The Stacks project

Lemma 85.7.3. With notation and hypotheses as in Lemma 85.7.2. For $K \in D(\mathcal{O})$ we have $(g'_ n)^*Rh_{total, *}K = Rh_{n, *}g_ n^*K$.

Proof. Recall that $g_ n^* = g_ n^{-1}$ because $g_ n^{-1}\mathcal{O} = \mathcal{O}_ n$ by the construction in Lemma 85.6.1. In particular $g_ n^*$ is exact and $Lg_ n^*$ is given by applying $g_ n^*$ to any representative complex of modules. Similarly for $g'_ n$. There is a canonical base change map $(g'_ n)^*Rh_{total, *}K \to Rh_{n, *}g_ n^*K$, see Cohomology on Sites, Remark 21.19.3. By Cohomology on Sites, Lemma 21.20.7 the image of this in $D(\mathcal{C}'_ n)$ is the map $(g'_ n)^{-1}Rh_{total, *}K_{ab} \to Rh_{n, *}g_ n^{-1}K_{ab}$ where $K_{ab}$ is the image of $K$ in $D(\mathcal{C}_{total})$. This we proved to be an isomorphism in Lemma 85.5.3 and the result follows. $\square$

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Your email address will not be published.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 0DH1. Beware of the difference between the letter 'O' and the digit '0'.