Lemma 73.11.2. Let $S$ be a scheme. The property $\mathcal{P}(f) =$“$f$ is quasi-separated” is fpqc local on the base on algebraic spaces over $S$.

Proof. A base change of a quasi-separated morphism is quasi-separated, see Morphisms of Spaces, Lemma 66.4.4. Hence the direct implication in Definition 73.10.1.

Let $\{ Y_ i \to Y\} _{i \in I}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $X_ i := Y_ i \times _ Y X \to Y_ i$ is quasi-separated. This means that each of the morphisms

$\Delta _ i : X_ i \longrightarrow X_ i \times _{Y_ i} X_ i = Y_ i \times _ Y (X \times _ Y X)$

is quasi-compact. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 72.9.3 hence $\{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\}$ is an fpqc covering of algebraic spaces. Moreover, each $\Delta _ i$ is the base change of the morphism $\Delta : X \to X \times _ Y X$. Hence it follows from Lemma 73.11.1 that $\Delta$ is quasi-compact, i.e., $f$ is quasi-separated. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).