Lemma 64.4.8. 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 63.12.1) 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 63.12.5) 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 63.10.2 the morphism $X' = W'/R' \to X$ is an open immersion. Hence $X'$ is an algebraic space by Spaces, Lemma 63.11.3. 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 64.4.5 (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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