The Stacks project

Lemma 103.11.5. Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be quasi-compact and quasi-separated morphisms of algebraic stacks. Let $\mathcal{F}$ be a quasi-coherent sheaf on $\mathcal{X}$. Then there exists a spectral sequence with $E_2$-page

\[ E_2^{p, q} = R^ pg_{\mathit{QCoh}, *}(R^ qf_{\mathit{QCoh}, *}\mathcal{F}) \]

converging to $R^{p + q}(g \circ f)_{\mathit{QCoh}, *}\mathcal{F}$.

Proof. By Cohomology on Sites, Lemma 21.14.7 the Leray spectral sequence with

\[ E_2^{p, q} = R^ pg_*(R^ qf_*\mathcal{F}) \]

converges to $R^{p + q}(g \circ f)_*\mathcal{F}$. By the results of Proposition 103.8.1 all the terms of this spectral sequence are objects of $\textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {Z})$. Applying the exact functor $Q_\mathcal {Z} : \textit{LQCoh}^{fbc}(\mathcal{O}_\mathcal {Z}) \to \mathit{QCoh}(\mathcal{O}_\mathcal {Z})$ we obtain a spectral sequence in $\mathit{QCoh}(\mathcal{O}_\mathcal {Z})$ covering to $R^{p + q}(g \circ f)_{\mathit{QCoh}, *}\mathcal{F}$. Hence the result follows if we can show that

\[ Q_\mathcal {Z}(R^ pg_*(R^ qf_*\mathcal{F})) = Q_\mathcal {Z}(R^ pg_*(Q_\mathcal {X}(R^ qf_*\mathcal{F})) \]

This follows from the fact that the kernel and cokernel of the map

\[ Q_\mathcal {X}(R^ qf_*\mathcal{F}) \longrightarrow R^ qf_*\mathcal{F} \]

are parasitic (Lemma 103.10.2) and that $R^ pg_*$ transforms parasitic modules into parasitic modules (Lemma 103.9.3). $\square$


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Your email address will not be published.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 0783. Beware of the difference between the letter 'O' and the digit '0'.