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. $\square$
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