The Stacks project

Maybe I am missing something, but it looks like the definition of combiatorically equivilant is just saying thet the covers are the same. i.e. $\mathcal{U}$ = $\mathcal{V}$ No?

@#10213: Yes, this is confirmed e.g. in the proof of Set Theory, Lemma 3.11.1: “Note that two families $\mathcal{U} = \{\varphi_i : U_i \to U\}_{i\in I}$ and $\mathcal{V} = \{V_j \to V\}_{j \in J}$ are combinatorially equivalent if and only if $Supp(\mathcal{U}) = Supp(\mathcal{V})$.”

The point is that the definition of a family of morphisms with fixed target, Definition 7.6.1, includes the indexing set $I$ as part of the data. This gives the flexibility to allow repetition of the same morphism, for instance $\mathcal{U}=\{U\xrightarrow{\operatorname{id}} U\}$ and $\mathcal{V}=\coprod_{n\in\mathbb{N}}\{U\xrightarrow{\operatorname{id}}U\}$ are two distinct families of morphisms with fixed target that are combinatorially equivalent. Definition 7.6.1 is the notion that is required to define a pretopology in the sense of SGA 4, Exp. II, Definition 1.3, named simply as “site” here in the Stacks Project, Definition 7.6.2.

Definition 7.6.2 is to Definition 7.6.1 what a topology (Definition 7.47.6) is to a sieve (Definition 7.47.1). The notion of a sieve does not allow for repetition of morphisms.