The Stacks project

Proof. Identical to the proof of Lemma 20.43.2. $\square$

Proof. This follows from Lemma 20.45.2 and the fact that on $X$ every direct summand of a finite free module is finite locally free. See Modules, Lemma 17.14.6. $\square$

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 20.26.15). Thus $\mathcal{E}'$ is locally a direct summand of a finite free module by Modules, Lemma 17.16.11. Thus locally the complex

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

Proof. Assume (1). By definition this means there exists an open covering $X = \bigcup U_ i$ such that $E|_{U_ i}$ is represented by a strictly perfect complex. Thus $E$ is pseudo-coherent (i.e., $m$-pseudo-coherent for all $m$) by Lemma 20.43.2. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_ i}$ has finite Tor dimension by Lemma 20.44.3. Thus (2) holds.

Assume (2). After replacing $X$ by the members of an open covering we may assume there exist integers $a \leq b$ such that $E$ has tor amplitude in $[a, b]$. Since $E$ is $m$-pseudo-coherent for all $m$ we conclude using Lemma 20.45.4. $\square$

Proof. This follows from Lemma 20.45.5, 20.44.4, and 20.43.3. (An alternative proof is to copy the proof of Lemma 20.43.3.) $\square$

Proof. First proof: Combine Lemmas 20.45.5, 20.43.4, and 20.44.6. Second proof (sketch): Say $K$ and $L$ are perfect. After replacing $X$ by the members of an open covering we may assume that $K$ and $L$ are represented by strictly perfect complexes $\mathcal{K}^\bullet $ and $\mathcal{L}^\bullet $. After replacing $X$ by the members of an open covering we may assume the map $K \to L$ is given by a map of complexes $\alpha : \mathcal{K}^\bullet \to \mathcal{L}^\bullet $, see Lemma 20.42.8. Then $M$ is isomorphic to the cone of $\alpha $ which is strictly perfect by Lemma 20.42.2. $\square$

Proof. Follows from Lemmas 20.45.5, 20.43.5, and 20.44.7. $\square$

Proof. Follows from Lemmas 20.45.5, 20.43.6, and 20.44.8. $\square$

Proof. Being a perfect complex is local on $X$. Thus it suffices to check that $Rj_*E$ is perfect when restricted to $U$ and $V = X \setminus j(T)$. We have $Rj_*E|_ U = E$ which is perfect. We have $Rj_*E|_ V = 0$ because $E|_{U \setminus T} = 0$. $\square$

Proof. We will use without further mention that formation of internal hom commutes with restriction to opens (Lemma 20.38.3). In particular we may check the first two statements locally on $X$. By Lemma 20.38.9 to see the final statement it suffices to check that the map (20.38.9.1)

is an isomorphism. This is local on $X$ as well. Hence it suffices to prove the lemma when $K$ is represented by a strictly perfect complex.

Assume $K$ is represented by the strictly perfect complex $\mathcal{E}^\bullet $. Then it follows from Lemma 20.42.9 that $K^\vee $ is represented by the complex whose terms are $(\mathcal{E}^{-n})^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{E}^{-n}, \mathcal{O}_ X)$ in degree $n$. Since $\mathcal{E}^{-n}$ is a direct summand of a finite free $\mathcal{O}_ X$-module, so is $(\mathcal{E}^{-n})^\vee $. Hence $K^\vee $ is represented by a strictly perfect complex too. It is also clear that $(K^\vee )^\vee = K$ as we have $((\mathcal{E}^{-n})^\vee )^\vee = \mathcal{E}^{-n}$. To see that (20.38.9.1) is an isomorphism, represent $M$ by a K-flat complex $\mathcal{F}^\bullet $. By Lemma 20.42.9 the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, M)$ is represented by the complex with terms

On the other hand, then object $K^\vee \otimes ^\mathbf {L} M$ is represented by the complex with terms

Thus the assertion that (20.38.9.1) is an isomorphism reduces to the assertion that the canonical map

is an isomorphism when $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_ X$-module and $\mathcal{F}$ is any $\mathcal{O}_ X$-module. This follows immediately from the corresponding statement when $\mathcal{E}$ is finite free. $\square$

Proof. By Lemma 20.45.11 we have $R\mathop{\mathrm{lim}}\nolimits E \otimes ^\mathbf {L}_{\mathcal{O}_ X} K_ n^\vee = R\mathop{\mathrm{lim}}\nolimits R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E)$ which fits into the distinguished triangle

Because $K$ similarly fits into the distinguished triangle $\bigoplus K_ n \to \bigoplus K_ n \to K$ it suffices to show that $\prod R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K_ n, E) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\bigoplus K_ n, E)$. This is a formal consequence of (20.38.0.1) and the fact that derived tensor product commutes with direct sums. $\square$