In Spaces, Section 65.12 we introduced the notion of a Zariski covering of an algebraic space by open subspaces. Here is the corresponding notion with open subspaces replaced by open immersions.
Definition 73.3.1. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. A Zariski covering of $X$ is a family of morphisms $\{ f_ i : X_ i \to X\} _{i \in I}$ of algebraic spaces over $S$ such that each $f_ i$ is an open immersion and such that i.e., the morphisms are jointly surjective.
\[ |X| = \bigcup \nolimits _{i \in I} |f_ i|(|X_ i|), \]
Although Zariski coverings are occasionally useful the corresponding topology on the category of algebraic spaces is really too coarse, and not particularly useful. Still, it does define a site.
Lemma 73.3.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
If $X' \to X$ is an isomorphism then $\{ X' \to X\} $ is a Zariski covering of $X$.
If $\{ X_ i \to X\} _{i\in I}$ is a Zariski covering and for each $i$ we have a Zariski covering $\{ X_{ij} \to X_ i\} _{j\in J_ i}$, then $\{ X_{ij} \to X\} _{i \in I, j\in J_ i}$ is a Zariski covering.
If $\{ X_ i \to X\} _{i\in I}$ is a Zariski covering and $X' \to X$ is a morphism of algebraic spaces then $\{ X' \times _ X X_ i \to X'\} _{i\in I}$ is a Zariski covering.
Proof. Omitted. $\square$
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