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

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