The Stacks project

Example 25.3.4 (Čech hypercoverings). Let $\mathcal{C}$ be a site with fibre products. Let $\{ U_ i \to X\} _{i \in I}$ be a covering of $\mathcal{C}$. Set $K_0 = \{ U_ i \to X\} _{i \in I}$. Then $K_0$ is a $0$-truncated simplicial object of $\text{SR}(\mathcal{C}, X)$. Hence we may form

\[ K = \text{cosk}_0 K_0. \]

Clearly $K$ passes condition (1) of Definition 25.3.3. Since all the morphisms $K_{n + 1} \to (\text{cosk}_ n \text{sk}_ n K)_{n + 1}$ are isomorphisms by Simplicial, Lemma 14.19.10 it also passes condition (2). Note that the terms $K_ n$ are the usual

\[ K_ n = \{ U_{i_0} \times _ X U_{i_1} \times _ X \ldots \times _ X U_{i_ n} \to X \} _{(i_0, i_1, \ldots , i_ n) \in I^{n + 1}} \]

A hypercovering of $X$ of this form is called a Čech hypercovering of $X$.

Comments (2)

Comment #1025 by correction_bot on

To see that is an isomorphism, perhaps it is useful to reference Lemma 14.17.11(3) (tag 018B), which shows is an isomorphism.

Comment #1031 by on

Many thanks for all the comments. I have incorporated all of your suggestions. The changes can be found in this commit.

There are also:

  • 3 comment(s) on Section 25.3: Hypercoverings

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Your email address will not be published.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 01G6. Beware of the difference between the letter 'O' and the digit '0'.