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
\[ \xymatrix{ \mathop{\mathrm{Spec}}(\Omega ) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(L) \ar[d]^ q \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X}. } \]
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 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
\[ \xymatrix{ & \mathop{\mathrm{Spec}}(\Omega '') \ar[ld]_ c \ar[d]^{q'} \ar[rd]^{c'} \\ \mathop{\mathrm{Spec}}(\Omega ) \ar[r]_ b \ar[d]_ a & \mathop{\mathrm{Spec}}(L) \ar[d]^ q & \mathop{\mathrm{Spec}}(\Omega ') \ar[l]^{b'} \ar[d]^{a'} \\ \mathop{\mathrm{Spec}}(K) \ar[r]^ p & \mathcal{X} & \mathop{\mathrm{Spec}}(K') \ar[l]_{p'} } \]
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 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
\[ \xymatrix{ \mathcal{X} \ar[d] \ar[r] & \mathcal{Y} \ar[d] \\ \mathcal{W} \ar[r] & \mathcal{Z} } \]
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 100.4.3. Let be a fibre product of algebraic stacks. Then the map of sets of points is surjective.
\[ \xymatrix{ \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Z} \ar[r] & \mathcal{Y} } \]
\[ |\mathcal{Z} \times _\mathcal {Y} \mathcal{X}| \longrightarrow |\mathcal{Z}| \times _{|\mathcal{Y}|} |\mathcal{X}| \]
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
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \times _{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \mathop{\mathrm{Spec}}(K) \ar@{=}[r] & \mathop{\mathrm{Spec}}(K) \ar[r]^-y & \mathcal{Y} } \]
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 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
\[ |R| \longrightarrow |U| \times _{|\mathcal{X}|} |U| \]
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 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
\[ \xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & \mathcal{X} } \]
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 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
\[ \xymatrix{ U \ar[d]_ x \ar[r]_ a & V \ar[d]^ y \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } \]
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 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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