Lemma 62.11.4. 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 62.10.2. 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 62.10.2 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Your email address will not be published.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.

## Comments (0)