The Stacks project

Lemma 20.37.11. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)$ be an inverse system of objects of $D(\mathcal{O}_ X)$. Let $\mathcal{B}$ be a set of opens of $X$. Assume

  1. every open of $X$ has a covering whose members are elements of $\mathcal{B}$,

  2. for all $U \in \mathcal{B}$ and all $q \in \mathbf{Z}$ we have

    1. $H^ p(U, H^ q(K_ n)) = 0$ for $p > 0$,

    2. the inverse system $H^0(U, H^ q(K_ n))$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits $.

Then $H^ q(R\mathop{\mathrm{lim}}\nolimits K_ n) = \mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ for $q \in \mathbf{Z}$.

Proof. Set $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. Let $U \in \mathcal{B}$. By Lemma 20.37.10 and (2)(a) we have $H^ q(U, K_ n) = H^0(U, H^ q(K_ n))$. By Lemma 20.37.1 and (2)(b) we have $H^ q(U, K) = \mathop{\mathrm{lim}}\nolimits H^0(U, H^ q(K_ n))$. Thus $H^ q(U, K)$ is the inverse limit the sections of the sheaves $H^ q(K_ n)$ over $U$. Since $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ is a sheaf we find using assumption (1) that $H^ q(K)$, which is the sheafification of the presheaf $U \mapsto H^ q(U, K)$, is equal to $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$. This proves the lemma. $\square$


Comments (2)

Comment #11607 by Alex Scheffelin on

I think using https://stacks.math.columbia.edu/tag/0D60 you can directly get that using that .


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Your email address will not be published.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 0BKU. Beware of the difference between the letter 'O' and the digit '0'.