Lemma 66.4.12. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\mathcal{P}$ be any of the separation axioms of Definition 66.4.2. The following are equivalent

1. $f$ is $\mathcal{P}$,

2. for every scheme $Z$ and morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is $\mathcal{P}$,

3. for every affine scheme $Z$ and every morphism $Z \to Y$ the base change $Z \times _ Y X \to Z$ of $f$ is $\mathcal{P}$,

4. for every affine scheme $Z$ and every morphism $Z \to Y$ the algebraic space $Z \times _ Y X$ is $\mathcal{P}$ (see Properties of Spaces, Definition 65.3.1),

5. there exists a scheme $V$ and a surjective étale morphism $V \to Y$ such that the base change $V \times _ Y X \to V$ has $\mathcal{P}$, and

6. there exists a Zariski covering $Y = \bigcup Y_ i$ such that each of the morphisms $f^{-1}(Y_ i) \to Y_ i$ has $\mathcal{P}$.

Proof. We will repeatedly use Lemma 66.4.4 without further mention. In particular, it is clear that (1) implies (2) and (2) implies (3).

Let us prove that (3) and (4) are equivalent. Note that if $Z$ is an affine scheme, then the morphism $Z \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is a separated morphism as a morphism of algebraic spaces over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. If $Z \times _ Y X \to Z$ is $\mathcal{P}$, then $Z \times _ Y X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is $\mathcal{P}$ as a composition (see Lemma 66.4.8). Hence the algebraic space $Z \times _ Y X$ is $\mathcal{P}$. Conversely, if the algebraic space $Z \times _ Y X$ is $\mathcal{P}$, then $Z \times _ Y X \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is $\mathcal{P}$, and hence by Lemma 66.4.10 we see that $Z \times _ Y X \to Z$ is $\mathcal{P}$.

Let us prove that (3) implies (5). Assume (3). Let $V$ be a scheme and let $V \to Y$ be étale surjective. We have to show that $V \times _ Y X \to V$ has property $\mathcal{P}$. In other words, we have to show that the morphism

$V \times _ Y X \longrightarrow (V \times _ Y X) \times _ V (V \times _ Y X) = V \times _ Y X \times _ Y X$

has the corresponding property (i.e., is a closed immersion, immersion, or quasi-compact). Let $V = \bigcup V_ j$ be an affine open covering of $V$. By assumption we know that each of the morphisms

$V_ j \times _ Y X \longrightarrow V_ j \times _ Y X \times _ Y X$

does have the corresponding property. Since being a closed immersion, immersion, quasi-compact immersion, or quasi-compact is Zariski local on the target, and since the $V_ j$ cover $V$ we get the desired conclusion.

Let us prove that (5) implies (1). Let $V \to Y$ be as in (5). Then we have the fibre product diagram

$\xymatrix{ V \times _ Y X \ar[r] \ar[d] & X \ar[d] \\ V \times _ Y X \times _ Y X \ar[r] & X \times _ Y X }$

By assumption the left vertical arrow is a closed immersion, immersion, quasi-compact immersion, or quasi-compact. It follows from Spaces, Lemma 64.5.6 that also the right vertical arrow is a closed immersion, immersion, quasi-compact immersion, or quasi-compact.

It is clear that (1) implies (6) by taking the covering $Y = Y$. Assume $Y = \bigcup Y_ i$ is as in (6). Choose schemes $V_ i$ and surjective étale morphisms $V_ i \to Y_ i$. Note that the morphisms $V_ i \times _ Y X \to V_ i$ have $\mathcal{P}$ as they are base changes of the morphisms $f^{-1}(Y_ i) \to Y_ i$. Set $V = \coprod V_ i$. Then $V \to Y$ is a morphism as in (5) (details omitted). Hence (6) implies (5) and we are done. $\square$

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).