then $H^ m(\mathcal{E}^\bullet )$ is a quotient of $\mathcal{E}^ n$ and the result is clear. If $n > m$, then $\mathcal{E}^{n - 1} \to \mathcal{E}^ n$ is surjective as $H^ n(E^\bullet ) = 0$. By Lemma 20.46.5 we can locally find a section of this surjection and write $\mathcal{E}^{n - 1} = \mathcal{E}' \oplus \mathcal{E}^ n$. Hence it suffices to prove the result for the complex $(\mathcal{E}')^\bullet $ which is the same as $\mathcal{E}^\bullet $ except has $\mathcal{E}'$ in degree $n - 1$ and $0$ in degree $n$. We win by induction on $n$.

Proof of (2). We may work locally on $X$. Hence we may assume there exists a strictly perfect complex $\mathcal{E}^\bullet $ and a map $\alpha : \mathcal{E}^\bullet \to K$ which induces an isomorphism on cohomology in degrees $> m$ and a surjection in degree $m$. As in the proof of (1) we can reduce to the case that $\mathcal{E}^ i = 0$ for $i > m + 1$. Then we see that $H^{m + 1}(K) \cong H^{m + 1}(\mathcal{E}^\bullet ) = \mathop{\mathrm{Coker}}(\mathcal{E}^ m \to \mathcal{E}^{m + 1})$ which is of finite presentation. $\square$

Proof. Use Lemma 20.47.9 to prove the implications in one direction and Lemma 20.47.8 for the other. $\square$