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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)