Lemma 100.4.1. The notion above does indeed define an equivalence relation on morphisms from spectra of fields into the algebraic stack \mathcal{X}.
100.4 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 94.5) 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 100.4.2. 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 66.4.1. Moreover, for a scheme we recover the usual notion of points, see Properties of Spaces, Lemma 66.4.2. 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 100.4.3. 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 100.4.4. 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 100.3).
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 67.5.2 this means we have to show that |T \times _\mathcal {Y} \mathcal{X}| \to |T| is surjective. Applying Lemma 100.4.3 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 67.5.2 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 100.4.5. Let \mathcal{X} be an algebraic stack. Let \mathcal{X} = [U/R] be a presentation of \mathcal{X}, see Algebraic Stacks, Definition 94.16.5. 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 94.17.2. Hence |U| \to |\mathcal{X}| is surjective by Lemma 100.4.4. Note that R = U \times _\mathcal {X} U, see Groupoids in Spaces, Lemma 78.22.2. Hence Lemma 100.4.3 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 100.4.6. The result of Lemma 100.4.5 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 100.3). 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 94.16.1. Then the image of |R| \to |U| \times |U| is an equivalence relation and |\mathcal{X}| = |U|/|R|. The proof of Lemma 100.4.5 works without change. (Of course in general [U/R] is not an algebraic stack, and in general f_{can} is not an isomorphism.)
Lemma 100.4.7. 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 67.37.5 and 67.37.7). 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 100.3.2. Hence the maps |U''| \to |U'| and |U''| \to |U| are continuous, open and surjective, see Morphisms of Spaces, Definition 67.5.2 and Lemma 67.30.6. 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 94.15.1 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 100.4.3. 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 67.30.6. 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 100.4.8. 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 100.4.7.
This definition does not conflict with the already existing topology on |\mathcal{X}| if \mathcal{X} is an algebraic space.
Lemma 100.4.9. 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 100.4.7 guarantees that |U| \to |\mathcal{X}| is continuous, surjective, and open. \square
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