The Stacks project

98.4 Points of algebraic stacks

Let $\mathcal{X}$ be an algebraic stack. Let $K, L$ be two fields and let $p : \mathop{\mathrm{Spec}}(K) \to \mathcal{X}$ and $q : \mathop{\mathrm{Spec}}(L) \to \mathcal{X}$ be morphisms. We say that $p$ and $q$ are equivalent if there exists a field $\Omega $ and a $2$-commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(L) \ar[d]^ q \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X}. } \]

Lemma 98.4.1. The notion above does indeed define an equivalence relation on morphisms from spectra of fields into the algebraic stack $\mathcal{X}$.

Proof. It is clear that the relation is reflexive and symmetric. Hence we have to prove that it is transitive. This comes down to the following: Given a diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r]_ b \ar[d]_ a & \mathop{\mathrm{Spec}}(L) \ar[d]^ q & \mathop{\mathrm{Spec}}(\Omega ') \ar[l]^{b'} \ar[d]^{a'} \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X} & \mathop{\mathrm{Spec}}(K') \ar[l]_{p'} } \]

with both squares $2$-commutative we have to show that $p$ is equivalent to $p'$. By the $2$-Yoneda lemma (see Algebraic Stacks, Section 92.5) the morphisms $p$, $p'$, and $q$ are given by objects $x$, $x'$, and $y$ in the fibre categories of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(K)$, $\mathop{\mathrm{Spec}}(K')$, and $\mathop{\mathrm{Spec}}(L)$. The $2$-commutativity of the squares means that there are isomorphisms $\alpha : a^*x \to b^*y$ and $\alpha ' : (a')^*x' \to (b')^*y$ in the fibre categories of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(\Omega )$ and $\mathop{\mathrm{Spec}}(\Omega ')$. Choose any field $\Omega ''$ and embeddings $\Omega \to \Omega ''$ and $\Omega ' \to \Omega ''$ agreeing on $L$. Then we can extend the diagram above to

\[ \xymatrix{ & \mathop{\mathrm{Spec}}(\Omega '') \ar[ld]_ c \ar[d]^{q'} \ar[rd]^{c'} \\ \mathop{\mathrm{Spec}}(\Omega ) \ar[r]_ b \ar[d]_ a & \mathop{\mathrm{Spec}}(L) \ar[d]^ q & \mathop{\mathrm{Spec}}(\Omega ') \ar[l]^{b'} \ar[d]^{a'} \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X} & \mathop{\mathrm{Spec}}(K') \ar[l]_{p'} } \]

with commutative triangles and

\[ (q')^*(\alpha ')^{-1} \circ (q')^*\alpha : (a \circ c)^*x \longrightarrow (a' \circ c')^*x' \]

is an isomorphism in the fibre category over $\mathop{\mathrm{Spec}}(\Omega '')$. Hence $p$ is equivalent to $p'$ as desired. $\square$

Definition 98.4.2. Let $\mathcal{X}$ be an algebraic stack. A point of $\mathcal{X}$ is an equivalence class of morphisms from spectra of fields into $\mathcal{X}$. The set of points of $\mathcal{X}$ is denoted $|\mathcal{X}|$.

This agrees with our definition of points of algebraic spaces, see Properties of Spaces, Definition 64.4.1. Moreover, for a scheme we recover the usual notion of points, see Properties of Spaces, Lemma 64.4.2. If $f : \mathcal{X} \to \mathcal{Y}$ is a morphism of algebraic stacks then there is an induced map $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ which maps a representative $x : \mathop{\mathrm{Spec}}(K) \to \mathcal{X}$ to the representative $f \circ x : \mathop{\mathrm{Spec}}(K) \to \mathcal{Y}$. This is well defined: namely $2$-isomorphic $1$-morphisms remain $2$-isomorphic after pre- or post-composing by a $1$-morphism because you can horizontally pre- or post-compose by the identity of the given $1$-morphism. This holds in any (strict) $(2, 1)$-category. If

\[ \xymatrix{ \mathcal{X} \ar[d] \ar[r] & \mathcal{Y} \ar[d] \\ \mathcal{W} \ar[r] & \mathcal{Z} } \]

is a $2$-commutative diagram of algebraic stacks, then the diagram of sets

\[ \xymatrix{ |\mathcal{X}| \ar[d] \ar[r] & |\mathcal{Y}| \ar[d] \\ |\mathcal{W}| \ar[r] & |\mathcal{Z}| } \]

is commutative. In particular, if $\mathcal{X} \to \mathcal{Y}$ is an equivalence then $|\mathcal{X}| \to |\mathcal{Y}|$ is a bijection.

Lemma 98.4.3. Let

\[ \xymatrix{ \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Z} \ar[r] & \mathcal{Y} } \]

be a fibre product of algebraic stacks. Then the map of sets of points

\[ |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \longrightarrow |\mathcal{Z}| \times _{|\mathcal{Y}|} |\mathcal{X}| \]

is surjective.

Proof. Namely, suppose given fields $K$, $L$ and morphisms $\mathop{\mathrm{Spec}}(K) \to \mathcal{X}$, $\mathop{\mathrm{Spec}}(L) \to \mathcal{Z}$, then the assumption that they agree as elements of $|\mathcal{Y}|$ means that there is a common extension $K \subset M$ and $L \subset M$ such that $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(K) \to \mathcal{X} \to \mathcal{Y}$ and $\mathop{\mathrm{Spec}}(M) \to \mathop{\mathrm{Spec}}(L) \to \mathcal{Z} \to \mathcal{Y}$ are $2$-isomorphic. And this is exactly the condition which says you get a morphism $\mathop{\mathrm{Spec}}(M) \to \mathcal{Z} \times _\mathcal {Y} \mathcal{X}$. $\square$

Lemma 98.4.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent:

  1. $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is surjective, and

  2. $f$ is surjective (in the sense of Section 98.3).

Proof. Assume (1). Let $T \to \mathcal{Y}$ be a morphism whose source is a scheme. To prove (2) we have to show that the morphism of algebraic spaces $T \times _\mathcal {Y} \mathcal{X} \to T$ is surjective. By Morphisms of Spaces, Definition 65.5.2 this means we have to show that $|T \times _\mathcal {Y} \mathcal{X}| \to |T|$ is surjective. Applying Lemma 98.4.3 we see that this follows from (1).

Conversely, assume (2). Let $y : \mathop{\mathrm{Spec}}(K) \to \mathcal{Y}$ be a morphism from the spectrum of a field into $\mathcal{Y}$. By assumption the morphism $\mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \to \mathop{\mathrm{Spec}}(K)$ of algebraic spaces is surjective. By Morphisms of Spaces, Definition 65.5.2 this means there exists a field extension $K \subset K'$ and a morphism $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X}$ such that the left square of the diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-y & \mathcal{Y} } \]

is commutative. This shows that $|X| \to |\mathcal{Y}|$ is surjective. $\square$

Here is a lemma explaining how to compute the set of points in terms of a presentation.

Lemma 98.4.5. Let $\mathcal{X}$ be an algebraic stack. Let $\mathcal{X} = [U/R]$ be a presentation of $\mathcal{X}$, see Algebraic Stacks, Definition 92.16.5. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|\mathcal{X}|$ is the quotient of $|U|$ by this equivalence relation.

Proof. The assumption means that we have a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces, and an equivalence $f : [U/R] \to \mathcal{X}$. We may assume $\mathcal{X} = [U/R]$. The induced morphism $p : U \to \mathcal{X}$ is smooth and surjective, see Algebraic Stacks, Lemma 92.17.2. Hence $|U| \to |\mathcal{X}|$ is surjective by Lemma 98.4.4. Note that $R = U \times _\mathcal {X} U$, see Groupoids in Spaces, Lemma 76.21.2. Hence Lemma 98.4.3 implies the map

\[ |R| \longrightarrow |U| \times _{|\mathcal{X}|} |U| \]

is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|\mathcal{X}|$. Combining these two statements we get the result of the lemma. $\square$

Remark 98.4.6. The result of Lemma 98.4.5 can be generalized as follows. Let $\mathcal{X}$ be an algebraic stack. Let $U$ be an algebraic space and let $f : U \to \mathcal{X}$ be a surjective morphism (which makes sense by Section 98.3). Let $R = U \times _\mathcal {X} U$, let $(U, R, s, t, c)$ be the groupoid in algebraic spaces, and let $f_{can} : [U/R] \to \mathcal{X}$ be the canonical morphism as constructed in Algebraic Stacks, Lemma 92.16.1. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|\mathcal{X}| = |U|/|R|$. The proof of Lemma 98.4.5 works without change. (Of course in general $[U/R]$ is not an algebraic stack, and in general $f_{can}$ is not an isomorphism.)

Lemma 98.4.7. 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 65.37.5 and 65.37.7). 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 98.3.2. Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open and surjective, see Morphisms of Spaces, Definition 65.5.2 and Lemma 65.30.6. 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 92.15.1 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 98.4.3. 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 65.30.6. 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. $\square$

Definition 98.4.8. Let $\mathcal{X}$ be an algebraic stack. The underlying topological space of $\mathcal{X}$ is the set of points $|\mathcal{X}|$ endowed with the topology constructed in Lemma 98.4.7.

This definition does not conflict with the already existing topology on $|\mathcal{X}|$ if $\mathcal{X}$ is an algebraic space.

Lemma 98.4.9. Let $\mathcal{X}$ be an algebraic stack. Every point of $|\mathcal{X}|$ has a fundamental system of quasi-compact open neighbourhoods. In particular $|\mathcal{X}|$ is locally quasi-compact in the sense of Topology, Definition 5.13.1.

Proof. This follows formally from the fact that there exists a scheme $U$ and a surjective, open, continuous map $U \to |\mathcal{X}|$ of topological spaces. Namely, if $U \to \mathcal{X}$ is surjective and smooth, then Lemma 98.4.7 guarantees that $|U| \to |\mathcal{X}|$ is continuous, surjective, and open. $\square$


Comments (2)

Comment #5391 by Julian von Abele on

To be clear, should we add the condition to Lemma 98.4.7 that the topology on the set of points of (the algebraic stack associated to) an algebraic space agrees with the definition in 64.4.6? This was not explicitly stated in the definition of the topology

Comment #5624 by on

On the one hand, you are right. But as I said in comment #4396 on Lemma 64.4.6 I think this time it is clear enough due to the typographic difference between algebraic spaces and algebraic stacks in the statement of the lemma. But if more people second your suggestion, then I will change it.


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