hcju/stacksqa huggingface dataset:
Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times_\mathcal{Z} \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times_\mathcal{Y} \mathcal{X}$. We know that $U \to V$ is flat and locally of finite presentation and that $U \to W$ is locally of finite presentation. Also, as $\mathcal{X} \to \mathcal{Y}$ is surjective we see that $U \to V$ is surjective (as a composition of surjective morphisms). Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma \ref{spaces-descent-lemma-locally-finite-presentation-fppf-local-source}.
We may think of $y$ as a morphism $U \to \mathcal{Y}$. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-base-change-surjective} and Lemmas \ref{lemma-base-change-finite-presentation} and \ref{lemma-base-change-flat} we see that $\mathcal{X} \times_\mathcal{Y} U \to U$ is surjective, flat, and locally of finite presentation. Let $V$ be a scheme and let $V \to \mathcal{X} \times_\mathcal{Y} U$ smooth and surjective. Then $V \to \mathcal{X} \times_\mathcal{Y} U$ is also surjective, flat, and locally of finite presentation (see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-smooth-flat} and \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation}). Hence also $V \to U$ is surjective, flat, and locally of finite presentation, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-composition-surjective} and Lemmas \ref{lemma-composition-finite-presentation}, and \ref{lemma-composition-flat}. Hence $\{V \to U\}$ is the desired fppf covering and $x : V \to \mathcal{X}$ is the desired object.
Apply Lemma \ref{lemma-surjective-flat-locally-finite-presentation} to the morphism $\coprod_{j \in J} \mathcal{X}_j \to \mathcal{X}$. (There is a slight set theoretic issue here -- due to our setup of things -- which we ignore.) To finish, note that a morphism $x_i : U_i \to \coprod_{j \in J} \mathcal{X}_j$ is given by a disjoint union decomposition $U_i = \coprod U_{i, j}$ and morphisms $U_{i, j} \to \mathcal{X}_j$. Then the fppf covering $\{U_{i, j} \to U\}$ and the morphisms $U_{i, j} \to \mathcal{X}_j$ do the job.
Choose a scheme $V$ and a smooth surjective morphism $V \to \mathcal{Y}$. Choose a scheme $U$ and a smooth surjective morphism $U \to V \times_\mathcal{Y} \mathcal{X}$. By assumption the morphism of schemes $U \to V$ is flat and locally of finite presentation. Hence $U \to V$ is open by Morphisms, Lemma \ref{morphisms-lemma-fppf-open}. By construction of the topology on $|\mathcal{Y}|$ the map $|V| \to |\mathcal{Y}|$ is open. The map $|U| \to |\mathcal{X}|$ is surjective. The result follows from these facts by elementary topology.
We have to show that given $\mathcal{Y}' \to \mathcal{Y}$ with $\mathcal{Y}'$ quasi-compact, we have $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$ is quasi-compact. Denote $\mathcal{Z}' = \mathcal{Z} \times_\mathcal{Y} \mathcal{Y}'$. Then $|\mathcal{Z}'| \to |\mathcal{Y}'|$ is open, see Lemma \ref{lemma-fppf-open}. Hence we can find a quasi-compact open substack $\mathcal{W} \subset \mathcal{Z}'$ mapping onto $\mathcal{Y}'$. Because $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}$ is quasi-compact, we know that $$ \mathcal{W} \times_\mathcal{Z} \mathcal{Z} \times_\mathcal{Y} \mathcal{X} = \mathcal{W} \times_\mathcal{Y} \mathcal{X} $$ is quasi-compact. And the map $\mathcal{W} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}' \times_\mathcal{Y} \mathcal{X}$ is surjective, hence we win. Some details omitted.
Consider the diagram $$ \xymatrix{ \mathcal{X} \ar[r]_\Delta \ar[rd] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times_\mathcal{Z} \mathcal{X} \ar[d] \\ & \mathcal{Y} \ar[r] & \mathcal{Y} \times_\mathcal{Z} \mathcal{Y} } $$ The square is cartesian. We have to show the bottom horizontal arrow is proper. We already know that it is representable by algebraic spaces and locally of finite type (Lemma \ref{lemma-properties-diagonal}). Since the right vertical arrow is surjective, flat, and locally of finite presentation it suffices to show the top right horizontal arrow is proper (Lemma \ref{lemma-check-property-after-fppf-base-change}). Since $h$ is separated, the composition of the top horizontal arrows is proper. \medskip\noindent Since $f$ is universally injective $\Delta$ is surjective (Lemma \ref{lemma-universally-injective}). Since the composition of $\Delta$ with the projection $\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}$ is the identity, we see that $\Delta$ is universally closed. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-image-universally-closed-separated} we conclude that $\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}$ is separated as $\mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}$ is separated. Here we use that implications between properties of morphisms of algebraic spaces can be transferred to the same implications between properties of morphisms of algebraic stacks representable by algebraic spaces; this is discussed in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Finally, we use the same principle to conlude that $\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}$ is proper from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-image-proper-is-proper}.
(In this proof the abuse of language introduced in Section \ref{section-conventions} really pays off.) Choose a morphism $\pi : \mathcal{X} \to X$ where $X$ is an algebraic space which turns $\mathcal{X}$ into a gerbe over $X$. It suffices to prove that $X$ is the sheafification of the presheaf $\mathcal{F}$ displayed in the lemma. It is clear that there is a map $c : \mathcal{F} \to X$. We will use Stacks, Lemma \ref{stacks-lemma-when-gerbe} properties (2)(a) and (2)(b) to see that the map $c^\# : \mathcal{F}^\# \to X$ is surjective and injective, hence an isomorphism, see Sites, Lemma \ref{sites-lemma-mono-epi-sheaves}. Surjective: Let $T$ be a scheme and let $f : T \to X$. By property (2)(a) there exists an fppf covering $\{h_i : T_i \to T\}$ and morphisms $x_i : T_i \to \mathcal{X}$ such that $f \circ h_i$ corresponds to $\pi \circ x_i$. Hence we see that $f|_{T_i}$ is in the image of $c$. Injective: Let $T$ be a scheme and let $x, y : T \to \mathcal{X}$ be morphisms such that $c \circ x = c \circ y$. By (2)(b) we can find a covering $\{T_i \to T\}$ and morphisms $x|_{T_i} \to y|_{T_i}$ in the fibre category $\mathcal{X}_{T_i}$. Hence the restrictions $x|_{T_i}, y|_{T_i}$ are equal in $\mathcal{F}(T_i)$. This proves that $x, y$ give the same section of $\mathcal{F}^\#$ over $T$ as desired.
Set $R = U \times_{x, \mathcal{X}, x} U$. The morphism $R \to U \times U$ factors through the diagonal $\Delta_U : U \to U \times U$ as it factors through $U \times_U U = U$. Hence $R = G$ because \begin{align*} G & = \mathit{Isom}_\mathcal{X}(x, x) \\ & = U \times_{x, \mathcal{X}} \mathcal{I}_\mathcal{X} \\ & = U \times_{x, \mathcal{X}} (\mathcal{X} \times_{\Delta, \mathcal{X} \times_S \mathcal{X}, \Delta} \mathcal{X}) \\ & = (U \times_{x, \mathcal{X}, x} U) \times_{U \times U, \Delta_U} U \\ & = R \times_{U \times U, \Delta_U} U \\ & = R \end{align*} for the fourth equality use Categories, Lemma \ref{categories-lemma-diagonal-2}. Let $t, s : R \to U$ be the projections. The composition law $c : R \times_{s, U, t} R \to R$ constructed on $R$ in Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack} agrees with the group law on $G$ (proof omitted). Thus Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack} shows we obtain a canonical fully faithful $1$-morphism $$ x_{can} : [U/G] \longrightarrow \mathcal{X} $$ of stacks in groupoids over $(\Sch/S)_{fppf}$. To see that it is an equivalence it suffices to show that it is essentially surjective. To do this it suffices to show that any object of $\mathcal{X}$ over a scheme $T$ comes fppf locally from $x$ via a morphism $T \to U$, see Stacks, Lemma \ref{stacks-lemma-characterize-essentially-surjective-when-ff}. However, this follows the condition that $\pi$ turns $\mathcal{X}$ into a gerbe over $U$, see property (2)(a) of Stacks, Lemma \ref{stacks-lemma-when-gerbe}. \medskip\noindent By Criteria for Representability, Lemma \ref{criteria-lemma-BG-algebraic} we conclude that $G \to U$ is flat and locally of finite presentation. Finally, $U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation by Criteria for Representability, Lemma \ref{criteria-lemma-flat-quotient-flat-presentation}.
Assume (2). By Lemma \ref{lemma-gerbe-descent} to prove (1) it suffices to show that $[U/G]$ is a gerbe over $U$. This is immediate from Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-group-quotient-gerbe}. \medskip\noindent Assume (1). Any base change of $\pi$ is a gerbe, see Lemma \ref{lemma-base-change-gerbe}. As a first step we choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Thus we may assume that $\pi : \mathcal{X} \to V$ is a gerbe over a scheme. This means that there exists an fppf covering $\{V_i \to V\}$ such that the fibre category $\mathcal{X}_{V_i}$ is nonempty, see Stacks, Lemma \ref{stacks-lemma-when-gerbe} (2)(a). Note that $U = \coprod V_i \to V$ is surjective, flat, and locally of finite presentation. Hence we may replace $V$ by $U$ and assume that $\pi : \mathcal{X} \to U$ is a gerbe over a scheme $U$ and that there exists an object $x$ of $\mathcal{X}$ over $U$. By Lemma \ref{lemma-gerbe-with-section} we see that $\mathcal{X} = [U/G]$ over $U$ for some flat and locally finitely presented group algebraic space $G$ over $U$.
By Properties of Stacks, Lemma \ref{stacks-properties-lemma-descent-surjective} and Lemmas \ref{lemma-descent-flat} and \ref{lemma-descent-finite-presentation} it suffices to prove to the lemma after replacing $\pi$ by a base change with a surjective, flat, locally finitely presented morphism $\mathcal{Y}' \to \mathcal{Y}$. By Lemma \ref{lemma-local-structure-gerbe} we may assume $\mathcal{Y} = U$ is an algebraic space and $\mathcal{X} = [U/G]$ over $U$. Then $U \to [U/G]$ is surjective, flat, and locally of finite presentation, see Lemma \ref{lemma-gerbe-with-section}. This implies that $\pi$ is surjective, flat, and locally of finite presentation by Properties of Stacks, Lemma \ref{stacks-properties-lemma-surjective-permanence} and Lemmas \ref{lemma-flat-permanence} and \ref{lemma-flat-finite-presentation-permanence}.
Proof of (1). Choose a scheme $Y$ and a surjective smooth morphism $Y \to \mathcal{Y}$. Set $\mathcal{X}' = \mathcal{X} \times_\mathcal{Y} Y$. By Lemma \ref{lemma-cartesian-square-inertia} we obtain cartesian squares $$ \xymatrix{ \mathcal{I}_{\mathcal{X}'} \ar[r] \ar[d] & \mathcal{X}' \ar[r] \ar[d] & Y \ar[d] \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r] & \mathcal{Y} } $$ By Lemmas \ref{lemma-descent-flat} and \ref{lemma-descent-finite-presentation} it suffices to prove that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is flat and locally of finite presentation. This follows from Proposition \ref{proposition-when-gerbe} (because $\mathcal{X}'$ is a gerbe over $Y$ by Lemma \ref{lemma-base-change-gerbe}). \medskip\noindent Proof of (2). With notation as above, note that we may assume that $\mathcal{X}' = [Y/G]$ for some group algebraic space $G$ flat and locally of finite presentation over $Y$, see Lemma \ref{lemma-local-structure-gerbe}. The base change of the morphism $\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ over $\mathcal{Y}$ by the morphism $Y \to \mathcal{Y}$ is the morphism $\Delta' : \mathcal{X}' \to \mathcal{X}' \times_Y \mathcal{X}'$. Hence it suffices to show that $\Delta'$ is surjective, flat, and locally of finite presentation (see Lemmas \ref{lemma-descent-flat} and \ref{lemma-descent-finite-presentation}). In other words, we have to show that $$ [Y/G] \longrightarrow [Y/G \times_Y G] $$ is surjective, flat, and locally of finite presentation. This is true because the base change by the surjective, flat, locally finitely presented morphism $Y \to [Y/G \times_Y G]$ is the morphism $G \to Y$. \medskip\noindent Proof of (3). Observe that the diagram $$ \xymatrix{ T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times_{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} } $$ is cartesian. Hence (3) follows from (2). \medskip\noindent Proof of (4). This is true because $$ \mathit{Isom}_\mathcal{X}(x_1, x_2) = (T \times_{x_1, \mathcal{X}, x_2} T) \times_{T \times T, \Delta_T} T $$ hence the morphism in (4) is a base change of the morphism in (3).
By Properties of Stacks, Lemma \ref{stacks-properties-lemma-unique-point-better} there exists a surjective, flat, locally finitely presented morphism $\Spec(k) \to \mathcal{Z}$ where $k$ is a field. Then $\mathcal{I}_Z \times_\mathcal{Z} \Spec(k) \to \Spec(k)$ is representable by algebraic spaces and locally of finite type (as a base change of $\mathcal{I}_\mathcal{Z} \to \mathcal{Z}$, see Lemmas \ref{lemma-inertia} and \ref{lemma-base-change-finite-type}). Therefore it is locally of finite presentation, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-noetherian-finite-type-finite-presentation}. Of course it is also flat as $k$ is a field. Hence we may apply Lemmas \ref{lemma-descent-flat} and \ref{lemma-descent-finite-presentation} to see that $\mathcal{I}_\mathcal{Z} \to \mathcal{Z}$ is flat and locally of finite presentation. We conclude that $\mathcal{Z}$ is a gerbe by Proposition \ref{proposition-when-gerbe}. Let $\pi : \mathcal{Z} \to Z$ be a morphism to an algebraic space such that $\mathcal{Z}$ is a gerbe over $Z$. Then $\pi$ is surjective, flat, and locally of finite presentation by Lemma \ref{lemma-gerbe-fppf}. Hence $\Spec(k) \to Z$ is surjective, flat, and locally of finite presentation as a composition, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-composition-surjective} and Lemmas \ref{lemma-composition-flat} and \ref{lemma-composition-finite-presentation}. Hence by Properties of Stacks, Lemma \ref{stacks-properties-lemma-unique-point-better} we see that $|Z|$ is a singleton and that $Z$ is locally Noetherian and reduced.
By Lemma \ref{lemma-base-change-gerbe} the assumption on $f$ is preserved under base change. Hence it suffices to show that the map $|\mathcal{X}| \to |\mathcal{Y}|$ is a homeomorphism of topological spaces. Let $k$ be a field and let $y$ be an object of $\mathcal{Y}$ over $\Spec(k)$. By Stacks, Lemma \ref{stacks-lemma-when-gerbe} property (2)(a) there exists an fppf covering $\{T_i \to \Spec(k)\}$ and objects $x_i$ of $\mathcal{X}$ over $T_i$ with $f(x_i) \cong y|_{T_i}$. Choose an $i$ such that $T_i \not = \emptyset$. Choose a morphism $\Spec(K) \to T_i$ for some field $K$. Then $k \subset K$ and $x_i|_K$ is an object of $\mathcal{X}$ lying over $y|_K$. Thus we see that $|\mathcal{Y}| \to |\mathcal{X}|$. is surjective. The map $|\mathcal{Y}| \to |\mathcal{X}|$ is also injective. Namely, if $x, x'$ are objects of $\mathcal{X}$ over $\Spec(k)$ whose images $f(x), f(x')$ become isomorphic (over an extension) in $\mathcal{Y}$, then Stacks, Lemma \ref{stacks-lemma-when-gerbe} property (2)(b) guarantees the existence of an extension of $k$ over which $x$ and $x'$ become isomorphic (details omitted). Hence $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous and bijective and it suffices to show that it is also open. This follows from Lemmas \ref{lemma-gerbe-fppf} and \ref{lemma-fppf-open}.
Consider the diagram $$ \xymatrix{ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times \mathcal{X} \ar[d] \\ & \mathcal{Y} \ar[r] & \mathcal{Y} \times \mathcal{Y} } $$ By Proposition \ref{proposition-when-gerbe-over} we find that the arrow on the top left is surjective. Since the composition of the top horizontal arrows is quasi-compact, we conclude that the top right arrow is quasi-compact by Lemma \ref{lemma-surjection-from-quasi-compact}. The square is cartesian and the right vertical arrow is surjective, flat, and locally of finite presentation. Thus we conclude by Lemma \ref{lemma-descent-quasi-compact}.
Let $\pi : \mathcal{X} \to X$ be a morphism from $\mathcal{X}$ into an algebraic space $X$ which turns $\mathcal{X}$ into a gerbe over $X$. Let $Z_x \to X$ be the residual space of $X$ at $x$, see Decent Spaces, Definition \ref{decent-spaces-definition-residual-space}. Let $\mathcal{Z} = \mathcal{X} \times_X Z_x$. By Lemma \ref{lemma-base-change-gerbe} the algebraic stack $\mathcal{Z}$ is a gerbe over $Z_x$. Hence $|\mathcal{Z}| = |Z_x|$ (Lemma \ref{lemma-gerbe-bijection-points}) is a singleton. Since $\mathcal{Z} \to Z_x$ is locally of finite presentation as a base change of $\pi$ (see Lemmas \ref{lemma-gerbe-fppf} and \ref{lemma-base-change-finite-presentation}) we see that $\mathcal{Z}$ is locally Noetherian, see Lemma \ref{lemma-locally-finite-type-locally-noetherian}. Thus the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ at $x$ exists and is equal to $\mathcal{Z}_x = \mathcal{Z}_{red}$ the reduction of the algebraic stack $\mathcal{Z}$. Namely, we have seen above that $|\mathcal{Z}_{red}|$ is a singleton mapping to $x \in |\mathcal{X}|$, it is reduced by construction, and it is locally Noetherian (as the reduction of a locally Noetherian algebraic stack is locally Noetherian, details omitted).
Let $T \subset |\mathcal{X}|$ be a nonempty closed subset. We are going to find (resp.\ choose) for every such $T$ a reduced locally closed substack $\mathcal{U}(T) \subset \mathcal{X}$ with $|\mathcal{U}(T)| \subset T$ open dense (resp.\ nonempty quasi-compact). Namely, by Properties of Stacks, Lemma \ref{stacks-properties-lemma-reduced-closed-substack} there exists a unique reduced closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $T = |\mathcal{X}'|$. Note that $\mathcal{I}_{\mathcal{X}'} = \mathcal{I}_\mathcal{X} \times_\mathcal{X} \mathcal{X}'$ by Lemma \ref{lemma-monomorphism-cartesian-square-inertia}. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is quasi-compact as a base change, see Lemma \ref{lemma-base-change-quasi-compact}. Therefore Proposition \ref{proposition-open-stratum} implies there exists a dense maximal (see proof proposition) open substack $\mathcal{U} \subset \mathcal{X}'$ which is a gerbe. In case (a) we set $\mathcal{U}(T) = \mathcal{U}$ (this is canonical) and in case (b) we simply choose a nonempty quasi-compact open $\mathcal{U}(T) \subset \mathcal{U}$, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-space-locally-quasi-compact} (we can do this for all $T$ simultaneously by the axiom of choice). \medskip\noindent By transfinite induction we construct for every ordinal $\alpha$ a closed subset $T_\alpha \subset |\mathcal{X}|$. For $\alpha = 0$ we set $T_0 = |\mathcal{X}|$. Given $T_\alpha$ set $$ T_{\alpha + 1} = T_\alpha \setminus |\mathcal{U}(T_\alpha)|. $$ If $\beta$ is a limit ordinal we set $$ T_\beta = \bigcap\nolimits_{\alpha < \beta} T_\alpha. $$ We claim that $T_\alpha = \emptyset$ for all $\alpha$ large enough. Namely, assume that $T_\alpha \not = \emptyset$ for all $\alpha$. Then we obtain an injective map from the class of ordinals into the set of subsets of $|\mathcal{X}|$ which is a contradiction. \medskip\noindent The claim implies the lemma. Namely, let $$ I = \{\alpha \mid \mathcal{U}_\alpha \not = \emptyset \}. $$ This is a well-ordered set by the claim. For $i = \alpha \in I$ we set $\mathcal{U}_i = \mathcal{U}_\alpha$. So $\mathcal{U}_i$ is a reduced locally closed substack and a gerbe, i.e., (1) holds. By construction $T_i = T\alpha$ if $i = \alpha \in I$, hence (3) holds. Also, (4) and (a) or (b) hold by our choice of $\mathcal{U}(T)$ as well. Finally, to see (2) let $x \in |\mathcal{X}|$. There exists a smallest ordinal $\beta$ with $x \not \in T_\beta$ (because the ordinals are well-ordered). In this case $\beta$ has to be a successor ordinal by the definition of $T_\beta$ for limit ordinals. Hence $\beta = \alpha + 1$ and $x \in |\mathcal{U}(T_\alpha)|$ and we win.
Choose an affine scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack}. Then $|U| \to |\mathcal{X}|$ is continuous, open, and surjective, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}. Hence the quasi-compact opens of $|\mathcal{X}|$ form a basis for the topology. For $W_1, W_2 \subset |\mathcal{X}|$ quasi-compact open, we may choose a quasi-compact opens $V_1, V_2$ of $U$ mapping to $W_1$ and $W_2$. Since $\Delta$ is quasi-compact, we see that $$ V_1 \times_\mathcal{X} V_2 = (V_1 \times V_2) \times_{\mathcal{X} \times \mathcal{X}, \Delta} \mathcal{X} $$ is quasi-compact. Then image of $|V_1 \times_\mathcal{X} V_2|$ in $|\mathcal{X}|$ is $W_1 \cap W_2$ by Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian}. Thus $W_1 \cap W_2$ is quasi-compact. To finish the proof, it suffices to show that $|\mathcal{X}|$ is sober, see Topology, Definition \ref{topology-definition-spectral-space}. \medskip\noindent Let $T \subset |\mathcal{X}|$ be an irreducible closed subset. We have to show $T$ has a unique generic point. Let $\mathcal{Z} \subset \mathcal{X}$ be the reduced induced closed substack corresponding to $T$, see Properties of Stacks, Definition \ref{stacks-properties-definition-reduced-induced-stack}. Since $\mathcal{Z} \to \mathcal{X}$ is a closed immersion, we see that $\Delta_\mathcal{Z}$ is quasi-compact: first show that $\mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ is quasi-compact as the composition of $\mathcal{Z} \to \mathcal{X}$ with $\Delta$, then write $\mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ as the composition of $\Delta_\mathcal{Z}$ and $\mathcal{Z} \times \mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ and use Lemma \ref{lemma-quasi-compact-permanence} and the fact that $\mathcal{Z} \times \mathcal{Z} \to \mathcal{X} \times \mathcal{X}$ is separated. Thus we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $\mathcal{X}$ is quasi-compact, $\Delta$ is quasi-compact, $\mathcal{X}$ is reduced, and $|\mathcal{X}|$ irreducible. We have to show $|\mathcal{X}|$ has a unique generic point. Since $\mathcal{I}_\mathcal{X} \to \mathcal{X}$ is a base change of $\Delta$, we see that $\mathcal{I}_\mathcal{X} \to \mathcal{X}$ is quasi-compact (Lemma \ref{lemma-base-change-quasi-compact}). Thus there exists a dense open substack $\mathcal{U} \subset \mathcal{X}$ which is a gerbe by Proposition \ref{proposition-open-stratum}. In other words, $|\mathcal{U}| \subset |\mathcal{X}|$ is open dense. Thus we may assume that $\mathcal{X}$ is a gerbe. Say $\mathcal{X} \to X$ turns $\mathcal{X}$ into a gerbe over the algebraic space $X$. Then $|\mathcal{X}| \cong |X|$ by Lemma \ref{lemma-gerbe-bijection-points}. In particular, $X$ is quasi-compact. By Lemma \ref{lemma-gerbe-diagonal-quasi-compact} we see that $X$ has quasi-compact diagonal, i.e., $X$ is a quasi-separated algebraic space. Then $|X|$ is spectral by Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-quasi-separated-spectral} which implies what we want is true.