The Stacks project

Lemma 21.11.4. Let $\mathcal{C}$ be a site. Let $\mathcal{G}$ be an abelian sheaf on $\mathcal{C}$. Let $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ be a covering of $\mathcal{C}$. The map

\[ \check{H}^1(\mathcal{U}, \mathcal{G}) \longrightarrow H^1(U, \mathcal{G}) \]

is injective and identifies $\check{H}^1(\mathcal{U}, \mathcal{G})$ via the bijection of Lemma 21.5.3 with the set of isomorphism classes of $\mathcal{G}|_ U$-torsors which restrict to trivial torsors over each $U_ i$.

Proof. To see this we construct an inverse map. Namely, let $\mathcal{F}$ be a $\mathcal{G}|_ U$-torsor on $\mathcal{C}/U$ whose restriction to $\mathcal{C}/U_ i$ is trivial. By Lemma 21.5.2 this means there exists a section $s_ i \in \mathcal{F}(U_ i)$. On $U_{i_0} \times _ U U_{i_1}$ there is a unique section $s_{i_0i_1}$ of $\mathcal{G}$ such that $s_{i_0i_1} \cdot s_{i_0}|_{U_{i_0} \times _ U U_{i_1}} = s_{i_1}|_{U_{i_0} \times _ U U_{i_1}}$. An easy computation shows that $s_{i_0i_1}$ is a Čech cocycle and that its class is well defined (i.e., does not depend on the choice of the sections $s_ i$). The inverse maps the isomorphism class of $\mathcal{F}$ to the cohomology class of the cocycle $(s_{i_0i_1})$. We omit the verification that this map is indeed an inverse. $\square$


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Your email address will not be published.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 0A6G. Beware of the difference between the letter 'O' and the digit '0'.