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 $K \subset M$ 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 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 $K \subset K'$ 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

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.
$\square$

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