The Stacks project

Lemma 20.37.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E \in D(\mathcal{O}_ X)$. Assume that for every $x \in X$ there exist a function $p(x, -) : \mathbf{Z} \to \mathbf{Z}$ and a fundamental system $\mathfrak {U}_ x$ of open neighbourhoods of $x$ such that

\[ H^ p(U, H^{m - p}(E)) = 0 \text{ for } U \in \mathfrak {U}_ x \text{ and } p > p(x, m) \]

Then the canonical map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ is an isomorphism in $D(\mathcal{O}_ X)$.

Proof. Set $K_ n = \tau _{\geq -n}E$ and $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. The canonical map $E \to K$ comes from the canonical maps $E \to K_ n = \tau _{\geq -n}E$. We have to show that $E \to K$ induces an isomorphism $H^ m(E) \to H^ m(K)$ of cohomology sheaves. In the rest of the proof we fix $m$. If $n \geq -m$, then the map $E \to \tau _{\geq -n}E = K_ n$ induces an isomorphism $H^ m(E) \to H^ m(K_ n)$. To finish the proof it suffices to show that for every $x \in X$ there exists an integer $n(x) \geq -m$ such that the map $H^ m(K)_ x \to H^ m(K_{n(x)})_ x$ is injective. Namely, then the composition

\[ H^ m(E)_ x \to H^ m(K)_ x \to H^ m(K_{n(x)})_ x \]

is a bijection and the second arrow is injective, hence the first arrow is bijective. Set

\[ n(x) = 1 + \max \{ -m, p(x, m - 1) - m, -1 + p(x, m) - m, -2 + p(x, m + 1) - m\} . \]

so that in any case $n(x) \geq -m$. Claim: the maps

\[ H^{m - 1}(U, K_{n + 1}) \to H^{m - 1}(U, K_ n) \quad \text{and}\quad H^ m(U, K_{n + 1}) \to H^ m(U, K_ n) \]

are isomorphisms for $n \geq n(x)$ and $U \in \mathfrak {U}_ x$. The claim implies conditions (1) and (2) of Lemma 20.37.5 are satisfied and hence implies the desired injectivity. Recall (Derived Categories, Remark 13.12.4) that we have distinguished triangles

\[ H^{-n - 1}(E)[n + 1] \to K_{n + 1} \to K_ n \to H^{-n - 1}(E)[n + 2] \]

Looking at the asssociated long exact cohomology sequence the claim follows if

\[ H^{m + n}(U, H^{-n - 1}(E)),\quad H^{m + n + 1}(U, H^{-n - 1}(E)),\quad H^{m + n + 2}(U, H^{-n - 1}(E)) \]

are zero for $n \geq n(x)$ and $U \in \mathfrak {U}_ x$. This follows from our choice of $n(x)$ and the assumption in the lemma. $\square$


Comments (0)


Post a comment

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0D62. Beware of the difference between the letter 'O' and the digit '0'.