The Stacks project

Lemma 21.31.10. With notation as above.

  1. For $X \in \mathop{\mathrm{Ob}}\nolimits (\textit{LC}_{qc})$ and an abelian sheaf $\mathcal{F}$ on $X$ we have $\epsilon _{X, *}a_ X^{-1}\mathcal{F} = \pi _ X^{-1}\mathcal{F}$ and $R^ i\epsilon _{X, *}(a_ X^{-1}\mathcal{F}) = 0$ for $i > 0$.

  2. For a proper morphism $f : X \to Y$ in $\textit{LC}_{qc}$ and abelian sheaf $\mathcal{F}$ on $X$ we have $a_ Y^{-1}(R^ if_*\mathcal{F}) = R^ if_{qc, *}(a_ X^{-1}\mathcal{F})$ for all $i$.

  3. For $X \in \mathop{\mathrm{Ob}}\nolimits (\textit{LC}_{qc})$ and $K$ in $D^+(X)$ the map $\pi _ X^{-1}K \to R\epsilon _{X, *}(a_ X^{-1}K)$ is an isomorphism.

  4. For a proper morphism $f : X \to Y$ in $\textit{LC}_{qc}$ and $K$ in $D^+(X)$ we have $a_ Y^{-1}(Rf_*K) = Rf_{qc, *}(a_ X^{-1}K)$.

Proof. By Lemma 21.31.9 the lemmas in Section 21.30 all apply to our current setting. To translate the results observe that the category $\mathcal{A}_ X$ of Lemma 21.30.2 is the essential image of $a_ X^{-1} : \textit{Ab}(X) \to \textit{Ab}(\textit{LC}_{qc}/X)$.

Part (1) is equivalent to $(V_ n)$ for all $n$ which holds by Lemma 21.30.8.

Part (2) follows by applying $\epsilon _ Y^{-1}$ to the conclusion of Lemma 21.30.3.

Part (3) follows from Lemma 21.30.8 part (1) because $\pi _ X^{-1}K$ is in $D^+_{\mathcal{A}'_ X}(\textit{LC}_{Zar}/X)$ and $a_ X^{-1} = \epsilon _ X^{-1} \circ a_ X^{-1}$.

Part (4) follows from Lemma 21.30.8 part (2) for the same reason. $\square$

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Your email address will not be published.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