The Stacks project

Remark 7.48.4. Shrinking the class of coverings. Let $\mathcal{C}$ be a site. Consider the set

\[ \mathcal{S} = P(\text{Arrows}(\mathcal{C})) \times \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) \]

where $P(\text{Arrows}(\mathcal{C}))$ is the power set of the set of morphisms, i.e., the set of all sets of morphisms. Let $\mathcal{S}_\tau \subset \mathcal{S}$ be the subset consisting of those $(T, U) \in \mathcal{S}$ such that (a) all $\varphi \in T$ have target $U$, (b) the collection $\{ \varphi \} _{\varphi \in T}$ is tautologically equivalent (see Definition 7.8.2) to some covering in $\text{Cov}(\mathcal{C})$. Clearly, considering the elements of $\mathcal{S}_\tau $ as the coverings, we do not get exactly the notion of a site as defined in Definition 7.6.2. The structure $(\mathcal{C}, \mathcal{S}_\tau )$ we get satisfies slightly modified conditions. The modified conditions are:

  1. $\text{Cov}(\mathcal{C}) \subset P(\text{Arrows}(\mathcal{C})) \times \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$,

  2. If $V \to U$ is an isomorphism then $(\{ V \to U\} , U) \in \text{Cov}(\mathcal{C})$.

  3. If $(T, U) \in \text{Cov}(\mathcal{C})$ and for $f : U' \to U$ in $T$ we are given $(T_ f, U') \in \text{Cov}(\mathcal{C})$, then setting $T' = \{ f \circ f' \mid f \in T,\ f' \in T_ f\} $, we get $(T', U) \in \text{Cov}(\mathcal{C})$.

  4. If $(T, U) \in \text{Cov}(\mathcal{C})$ and $g : V \to U$ is a morphism of $\mathcal{C}$ then

    1. $U' \times _{f, U, g} V$ exists for $f : U' \to U$ in $T$, and

    2. setting $T' = \{ \text{pr}_2 : U' \times _{f, U, g} V \to V \mid f : U' \to U \in T\} $ for some choice of fibre products we get $(T', V) \in \text{Cov}(\mathcal{C})$.

And it is easy to verify that, given a structure satisfying (0') – (3') above, then after suitably enlarging $\text{Cov}(\mathcal{C})$ (compare Sets, Section 3.11) we get a site. Obviously there is little difference between this notion and the actual notion of a site, at least from the point of view of the topology. There are two benefits: because of condition (0') above the coverings automatically form a set, and because of (0') the totality of all structures of this type forms a set as well. The price you pay for this is that you have to keep writing “tautologically equivalent” everywhere.

Comments (2)

Comment #1709 by Keenan Kidwell on

To view elements of the power set of all arrows in as families of morphisms with fixed target (to which we can then apply the notion of tautological equivalence), we want to have an index set. This is undoubtedly a dumb question (perhaps worse than dumb, yet I ask it still): we're taking the index set for such a to be itself, right? That is, the indices are the arrows in ?

There are also:

  • 2 comment(s) on Section 7.48: The topology defined by a site

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 00ZF. Beware of the difference between the letter 'O' and the digit '0'.