Lemma 99.4.1. The notion above does indeed define an equivalence relation on morphisms from spectra of fields into the algebraic stack $\mathcal{X}$.

## 99.4 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

**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

with both squares $2$-commutative we have to show that $p$ is equivalent to $p'$. By the $2$-Yoneda lemma (see Algebraic Stacks, Section 93.5) 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

with commutative triangles and

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

Definition 99.4.2. 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 65.4.1. Moreover, for a scheme we recover the usual notion of points, see Properties of Spaces, Lemma 65.4.2. 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

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

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

Lemma 99.4.3. Let

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

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 $M/K$ and $M/L$ 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 99.4.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent:

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

$f$ is surjective (in the sense of Section 99.3).

**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 66.5.2 this means we have to show that $|T \times _\mathcal {Y} \mathcal{X}| \to |T|$ is surjective. Applying Lemma 99.4.3 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 66.5.2 this means there exists a field extension $K'/K$ 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

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 99.4.5. Let $\mathcal{X}$ be an algebraic stack. Let $\mathcal{X} = [U/R]$ be a presentation of $\mathcal{X}$, see Algebraic Stacks, Definition 93.16.5. 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 93.17.2. Hence $|U| \to |\mathcal{X}|$ is surjective by Lemma 99.4.4. Note that $R = U \times _\mathcal {X} U$, see Groupoids in Spaces, Lemma 77.22.2. Hence Lemma 99.4.3 implies the map

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 99.4.6. The result of Lemma 99.4.5 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 99.3). 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 93.16.1. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|\mathcal{X}| = |U|/|R|$. The proof of Lemma 99.4.5 works without change. (Of course in general $[U/R]$ is not an algebraic stack, and in general $f_{can}$ is not an isomorphism.)

Lemma 99.4.7. 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 66.37.5 and 66.37.7). 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

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 99.3.2. Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open and surjective, see Morphisms of Spaces, Definition 66.5.2 and Lemma 66.30.6. 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 93.15.1 we can find a $2$-commutative diagram

with surjective smooth vertical arrows. Consider the associated commutative diagram

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

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 99.4.3. 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 66.30.6. 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 99.4.8. 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 99.4.7.

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

Lemma 99.4.9. 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 99.4.7 guarantees that $|U| \to |\mathcal{X}|$ is continuous, surjective, and open.
$\square$

## Post a comment

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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #5391 by Julian von Abele on

Comment #5624 by Johan on