The Stacks project

Lemma 63.4.7. Let $f : X \to Y$ be a locally quasi-finite morphism of schemes. Let $X = \bigcup _{i \in I} X_ i$ be an open covering. Then there exists an exact complex

\[ \ldots \to \bigoplus \nolimits _{i_0, i_1, i_2} f_{i_0i_1i_2, !} \mathcal{F}|_{X_{i_0i_1i_2}} \to \bigoplus \nolimits _{i_0, i_1} f_{i_0i_1, !} \mathcal{F}|_{X_{i_0i_1}} \to \bigoplus \nolimits _{i_0} f_{i_0, !} \mathcal{F}|_{X_{i_0}} \to f_!\mathcal{F} \to 0 \]

functorial in $\mathcal{F} \in \textit{Ab}(X_{\acute{e}tale})$, see proof for details.

Proof. Here as usual we set $X_{i_0 \ldots i_ p} = X_{i_0} \cap \ldots \cap X_{i_ p}$ and we denote $f_{i_0 \ldots i_ p}$ the restriction of $f$ to $X_{i_0 \ldots i_ p}$. The maps in the complex are the maps constructed in Remark 63.4.6 with sign rules as in the Čech complex. Exactness follows easily from the description of stalks in Lemma 63.4.5. Details omitted. $\square$

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Your email address will not be published.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 0F5H. Beware of the difference between the letter 'O' and the digit '0'.