Lemma 66.4.8. Let S be a scheme. Let X be an algebraic space over S.
The rule X' \mapsto |X'| defines an inclusion preserving bijection between open subspaces X' (see Spaces, Definition 65.12.1) of X, and opens of the topological space |X|.
A family \{ X_ i \subset X\} _{i \in I} of open subspaces of X is a Zariski covering (see Spaces, Definition 65.12.5) if and only if |X| = \bigcup |X_ i|.
In other words, the small Zariski site X_{Zar} of X is canonically identified with a site associated to the topological space |X| (see Sites, Example 7.6.4).
Proof.
In order to prove (1) let us construct the inverse of the rule. Namely, suppose that W \subset |X| is open. Choose a presentation X = U/R corresponding to the surjective étale map p : U \to X and étale maps s, t : R \to U. By construction we see that |p|^{-1}(W) is an open of U. Denote W' \subset U the corresponding open subscheme. It is clear that R' = s^{-1}(W') = t^{-1}(W') is a Zariski open of R which defines an étale equivalence relation on W'. By Spaces, Lemma 65.10.2 the morphism X' = W'/R' \to X is an open immersion. Hence X' is an algebraic space by Spaces, Lemma 65.11.3. By construction |X'| = W, i.e., X' is a subspace of X corresponding to W. Thus (1) is proved.
To prove (2), note that if \{ X_ i \subset X\} _{i \in I} is a collection of open subspaces, then it is a Zariski covering if and only if the U = \bigcup U \times _ X X_ i is an open covering. This follows from the definition of a Zariski covering and the fact that the morphism U \to X is surjective as a map of presheaves on (\mathit{Sch}/S)_{fppf}. On the other hand, we see that |X| = \bigcup |X_ i| if and only if U = \bigcup U \times _ X X_ i by Lemma 66.4.5 (and the fact that the projections U \times _ X X_ i \to X_ i are surjective and étale). Thus the equivalence of (2) follows.
\square
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