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

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:

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

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

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 $(T', U) \in \text{Cov}(\mathcal{C})$.

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

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

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

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