The Stacks Project


Tag 04XL

Chapter 77: Properties of Algebraic Stacks > Section 77.4: Points of algebraic stacks

Lemma 77.4.7. There exists a unique topology on the sets of points of algebraic stacks with the following properties:

  1. for every morphism of algebraic stacks $\mathcal{X} \to \mathcal{Y}$ the map $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous, and
  2. 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 49.34.5 and 49.34.7). 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 77.3.2. Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open and surjective, see Morphisms of Spaces, Definition 49.5.2 and Lemma 49.28.6. 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 71.15.1 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 77.4.3. 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 49.28.6. 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 926–937 (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}

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