The Stacks project

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21.42 Strictly perfect complexes

This section is the analogue of Cohomology, Section 20.40.

Definition 21.42.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{E}^\bullet $ be a complex of $\mathcal{O}$-modules. We say $\mathcal{E}^\bullet $ is strictly perfect if $\mathcal{E}^ i$ is zero for all but finitely many $i$ and $\mathcal{E}^ i$ is a direct summand of a finite free $\mathcal{O}$-module for all $i$.

Let $U$ be an object of $\mathcal{C}$. We will often say “Let $\mathcal{E}^\bullet $ be a strictly perfect complex of $\mathcal{O}_ U$-modules” to mean $\mathcal{E}^\bullet $ is a strictly perfect complex of modules on the ringed site $(\mathcal{C}/U, \mathcal{O}_ U)$, see Modules on Sites, Definition 18.19.1.

Lemma 21.42.2. The cone on a morphism of strictly perfect complexes is strictly perfect.

Proof. This is immediate from the definitions. $\square$

Lemma 21.42.3. The total complex associated to the tensor product of two strictly perfect complexes is strictly perfect.

Proof. Omitted. $\square$

Lemma 21.42.4. Let $(f, f^\sharp ) : (\mathcal{C}, \mathcal{O}_\mathcal {C}) \to (\mathcal{D}, \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi. If $\mathcal{F}^\bullet $ is a strictly perfect complex of $\mathcal{O}_\mathcal {D}$-modules, then $f^*\mathcal{F}^\bullet $ is a strictly perfect complex of $\mathcal{O}_\mathcal {C}$-modules.

Proof. We have seen in Modules on Sites, Lemma 18.17.2 that the pullback of a finite free module is finite free. The functor $f^*$ is additive functor hence preserves direct summands. The lemma follows. $\square$

Lemma 21.42.5. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U$ be an object of $\mathcal{C}$. Given a solid diagram of $\mathcal{O}_ U$-modules

\[ \xymatrix{ \mathcal{E} \ar@{..>}[dr] \ar[r] & \mathcal{F} \\ & \mathcal{G} \ar[u]_ p } \]

with $\mathcal{E}$ a direct summand of a finite free $\mathcal{O}_ U$-module and $p$ surjective, then there exists a covering $\{ U_ i \to U\} $ such that a dotted arrow making the diagram commute exists over each $U_ i$.

Proof. We may assume $\mathcal{E} = \mathcal{O}_ U^{\oplus n}$ for some $n$. In this case finding the dotted arrow is equivalent to lifting the images of the basis elements in $\Gamma (U, \mathcal{F})$. This is locally possible by the characterization of surjective maps of sheaves (Sites, Section 7.11). $\square$

Lemma 21.42.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U$ be an object of $\mathcal{C}$.

  1. Let $\alpha : \mathcal{E}^\bullet \to \mathcal{F}^\bullet $ be a morphism of complexes of $\mathcal{O}_ U$-modules with $\mathcal{E}^\bullet $ strictly perfect and $\mathcal{F}^\bullet $ acyclic. Then there exists a covering $\{ U_ i \to U\} $ such that each $\alpha |_{U_ i}$ is homotopic to zero.

  2. Let $\alpha : \mathcal{E}^\bullet \to \mathcal{F}^\bullet $ be a morphism of complexes of $\mathcal{O}_ U$-modules with $\mathcal{E}^\bullet $ strictly perfect, $\mathcal{E}^ i = 0$ for $i < a$, and $H^ i(\mathcal{F}^\bullet ) = 0$ for $i \geq a$. Then there exists a covering $\{ U_ i \to U\} $ such that each $\alpha |_{U_ i}$ is homotopic to zero.

Proof. The first statement follows from the second, hence we only prove (2). We will prove this by induction on the length of the complex $\mathcal{E}^\bullet $. If $\mathcal{E}^\bullet \cong \mathcal{E}[-n]$ for some direct summand $\mathcal{E}$ of a finite free $\mathcal{O}$-module and integer $n \geq a$, then the result follows from Lemma 21.42.5 and the fact that $\mathcal{F}^{n - 1} \to \mathop{\mathrm{Ker}}(\mathcal{F}^ n \to \mathcal{F}^{n + 1})$ is surjective by the assumed vanishing of $H^ n(\mathcal{F}^\bullet )$. If $\mathcal{E}^ i$ is zero except for $i \in [a, b]$, then we have a split exact sequence of complexes

\[ 0 \to \mathcal{E}^ b[-b] \to \mathcal{E}^\bullet \to \sigma _{\leq b - 1}\mathcal{E}^\bullet \to 0 \]

which determines a distinguished triangle in $K(\mathcal{O}_ U)$. Hence an exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}( \sigma _{\leq b - 1}\mathcal{E}^\bullet , \mathcal{F}^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}(\mathcal{E}^\bullet , \mathcal{F}^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}(\mathcal{E}^ b[-b], \mathcal{F}^\bullet ) \]

by the axioms of triangulated categories. The composition $\mathcal{E}^ b[-b] \to \mathcal{F}^\bullet $ is homotopic to zero on the members of a covering of $U$ by the above, whence we may assume our map comes from an element in the left hand side of the displayed exact sequence above. This element is zero on the members of a covering of $U$ by induction hypothesis. $\square$

Lemma 21.42.7. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U$ be an object of $\mathcal{C}$. Given a solid diagram of complexes of $\mathcal{O}_ U$-modules

\[ \xymatrix{ \mathcal{E}^\bullet \ar@{..>}[dr] \ar[r]_\alpha & \mathcal{F}^\bullet \\ & \mathcal{G}^\bullet \ar[u]_ f } \]

with $\mathcal{E}^\bullet $ strictly perfect, $\mathcal{E}^ j = 0$ for $j < a$ and $H^ j(f)$ an isomorphism for $j > a$ and surjective for $j = a$, then there exists a covering $\{ U_ i \to U\} $ and for each $i$ a dotted arrow over $U_ i$ making the diagram commute up to homotopy.

Proof. Our assumptions on $f$ imply the cone $C(f)^\bullet $ has vanishing cohomology sheaves in degrees $\geq a$. Hence Lemma 21.42.6 guarantees there is a covering $\{ U_ i \to U\} $ such that the composition $\mathcal{E}^\bullet \to \mathcal{F}^\bullet \to C(f)^\bullet $ is homotopic to zero over $U_ i$. Since

\[ \mathcal{G}^\bullet \to \mathcal{F}^\bullet \to C(f)^\bullet \to \mathcal{G}^\bullet [1] \]

restricts to a distinguished triangle in $K(\mathcal{O}_{U_ i})$ we see that we can lift $\alpha |_{U_ i}$ up to homotopy to a map $\alpha _ i : \mathcal{E}^\bullet |_{U_ i} \to \mathcal{G}^\bullet |_{U_ i}$ as desired. $\square$

Lemma 21.42.8. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U$ be an object of $\mathcal{C}$. Let $\mathcal{E}^\bullet $, $\mathcal{F}^\bullet $ be complexes of $\mathcal{O}_ U$-modules with $\mathcal{E}^\bullet $ strictly perfect.

  1. For any element $\alpha \in \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ there exists a covering $\{ U_ i \to U\} $ such that $\alpha |_{U_ i}$ is given by a morphism of complexes $\alpha _ i : \mathcal{E}^\bullet |_{U_ i} \to \mathcal{F}^\bullet |_{U_ i}$.

  2. Given a morphism of complexes $\alpha : \mathcal{E}^\bullet \to \mathcal{F}^\bullet $ whose image in the group $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ is zero, there exists a covering $\{ U_ i \to U\} $ such that $\alpha |_{U_ i}$ is homotopic to zero.

Proof. Proof of (1). By the construction of the derived category we can find a quasi-isomorphism $f : \mathcal{F}^\bullet \to \mathcal{G}^\bullet $ and a map of complexes $\beta : \mathcal{E}^\bullet \to \mathcal{G}^\bullet $ such that $\alpha = f^{-1}\beta $. Thus the result follows from Lemma 21.42.7. We omit the proof of (2). $\square$

Lemma 21.42.9. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{E}^\bullet $, $\mathcal{F}^\bullet $ be complexes of $\mathcal{O}$-modules with $\mathcal{E}^\bullet $ strictly perfect. Then the internal hom $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ is represented by the complex $\mathcal{H}^\bullet $ with terms

\[ \mathcal{H}^ n = \bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{F}^ p) \]

and differential as described in Section 21.34.

Proof. Choose a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{I}^\bullet $ into a K-injective complex. Let $(\mathcal{H}')^\bullet $ be the complex with terms

\[ (\mathcal{H}')^ n = \prod \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}^{-q}, \mathcal{I}^ p) \]

which represents $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ by the construction in Section 21.34. It suffices to show that the map

\[ \mathcal{H}^\bullet \longrightarrow (\mathcal{H}')^\bullet \]

is a quasi-isomorphism. Given an object $U$ of $\mathcal{C}$ we have by inspection

\[ H^0(\mathcal{H}^\bullet (U)) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}(\mathcal{E}^\bullet |_ U, \mathcal{K}^\bullet |_ U) \to H^0((\mathcal{H}')^\bullet (U)) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(\mathcal{E}^\bullet |_ U, \mathcal{K}^\bullet |_ U) \]

By Lemma 21.42.8 the sheafification of $U \mapsto H^0(\mathcal{H}^\bullet (U))$ is equal to the sheafification of $U \mapsto H^0((\mathcal{H}')^\bullet (U))$. A similar argument can be given for the other cohomology sheaves. Thus $\mathcal{H}^\bullet $ is quasi-isomorphic to $(\mathcal{H}')^\bullet $ which proves the lemma. $\square$

Lemma 21.42.10. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{E}^\bullet $, $\mathcal{F}^\bullet $ be complexes of $\mathcal{O}$-modules with

  1. $\mathcal{F}^ n = 0$ for $n \ll 0$,

  2. $\mathcal{E}^ n = 0$ for $n \gg 0$, and

  3. $\mathcal{E}^ n$ isomorphic to a direct summand of a finite free $\mathcal{O}$-module.

Then the internal hom $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ is represented by the complex $\mathcal{H}^\bullet $ with terms

\[ \mathcal{H}^ n = \bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{F}^ p) \]

and differential as described in Section 21.34.

Proof. Choose a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{I}^\bullet $ where $\mathcal{I}^\bullet $ is a bounded below complex of injectives. Note that $\mathcal{I}^\bullet $ is K-injective (Derived Categories, Lemma 13.29.4). Hence the construction in Section 21.34 shows that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}^\bullet , \mathcal{F}^\bullet )$ is represented by the complex $(\mathcal{H}')^\bullet $ with terms

\[ (\mathcal{H}')^ n = \prod \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{I}^ p) = \bigoplus \nolimits _{n = p + q} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{I}^ p) \]

(equality because there are only finitely many nonzero terms). Note that $\mathcal{H}^\bullet $ is the total complex associated to the double complex with terms $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{F}^ p)$ and similarly for $(\mathcal{H}')^\bullet $. The natural map $(\mathcal{H}')^\bullet \to \mathcal{H}^\bullet $ comes from a map of double complexes. Thus to show this map is a quasi-isomorphism, we may use the spectral sequence of a double complex (Homology, Lemma 12.22.6)

\[ {}'E_1^{p, q} = H^ p(\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}^{-q}, \mathcal{F}^\bullet )) \]

converging to $H^{p + q}(\mathcal{H}^\bullet )$ and similarly for $(\mathcal{H}')^\bullet $. To finish the proof of the lemma it suffices to show that $\mathcal{F}^\bullet \to \mathcal{I}^\bullet $ induces an isomorphism

\[ H^ p(\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}, \mathcal{F}^\bullet )) \longrightarrow H^ p(\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{E}, \mathcal{I}^\bullet )) \]

on cohomology sheaves whenever $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}$-module. Since this is clear when $\mathcal{E}$ is finite free the result follows. $\square$


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