## Tag `04XL`

Chapter 85: Properties of Algebraic Stacks > Section 85.4: Points of algebraic stacks

Lemma 85.4.7. There exists a unique topology on the sets of points of algebraic stacks with the following properties:

- for every morphism of algebraic stacks $\mathcal{X} \to \mathcal{Y}$ the map $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous, and
- for every morphism $U \to \mathcal{X}$ which is flat and locally of finite presentation with $U$ an algebraic space the map of topological spaces $|U| \to |\mathcal{X}|$ is continuous and open.

Proof.Choose a morphism $p : U \to \mathcal{X}$ which is surjective, flat, and locally of finite presentation with $U$ an algebraic space. Such exist by the definition of an algebraic stack, as a smooth morphism is flat and locally of finite presentation (see Morphisms of Spaces, Lemmas 55.36.5 and 55.36.7). We define a topology on $|\mathcal{X}|$ by the rule: $W \subset |\mathcal{X}|$ is open if and only if $|p|^{-1}(W)$ is open in $|U|$. To show that this is independent of the choice of $p$, let $p' : U' \to \mathcal{X}$ be another morphism which is surjective, flat, locally of finite presentation from an algebraic space to $\mathcal{X}$. Set $U'' = U \times_\mathcal{X} U'$ so that we have a $2$-commutative diagram $$ \xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & \mathcal{X} } $$ As $U \to \mathcal{X}$ and $U' \to \mathcal{X}$ are surjective, flat, locally of finite presentation we see that $U'' \to U'$ and $U'' \to U$ are surjective, flat and locally of finite presentation, see Lemma 85.3.2. Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open and surjective, see Morphisms of Spaces, Definition 55.5.2 and Lemma 55.29.6. This clearly implies that our definition is independent of the choice of $p : U \to \mathcal{X}$.Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. By Algebraic Stacks, Lemma 79.15.1 we can find a $2$-commutative diagram $$ \xymatrix{ U \ar[d]_x \ar[r]_a & V \ar[d]^y \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } $$ with surjective smooth vertical arrows. Consider the associated commutative diagram $$ \xymatrix{ |U| \ar[d]_{|x|} \ar[r]_{|a|} & |V| \ar[d]^{|y|} \\ |\mathcal{X}| \ar[r]^{|f|} & |\mathcal{Y}| } $$ of sets. If $W \subset |\mathcal{Y}|$ is open, then by the definition above this means exactly that $|y|^{-1}(W)$ is open in $|V|$. Since $|a|$ is continuous we conclude that $|a|^{-1}|y|^{-1}(W) = |x|^{-1}|f|^{-1}(W)$ is open in $|W|$ which means by definition that $|f|^{-1}(W)$ is open in $|\mathcal{X}|$. Thus $|f|$ is continuous.

Finally, we have to show that if $U$ is an algebraic space, and $U \to \mathcal{X}$ is flat and locally of finite presentation, then $|U| \to |\mathcal{X}|$ is open. Let $V \to \mathcal{X}$ be surjective, flat, and locally of finite presentation with $V$ an algebraic space. Consider the commutative diagram $$ \xymatrix{ |U \times_\mathcal{X} V| \ar[r]_e \ar[rd]_f & |U| \times_{|\mathcal{X}|} |V| \ar[d]_c \ar[r]_d & |V| \ar[d]^b \\ & |U| \ar[r]^a & |\mathcal{X}| } $$ Now the morphism $U \times_\mathcal{X} V \to U$ is surjective, i.e, $f : |U \times_\mathcal{X} V| \to |U|$ is surjective. The left top horizontal arrow is surjective, see Lemma 85.4.3. The morphism $U \times_\mathcal{X} V \to V$ is flat and locally of finite presentation, hence $d \circ e : |U \times_\mathcal{X} V| \to |V|$ is open, see Morphisms of Spaces, Lemma 55.29.6. Pick $W \subset |U|$ open. The properties above imply that $b^{-1}(a(W)) = (d \circ e)(f^{-1}(W))$ is open, which by construction means that $a(W)$ is open as desired. $\square$

The code snippet corresponding to this tag is a part of the file `stacks-properties.tex` and is located in lines 934–945 (see updates for more information).

```
\begin{lemma}
\label{lemma-topology-points}
There exists a unique topology on the sets of points
of algebraic stacks with the following properties:
\begin{enumerate}
\item for every morphism of algebraic stacks $\mathcal{X} \to \mathcal{Y}$
the map $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous, and
\item for every morphism $U \to \mathcal{X}$ which is flat and locally
of finite presentation with $U$ an algebraic space
the map of topological spaces $|U| \to |\mathcal{X}|$ is continuous and open.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a morphism $p : U \to \mathcal{X}$ which is
surjective, flat, and locally of finite presentation
with $U$ an algebraic space. Such exist by the definition of an algebraic
stack, as a smooth morphism is flat and locally of finite presentation
(see
Morphisms of Spaces,
Lemmas \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation} and
\ref{spaces-morphisms-lemma-smooth-flat}).
We define a topology on $|\mathcal{X}|$ by the rule:
$W \subset |\mathcal{X}|$ is open if and only if $|p|^{-1}(W)$ is open
in $|U|$. To show that this is independent of the choice of $p$, let
$p' : U' \to \mathcal{X}$ be another morphism which is surjective, flat,
locally of finite presentation from an algebraic space to
$\mathcal{X}$. Set $U'' = U \times_\mathcal{X} U'$
so that we have a $2$-commutative diagram
$$
\xymatrix{
U'' \ar[r] \ar[d] & U' \ar[d] \\
U \ar[r] & \mathcal{X}
}
$$
As $U \to \mathcal{X}$ and $U' \to \mathcal{X}$ are surjective, flat,
locally of finite presentation we see that $U'' \to U'$ and $U'' \to U$
are surjective, flat and locally of finite presentation, see
Lemma \ref{lemma-property-spaces-too}.
Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open
and surjective, see
Morphisms of Spaces,
Definition \ref{spaces-morphisms-definition-surjective} and
Lemma \ref{spaces-morphisms-lemma-fppf-open}.
This clearly implies that our definition is independent of the choice
of $p : U \to \mathcal{X}$.
\medskip\noindent
Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
By
Algebraic Stacks, Lemma \ref{algebraic-lemma-lift-morphism-presentations}
we can find a $2$-commutative diagram
$$
\xymatrix{
U \ar[d]_x \ar[r]_a & V \ar[d]^y \\
\mathcal{X} \ar[r]^f & \mathcal{Y}
}
$$
with surjective smooth vertical arrows.
Consider the associated commutative diagram
$$
\xymatrix{
|U| \ar[d]_{|x|} \ar[r]_{|a|} & |V| \ar[d]^{|y|} \\
|\mathcal{X}| \ar[r]^{|f|} & |\mathcal{Y}|
}
$$
of sets. If $W \subset |\mathcal{Y}|$ is open, then by the definition
above this means exactly that $|y|^{-1}(W)$ is open in $|V|$. Since
$|a|$ is continuous we conclude that
$|a|^{-1}|y|^{-1}(W) = |x|^{-1}|f|^{-1}(W)$ is open in $|W|$ which means
by definition that $|f|^{-1}(W)$ is open in $|\mathcal{X}|$.
Thus $|f|$ is continuous.
\medskip\noindent
Finally, we have to show that if $U$ is an algebraic space, and
$U \to \mathcal{X}$ is flat and locally of finite presentation, then
$|U| \to |\mathcal{X}|$ is open. Let $V \to \mathcal{X}$ be surjective,
flat, and locally of finite presentation with $V$ an algebraic space.
Consider the commutative diagram
$$
\xymatrix{
|U \times_\mathcal{X} V| \ar[r]_e \ar[rd]_f &
|U| \times_{|\mathcal{X}|} |V| \ar[d]_c \ar[r]_d &
|V| \ar[d]^b \\
& |U| \ar[r]^a & |\mathcal{X}|
}
$$
Now the morphism $U \times_\mathcal{X} V \to U$ is surjective, i.e,
$f : |U \times_\mathcal{X} V| \to |U|$ is surjective.
The left top horizontal arrow is surjective, see
Lemma \ref{lemma-points-cartesian}.
The morphism $U \times_\mathcal{X} V \to V$ is flat and locally of finite
presentation, hence $d \circ e : |U \times_\mathcal{X} V| \to |V|$ is open,
see
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fppf-open}.
Pick $W \subset |U|$ open. The properties above imply that
$b^{-1}(a(W)) = (d \circ e)(f^{-1}(W))$ is open, which by construction means
that $a(W)$ is open as desired.
\end{proof}
```

## Comments (0)

## Add a comment on tag `04XL`

Your email address will not be published. Required fields are marked.

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.

There are no comments yet for this tag.