The construction of Section 20.25 isn't the “correct” one for unbounded complexes. The problem is that in the Stacks project we use direct sums in the totalization of a double complex and we would have to replace this by a product. Instead of doing so in this section we assume the covering is finite and we use the alternating Čech complex.
Let $(X, \mathcal{O}_ X)$ be a ringed space. Let ${\mathcal F}^\bullet $ be a complex of presheaves of $\mathcal{O}_ X$-modules. Let ${\mathcal U} : X = \bigcup _{i \in I} U_ i$ be a finite open covering of $X$. Since the alternating Čech complex $\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F})$ (Section 20.23) is functorial in the presheaf $\mathcal{F}$ we obtain a double complex $\check{\mathcal{C}}^\bullet _{alt}(\mathcal{U}, \mathcal{F}^\bullet )$. In this section we work with the associated total complex. The construction of $\text{Tot}(\check{\mathcal{C}}^\bullet _{alt}({\mathcal U}, {\mathcal F}^\bullet ))$ is functorial in ${\mathcal F}^\bullet $. As well there is a functorial transformation
of complexes defined by the following rule: The section $s\in \Gamma (X, {\mathcal F}^ n)$ is mapped to the element $\alpha = \{ \alpha _{i_0\ldots i_ p}\} $ with $\alpha _{i_0} = s|_{U_{i_0}}$ and $\alpha _{i_0\ldots i_ p} = 0$ for $p > 0$.
Lemma 20.40.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ be a finite open covering. For a complex $\mathcal{F}^\bullet $ of $\mathcal{O}_ X$-modules there is a canonical map functorial in $\mathcal{F}^\bullet $ and compatible with (20.40.0.1).
\[ \text{Tot}(\check{\mathcal{C}}^\bullet _{alt}(\mathcal{U}, \mathcal{F}^\bullet )) \longrightarrow R\Gamma (X, \mathcal{F}^\bullet ) \]
Proof. Let ${\mathcal I}^\bullet $ be a K-injective complex whose terms are injective $\mathcal{O}_ X$-modules. The map (20.40.0.1) for $\mathcal{I}^\bullet $ is a map $\Gamma (X, {\mathcal I}^\bullet ) \to \text{Tot}(\check{\mathcal{C}}^\bullet _{alt}({\mathcal U}, {\mathcal I}^\bullet ))$. This is a quasi-isomorphism of complexes of abelian groups as follows from Homology, Lemma 12.25.4 applied to the double complex $\check{\mathcal{C}}^\bullet _{alt}({\mathcal U}, {\mathcal I}^\bullet )$ using Lemmas 20.11.1 and 20.23.6. Suppose ${\mathcal F}^\bullet \to {\mathcal I}^\bullet $ is a quasi-isomorphism of ${\mathcal F}^\bullet $ into a K-injective complex whose terms are injectives (Injectives, Theorem 19.12.6). Since $R\Gamma (X, {\mathcal F}^\bullet )$ is represented by the complex $\Gamma (X, {\mathcal I}^\bullet )$ we obtain the map of the lemma using
We omit the verification of functoriality and compatibilities. $\square$
Lemma 20.40.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{U} : X = \bigcup _{i \in I} U_ i$ be a finite open covering. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}_ X$-modules. Let $\mathcal{B}$ be a set of open subsets of $X$. Assume
every open in $X$ has a covering whose members are elements of $\mathcal{B}$,
we have $U_{i_0\ldots i_ p} \in \mathcal{B}$ for all $i_0, \ldots , i_ p \in I$,
for every $U \in \mathcal{B}$ and $p > 0$ we have
$H^ p(U, \mathcal{F}^ q) = 0$,
$H^ p(U, \mathop{\mathrm{Coker}}(\mathcal{F}^{q - 1} \to \mathcal{F}^ q)) = 0$, and
$H^ p(U, H^ q(\mathcal{F})) = 0$.
Then the map
of Lemma 20.40.1 is an isomorphism in $D(\textit{Ab})$.
Proof. First assume $\mathcal{F}^\bullet $ is bounded below. In this case the map
is a quasi-isomorphism by Lemma 20.23.6. Namely, the map of double complexes $\check{\mathcal{C}}^\bullet _{alt}(\mathcal{U}, \mathcal{F}^\bullet ) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}^\bullet )$ induces an isomorphism between the first pages of the second spectral sequences associated to these complexes (by Homology, Lemma 12.25.1) and these spectral sequences converge (Homology, Lemma 12.25.3). Thus the conclusion in this case by Lemma 20.25.2 and assumption (3)(a).
In general, by assumption (3)(c) we may choose a resolution $\mathcal{F}^\bullet \to \mathcal{I}^\bullet = \mathop{\mathrm{lim}}\nolimits \mathcal{I}_ n^\bullet $ as in Lemma 20.38.1. Then the map of the lemma becomes
Here the arrow is in the derived category, but the equality on the right holds on the level of complexes. Note that (3)(b) shows that $\tau _{\geq -n}\mathcal{F}^\bullet $ is a bounded below complex satisfying the hypothesis of the lemma. Thus the case of bounded below complexes shows each of the maps
is a quasi-isomorphism. The cohomologies of the complexes on the left hand side in given degree are eventually constant (as the alternating Čech complex is finite). Hence the same is true on the right hand side. Thus the cohomology of the limit on the right hand side is this constant value by Homology, Lemma 12.31.7 (or the stronger More on Algebra, Lemma 15.86.3) and we win. $\square$
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