Lemma 61.11.4. Let $f : X \to Y$ be a finite type separated morphism of quasi-compact and quasi-separated schemes. Let $\Lambda$ be a torsion ring. The functor $Rf^!$ sends $D^+(Y_{\acute{e}tale}, \Lambda )$ into $D^+(X_{\acute{e}tale}, \Lambda )$. More precisely, there exists an integer $N \geq 0$ such that if $K \in D(Y_{\acute{e}tale}, \Lambda )$ has $H^ i(K) = 0$ for $i < a$ then $H^ i(Rf^!K) = 0$ for $i < a - N$.

Proof. Let $N$ be the integer found in Lemma 61.10.2. By construction, for $K \in D(Y_{\acute{e}tale}, \Lambda )$ and $L \in \in D(X_{\acute{e}tale}, \Lambda )$ we have $\mathop{\mathrm{Hom}}\nolimits _ X(L, Rf^!K) = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_!L, K)$. Suppose $H^ i(K) = 0$ for $i < a$. Then we take $L = \tau _{\leq a - N - 1}Rf^!K$. By Lemma 61.10.2 the complex $Rf_!L$ has vanishing cohomology sheaves in degrees $\leq a - 1$. Hence $\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_!L, K) = 0$ by Derived Categories, Lemma 13.27.3. Hence the canonical map $\tau _{\leq a - N - 1}Rf^!K \to Rf^!K$ is zero which implies $H^ i(Rf^!K) = 0$ for $i \leq a - N - 1$. $\square$

## 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Your email address will not be published.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).

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 0GLA. Beware of the difference between the letter 'O' and the digit '0'.