The Stacks project

Proof. Combine Remark 101.16.3 with Morphisms of Spaces, Lemma 67.28.2. $\square$

Proof. Combine Remark 101.16.4 with Morphisms of Spaces, Lemma 67.28.3. $\square$

Proof. Combine Remark 101.16.5 with Morphisms of Spaces, Lemma 67.28.5. $\square$

Proof. Assume $f : \mathcal{X} \to \mathcal{Y}$ locally of finite type and $\mathcal{Y}$ locally Noetherian. This means there exists a diagram as in Lemma 101.16.1 with $h$ locally of finite type and surjective vertical arrow $a$. By Morphisms of Spaces, Lemma 67.28.7 $h$ is locally of finite presentation. Hence $\mathcal{X} \to \mathcal{Y}$ is locally of finite presentation by definition. This proves (1). If $f$ is of finite type and quasi-separated then it is also quasi-compact and quasi-separated and (2) follows immediately. $\square$

Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y} \times _\mathcal {Z} W$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X} \times _\mathcal {Y} V$. The lemma follows upon applying Morphisms of Spaces, Lemma 67.28.9 to the morphisms $U \to V \to W$. $\square$

Proof. Note that $\Delta $ is a morphism over $\mathcal{X}$ (via the second projection). If $f$ is locally of finite type, then $\mathcal{X}$ is of finite presentation over $\mathcal{X}$ and $\text{pr}_2 : \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is locally of finite type by Lemma 101.17.3. Thus the first statement holds by Lemma 101.27.6. The second statement follows from the first and the definitions (because $f$ being quasi-separated means by definition that $\Delta _ f$ is quasi-compact and quasi-separated). $\square$

Proof. In view of Properties of Stacks, Definition 100.9.1 this follows from Morphisms of Spaces, Lemma 67.28.11. $\square$

Proof. Choose an algebraic space $W$ and a morphism $W \to \mathcal{Z}$ which is surjective, flat, and locally of finite presentation. By Properties of Stacks, Lemma 100.5.2 and Lemmas 101.25.2 and 101.27.2 the composition $W \to \mathcal{Y}$ is also surjective, flat, and locally of finite presentation. Denote $V = W \times _\mathcal {Z} \mathcal{W} = V \times _\mathcal {Y} \mathcal{X}$. By Properties of Stacks, Lemma 100.3.3 we see that $f$ has $\mathcal{P}$ if and only if $V \to W$ does and that $\mathcal{W} \to \mathcal{Z}$ has $\mathcal{P}$ if and only if $V \to W$ does. The lemma follows. $\square$

Proof. Assume $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ has $\mathcal{P}$. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Observe that $W \times _\mathcal {Z} \mathcal{Z} \times _\mathcal {Y} \mathcal{X} = W \times _\mathcal {Y} \mathcal{X}$. Thus by the very definition of what it means for $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ to have $\mathcal{P}$ (see Definition 101.16.2 and Lemma 101.16.1) we see that $W \times _\mathcal {Y} \mathcal{X} \to W$ has $\mathcal{P}$. On the other hand, $W \to \mathcal{Z}$ is surjective, flat, and locally of finite presentation (Morphisms of Spaces, Lemmas 67.37.7 and 67.37.5) hence $W \to \mathcal{Y}$ is surjective, flat, and locally of finite presentation (by Properties of Stacks, Lemma 100.5.2 and Lemmas 101.25.2 and 101.27.2). Thus we may replace $\mathcal{Z}$ by $W$.

Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We have to show that $U \to V$ has $\mathcal{P}$. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times _\mathcal {Y} U$, $V' = W \times _\mathcal {Y} V$, $\mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}$, and $\mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times _{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by Lemma 101.16.1 used in the definition of $\mathcal{X}' \to \mathcal{Y}' = W$ having $\mathcal{P}$ we see that $U' \to V'$ has $\mathcal{P}$. Then, since $V' \to V$ is surjective, flat, and locally of finite presentation as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ has $\mathcal{P}$ as $\mathcal{P}$ is local in the fppf topology on the target. $\square$

Proof. The property “locally of finite presentation” satisfies the conditions of Lemma 101.27.10. Smooth local on the source-and-target we have seen in the introduction to this section and fppf local on the target is Descent on Spaces, Lemma 74.11.10. $\square$

Proof. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times _\mathcal {Z} \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We know that $U \to V$ is flat and locally of finite presentation and that $U \to W$ is locally of finite presentation. Also, as $\mathcal{X} \to \mathcal{Y}$ is surjective we see that $U \to V$ is surjective (as a composition of surjective morphisms). Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma 74.16.1. $\square$

Proof. We may think of $y$ as a morphism $U \to \mathcal{Y}$. By Properties of Stacks, Lemma 100.5.3 and Lemmas 101.27.3 and 101.25.3 we see that $\mathcal{X} \times _\mathcal {Y} U \to U$ is surjective, flat, and locally of finite presentation. Let $V$ be a scheme and let $V \to \mathcal{X} \times _\mathcal {Y} U$ smooth and surjective. Then $V \to \mathcal{X} \times _\mathcal {Y} U$ is also surjective, flat, and locally of finite presentation (see Morphisms of Spaces, Lemmas 67.37.7 and 67.37.5). Hence also $V \to U$ is surjective, flat, and locally of finite presentation, see Properties of Stacks, Lemma 100.5.2 and Lemmas 101.27.2, and 101.25.2. Hence $\{ V \to U\} $ is the desired fppf covering and $x : V \to \mathcal{X}$ is the desired object. $\square$

Proof. Apply Lemma 101.27.13 to the morphism $\coprod _{j \in J} \mathcal{X}_ j \to \mathcal{X}$. (There is a slight set theoretic issue here – due to our setup of things – which we ignore.) To finish, note that a morphism $x_ i : U_ i \to \coprod _{j \in J} \mathcal{X}_ j$ is given by a disjoint union decomposition $U_ i = \coprod U_{i, j}$ and morphisms $U_{i, j} \to \mathcal{X}_ j$. Then the fppf covering $\{ U_{i, j} \to U\} $ and the morphisms $U_{i, j} \to \mathcal{X}_ j$ do the job. $\square$

Proof. Choose a scheme $V$ and a smooth surjective morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a smooth surjective morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. By assumption the morphism of schemes $U \to V$ is flat and locally of finite presentation. Hence $U \to V$ is open by Morphisms, Lemma 29.25.10. By construction of the topology on $|\mathcal{Y}|$ the map $|V| \to |\mathcal{Y}|$ is open. The map $|U| \to |\mathcal{X}|$ is surjective. The result follows from these facts by elementary topology. $\square$

Proof. We have to show that given $\mathcal{Y}' \to \mathcal{Y}$ with $\mathcal{Y}'$ quasi-compact, we have $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ is quasi-compact. Denote $\mathcal{Z}' = \mathcal{Z} \times _\mathcal {Y} \mathcal{Y}'$. Then $|\mathcal{Z}'| \to |\mathcal{Y}'|$ is open, see Lemma 101.27.15. Hence we can find a quasi-compact open substack $\mathcal{W} \subset \mathcal{Z}'$ mapping onto $\mathcal{Y}'$. Because $\mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z}$ is quasi-compact, we know that

is quasi-compact. And the map $\mathcal{W} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ is surjective, hence we win. Some details omitted. $\square$

Proof. Consider the diagram

The square is cartesian. We have to show the bottom horizontal arrow is proper. We already know that it is representable by algebraic spaces and locally of finite type (Lemma 101.3.3). Since the right vertical arrow is surjective, flat, and locally of finite presentation it suffices to show the top right horizontal arrow is proper (Lemma 101.27.9). Since $h$ is separated, the composition of the top horizontal arrows is proper.

Since $f$ is universally injective $\Delta $ is surjective (Lemma 101.14.5). Since the composition of $\Delta $ with the projection $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X}$ is the identity, we see that $\Delta $ is universally closed. By Morphisms of Spaces, Lemma 67.9.8 we conclude that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is separated as $\mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is separated. Here we use that implications between properties of morphisms of algebraic spaces can be transferred to the same implications between properties of morphisms of algebraic stacks representable by algebraic spaces; this is discussed in Properties of Stacks, Section 100.3. Finally, we use the same principle to conlude that $\mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$ is proper from Morphisms of Spaces, Lemma 67.40.7. $\square$