Lemma 67.4.12. 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 67.4.2. 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 66.3.1),

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 67.4.4 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 67.4.8). 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 67.4.10 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 65.5.6 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. $\square$

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