Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}_ X$-modules. The category $\textit{Mod}(\mathcal{O}_ X)$ has enough injectives, hence we can use Derived Categories, Lemma 13.29.3 produce a diagram
in the category of complexes of $\mathcal{O}_ X$-modules such that
the vertical arrows are quasi-isomorphisms,
$\mathcal{I}_ n^\bullet $ is a bounded below complex of injectives,
the arrows $\mathcal{I}_{n + 1}^\bullet \to \mathcal{I}_ n^\bullet $ are termwise split surjections.
The category of $\mathcal{O}_ X$-modules has limits (they are computed on the level of presheaves), hence we can form the termwise limit $\mathcal{I}^\bullet = \mathop{\mathrm{lim}}\nolimits _ n \mathcal{I}_ n^\bullet $. By Derived Categories, Lemmas 13.31.4 and 13.31.8 this is a K-injective complex. In general the canonical map
may not be a quasi-isomorphism. In the following lemma we describe some conditions under which it is.
Lemma 20.38.1. In the situation described above. Denote $\mathcal{H}^ m = H^ m(\mathcal{F}^\bullet )$ the $m$th cohomology sheaf. Let $\mathcal{B}$ be a set of open subsets of $X$. Let $d \in \mathbf{N}$. Assume
every open in $X$ has a covering whose members are elements of $\mathcal{B}$,
for every $U \in \mathcal{B}$ we have $H^ p(U, \mathcal{H}^ q) = 0$ for $p > d$ and $q < 0$1.
Then (20.38.0.1) is a quasi-isomorphism.
Proof. By Derived Categories, Lemma 13.34.4 it suffices to show that the canonical map $\mathcal{F}^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} \mathcal{F}^\bullet $ is an isomorphism. This is Lemma 20.37.9. $\square$
Here is a technical lemma about the cohomology sheaves of the inverse limit of a system of complexes of sheaves. In some sense this lemma is the wrong thing to try to prove as one should take derived limits and not actual inverse limits.
Lemma 20.38.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(\mathcal{F}_ n^\bullet )$ be an inverse system of complexes of $\mathcal{O}_ X$-modules. Let $m \in \mathbf{Z}$. Assume there exist a set $\mathcal{B}$ of open subsets of $X$ and an integer $n_0$ such that
every open in $X$ has a covering whose members are elements of $\mathcal{B}$,
for every $U \in \mathcal{B}$
the systems of abelian groups $\mathcal{F}_ n^{m - 2}(U)$ and $\mathcal{F}_ n^{m - 1}(U)$ have vanishing $R^1\mathop{\mathrm{lim}}\nolimits $ (for example these have the Mittag-Leffler condition),
the system of abelian groups $H^{m - 1}(\mathcal{F}_ n^\bullet (U))$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits $ (for example it has the Mittag-Leffler condition), and
we have $H^ m(\mathcal{F}_ n^\bullet (U)) = H^ m(\mathcal{F}_{n_0}^\bullet (U))$ for all $n \geq n_0$.
Then the maps $H^ m(\mathcal{F}^\bullet ) \to \mathop{\mathrm{lim}}\nolimits H^ m(\mathcal{F}_ n^\bullet ) \to H^ m(\mathcal{F}_{n_0}^\bullet )$ are isomorphisms of sheaves where $\mathcal{F}^\bullet = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n^\bullet $ is the termwise inverse limit.
Proof. Let $U \in \mathcal{B}$. Note that $H^ m(\mathcal{F}^\bullet (U))$ is the cohomology of
in the third spot from the left. By assumptions (2)(a) and (2)(b) we may apply More on Algebra, Lemma 15.86.3 to conclude that
By assumption (2)(c) we conclude
for all $n \geq n_0$. By assumption (1) we conclude that the sheafification of $U \mapsto H^ m(\mathcal{F}^\bullet (U))$ is equal to the sheafification of $U \mapsto H^ m(\mathcal{F}_ n^\bullet (U))$ for all $n \geq n_0$. Thus the inverse system of sheaves $H^ m(\mathcal{F}_ n^\bullet )$ is constant for $n \geq n_0$ with value $H^ m(\mathcal{F}^\bullet )$ which proves the lemma. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)