The Stacks project

Proof. The implication $(\gamma ) \Leftrightarrow (\alpha ) + (\beta )$ is immediate. The implications in the diamond on the left are clear from the definitions.

Assume $(\zeta )$, i.e., that $X$ is Zariski locally quasi-separated. Then $(\epsilon )$ holds by Properties of Spaces, Lemma 66.6.6.

Assume $(\epsilon )$. By Lemma 68.4.7 there exists a Zariski open covering $X = \bigcup X_ i$ such that for each $i$ there exists a scheme $U_ i$ and a quasi-compact surjective étale morphism $U_ i \to X_ i$. Choose an $i$ and an affine open subscheme $W \subset U_ i$. It suffices to show that $W \to X$ has universally bounded fibres, since then the family of all these morphisms $W \to X$ covers $X$. To do this we consider the diagram

Since $W \to X$ factors through $X_ i$ we see that $W \times _ X U_ i = W \times _{X_ i} U_ i$, and hence $q$ is quasi-compact. Since $W$ is affine this implies that the scheme $W \times _ X U_ i$ is quasi-compact. Thus we may apply Morphisms, Lemma 29.57.9 and we conclude that $p$ has universally bounded fibres. From Lemma 68.3.4 we conclude that $W \to X$ has universally bounded fibres as well.

Assume $(\delta )$. Let $U$ be an affine scheme, and let $U \to X$ be an étale morphism. By assumption the fibres of the morphism $U \to X$ are universally bounded. Thus also the fibres of both projections $R = U \times _ X U \to U$ are universally bounded, see Lemma 68.3.3. And by Lemma 68.3.2 also the fibres of $R \to X$ are universally bounded. Hence for any $x \in X$ the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite, see Lemma 68.3.6. In other words, the equivalent conditions of Lemma 68.4.5 hold. This proves that $(\delta ) \Rightarrow (\gamma )$. $\square$

Proof. Let $X$ be an algebraic space over $S$, and let $X = \bigcup X_ i$ is a Zariski open covering such that each $X_ i$ has $\mathcal{P}$.

The case $\mathcal{P} = (\alpha )$. The condition $(\alpha )$ for $X_ i$ means that for every $x \in |X_ i|$ and every affine scheme $U$, and étale morphism $\varphi : U \to X_ i$ the fibre of $\varphi : |U| \to |X_ i|$ over $x$ is finite. Consider $x \in X$, an affine scheme $U$ and an étale morphism $U \to X$. Since $X = \bigcup X_ i$ is a Zariski open covering there exits a finite affine open covering $U = U_1 \cup \ldots \cup U_ n$ such that each $U_ j \to X$ factors through some $X_{i_ j}$. By assumption the fibres of $|U_ j | \to |X_{i_ j}|$ over $x$ are finite for $j = 1, \ldots , n$. Clearly this means that the fibre of $|U| \to |X|$ over $x$ is finite. This proves the result for $(\alpha )$.

The case $\mathcal{P} = (\beta )$. The condition $(\beta )$ for $X_ i$ means that every $x \in |X_ i|$ is represented by a monomorphism from the spectrum of a field towards $X_ i$. Hence the same follows for $X$ as $X_ i \to X$ is a monomorphism and $X = \bigcup X_ i$.

The case $\mathcal{P} = (\gamma )$. Note that $(\gamma ) = (\alpha ) + (\beta )$ by Lemma 68.5.1 hence the lemma for $(\gamma )$ follows from the cases treated above.

The case $\mathcal{P} = (\delta )$. The condition $(\delta )$ for $X_ i$ means there exist schemes $U_{ij}$ and étale morphisms $U_{ij} \to X_ i$ with universally bounded fibres which cover $X_ i$. These schemes also give an étale surjective morphism $\coprod U_{ij} \to X$ and $U_{ij} \to X$ still has universally bounded fibres.

The case $\mathcal{P} = (\epsilon )$. The condition $(\epsilon )$ for $X_ i$ means we can find a set $J_ i$ and morphisms $\varphi _{ij} : U_{ij} \to X_ i$ such that each $\varphi _{ij}$ is étale, both projections $U_{ij} \times _{X_ i} U_{ij} \to U_{ij}$ are quasi-compact, and $\coprod _{j \in J_ i} U_{ij} \to X_ i$ is surjective. In this case the compositions $U_{ij} \to X_ i \to X$ are étale (combine Morphisms, Lemmas 29.36.3 and 29.36.9 and Spaces, Lemmas 65.5.4 and 65.5.3 ). Since $X_ i \subset X$ is a subspace we see that $U_{ij} \times _{X_ i} U_{ij} = U_{ij} \times _ X U_{ij}$, and hence the condition on fibre products is preserved. And clearly $\coprod _{i, j} U_{ij} \to X$ is surjective. Hence $X$ satisfies $(\epsilon )$.

The case $\mathcal{P} = (\zeta )$. The condition $(\zeta )$ for $X_ i$ means that $X_ i$ is Zariski locally quasi-separated. It is immediately clear that this means $X$ is Zariski locally quasi-separated.

For $(\theta )$, see Properties of Spaces, Lemma 66.13.1. $\square$

Proof. Assume $f : X \to Y$ is a representable morphism of algebraic spaces, and assume that $Y$ has $\mathcal{P}$. Let $x \in |X|$, and set $y = f(x) \in |Y|$.

The case $\mathcal{P} = (\beta )$. Condition $(\beta )$ for $Y$ means there exists a monomorphism $\mathop{\mathrm{Spec}}(k) \to Y$ representing $y$. The fibre product $X_ y = \mathop{\mathrm{Spec}}(k) \times _ Y X$ is a scheme, and $x$ corresponds to a point of $X_ y$, i.e., to a monomorphism $\mathop{\mathrm{Spec}}(k') \to X_ y$. As $X_ y \to X$ is a monomorphism also we see that $x$ is represented by the monomorphism $\mathop{\mathrm{Spec}}(k') \to X_ y \to X$. In other words $(\beta )$ holds for $X$.

The case $\mathcal{P} = (\gamma )$. Since $(\gamma ) \Rightarrow (\beta )$ we have seen in the preceding paragraph that $y$ and $x$ can be represented by monomorphisms as in the following diagram

Also, by definition of property $(\gamma )$ via Lemma 68.4.5 (2) there exist schemes $V_ i$ and étale morphisms $V_ i \to Y$ such that $\coprod V_ i \to Y$ is surjective and for each $i$, setting $R_ i = V_ i \times _ Y V_ i$ the fibres of both

over $y$ are finite. This means that the schemes $(V_ i)_ y$ and $(R_ i)_ y$ are finite schemes over $y = \mathop{\mathrm{Spec}}(k)$. As $X \to Y$ is representable, the fibre products $U_ i = V_ i \times _ Y X$ are schemes. The morphisms $U_ i \to X$ are étale, and $\coprod U_ i \to X$ is surjective. Finally, for each $i$ we have

and

hence these are finite over $k'$ as base changes of the finite schemes $(V_ i)_ y$ and $(R_ i)_ y$. This implies that $(\gamma )$ holds for $X$, again via the second condition of Lemma 68.4.5.

The case $\mathcal{P} = (\delta )$. Let $V \to Y$ be an étale morphism with $V$ an affine scheme. Since $Y$ has property $(\delta )$ this morphism has universally bounded fibres. By Lemma 68.3.3 the base change $V \times _ Y X \to X$ also has universally bounded fibres. Hence the first part of Lemma 68.4.6 applies and we see that $Y$ also has property $(\delta )$.

The case $\mathcal{P} = (\epsilon )$. We will repeatedly use Spaces, Lemma 65.5.5. Let $V_ i \to Y$ be as in Lemma 68.4.7 (2). Set $U_ i = X \times _ Y V_ i$. The morphisms $U_ i \to X$ are étale, and $\coprod U_ i \to X$ is surjective. Because $U_ i \times _ X U_ i = X \times _ Y (V_ i \times _ Y V_ i)$ we see that the projections $U_ i \times _ Y U_ i \to U_ i$ are base changes of the projections $V_ i \times _ Y V_ i \to V_ i$, and so quasi-compact as well. Hence $X$ satisfies Lemma 68.4.7 (2).

The case $\mathcal{P} = (\theta )$. In this case the result is Categories, Lemma 4.8.3. $\square$