Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section 100.3 we have defined what it means for $f$ to be a universal homeomorphism. Here is another characterization.
Lemma 101.15.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent
$f$ is a universal homeomorphism (Properties of Stacks, Section 100.3), and
for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map of topological spaces $|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$ is a homeomorphism.
Proof. Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times _\mathcal {Y} \mathcal{X} \to V$ of algebraic spaces is a universal homeomorphism, in particular the map $|V \times _\mathcal {Y} \mathcal{X}| \to |V|$ is a homeomorphism. By Properties of Stacks, Section 100.4 in the commutative diagram
the horizontal arrows are open and surjective, and moreover
is surjective. Hence as the left vertical arrow is a homeomorphism it follows that the right vertical arrow is a homeomorphism. This proves (2). The implication (2) $\Rightarrow $ (1) follows from the definitions. $\square$
Thus we may use the following natural definition.
Definition 101.15.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is a universal homeomorphism if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map of topological spaces is a homeomorphism.
\[ |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}| \]
Lemma 101.15.3. The base change of a universal homeomorphism of algebraic stacks by any morphism of algebraic stacks is a universal homeomorphism.
Proof. This is immediate from the definition. $\square$
Lemma 101.15.4. The composition of a pair of universal homeomorphisms of algebraic stacks is a universal homeomorphism.
Proof. Omitted. $\square$
Lemma 101.15.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times _\mathcal {Y} \mathcal{X} \to W$ is a universal homeomorphism, then $f$ is a universal homeomorphism.
Proof. Assume $g : W \times _\mathcal {Y} \mathcal{X} \to W$ is a universal homeomorphism. Then $g$ is universally injective, hence $f$ is universally injective by Lemma 101.14.8. On the other hand, let $\mathcal{Z} \to \mathcal{Y}$ be a morphism with $\mathcal{Z}$ an algebraic stack. Choose a scheme $U$ and a surjective smooth morphism $U \to W \times _\mathcal {Y} \mathcal{Z}$. Consider the diagram
The middle vertical arrow induces a homeomorphism on topological space by assumption on $g$. The morphism $U \to \mathcal{Z}$ and $U \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ are surjective, flat, and locally of finite presentation hence induce open maps on topological spaces. We conclude that $|\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \to |\mathcal{Z}|$ is open. Surjectivity is easy to prove; we omit the proof. $\square$
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