hcju/stacksqa huggingface dataset:
Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks with $\mathcal{Z}$ quasi-compact. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack} there exists a quasi-compact scheme $U$ and a surjective smooth morphism $U \to \mathcal{Z}$. Since $f$ is representable by algebraic spaces and quasi-compact we see by definition that $U \times_\mathcal{Y} \mathcal{X}$ is an algebraic space, and that $U \times_\mathcal{Y} \mathcal{X} \to U$ is quasi-compact. Hence $U \times_Y X$ is a quasi-compact algebraic space. The morphism $U \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ is smooth and surjective (as the base change of the smooth and surjective morphism $U \to \mathcal{Z}$). Hence $\mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ is quasi-compact by another application of Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack} \medskip\noindent Assume (2). Let $Z \to \mathcal{Y}$ be a morphism, where $Z$ is a scheme. We have to show that the morphism of algebraic spaces $p : Z \times_\mathcal{Y} \mathcal{X} \to Z$ is quasi-compact. Let $U \subset Z$ be affine open. Then $p^{-1}(U) = U \times_\mathcal{Y} \mathcal{Z}$ and the algebraic space $U \times_\mathcal{Y} \mathcal{Z}$ is quasi-compact by assumption (2). Hence $p$ is quasi-compact, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-local}.
Let $\mathcal{T}$ be a quasi-compact algebraic stack, and let $\mathcal{T} \to \mathcal{Z}$ be a morphism. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-base-change-surjective} the morphism $\mathcal{T} \times_\mathcal{Z} \mathcal{X} \to \mathcal{T} \times_\mathcal{Z} \mathcal{Y}$ is surjective and by assumption $\mathcal{T} \times_\mathcal{Z} \mathcal{X}$ is quasi-compact. Hence $\mathcal{T} \times_\mathcal{Z} \mathcal{Y}$ is quasi-compact by Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack}.
This is true because $f$ equals the composition $(1, f) : \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y}$. The first map is quasi-compact by Lemma \ref{lemma-section-immersion} because it is a section of the quasi-separated morphism $\mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{X}$ (a base change of $g$, see Lemma \ref{lemma-base-change-separated}). The second map is quasi-compact as it is the base change of $f$, see Lemma \ref{lemma-base-change-quasi-compact}. And compositions of quasi-compact morphisms are quasi-compact, see Lemma \ref{lemma-composition-quasi-compact}.
Part (1) follows from Lemma \ref{lemma-quasi-compact-permanence}. Part (2) follows from (1) and Lemma \ref{lemma-compose-after-separated}. For (3) let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Z} \to \mathcal{Y}$ be morphisms of quasi-compact and quasi-separated algebraic stacks. Then $\mathcal{X} \times_\mathcal{Y} \mathcal{Z} \to \mathcal{Z}$ is quasi-compact and quasi-separated as a base change of $\mathcal{X} \to \mathcal{Y}$ using (2) and Lemmas \ref{lemma-base-change-quasi-compact} and \ref{lemma-base-change-separated}. Hence $\mathcal{X} \times_\mathcal{Y} \mathcal{Z}$ is quasi-compact and quasi-separated as an algebraic stack quasi-compact and quasi-separated over $\mathcal{Z}$, see Lemmas \ref{lemma-separated-over-separated} and \ref{lemma-composition-quasi-compact}.
Choose an affine scheme $V$ and a point $v \in V$ and a smooth morphism $V \to \mathcal{Y}$ sending $v$ to $y$. Consider the base change diagram $$ \xymatrix{ V \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d]_g & \mathcal{X} \ar[d]^f \\ V \ar[r] & \mathcal{Y} } $$ Recall that $|V \times_\mathcal{Y} \mathcal{X}| \to |V| \times_{|\mathcal{Y}|} |\mathcal{X}|$ is surjective (Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian}). Because $|V| \to |\mathcal{Y}|$ is open (Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}) we conclude that $v$ is in the closure of the image of $|g|$. Thus it suffices to prove the lemma for the quasi-compact morphism $g$ (Lemma \ref{lemma-base-change-quasi-compact}) which we do in the next paragraph. \medskip\noindent Assume $\mathcal{Y} = Y$ is an affine scheme. Then $\mathcal{X}$ is quasi-compact as $f$ is quasi-compact (Definition \ref{definition-quasi-compact}). Choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Then the image of $|f|$ is the image of $W \to Y$. By Morphisms, Lemma \ref{morphisms-lemma-reach-points-scheme-theoretic-image} we can choose a diagram $$ \xymatrix{ \Spec(K) \ar[r] \ar[d] & W \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \Spec(A) \ar[r] & Y \ar[r] & Y } $$ such that the closed point of $\Spec(A)$ maps to $y$. Composing with $W \to \mathcal{X}$ we obtain a solution.
Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Then $U = \mathcal{X} \times_\mathcal{Y} V$ is a scheme and $V \to U$ is an immersion, see Properties of Stacks, Definition \ref{stacks-properties-definition-immersion}. Recall that $\mathcal{Y}$ is locally Noetherian if and only if $V$ is locally Noetherian. In this case $U$ is locally Noetherian too (Morphisms, Lemmas \ref{morphisms-lemma-immersion-locally-finite-type} and \ref{morphisms-lemma-finite-type-noetherian}) and $U \to V$ is quasi-compact (Properties, Lemma \ref{properties-lemma-immersion-into-noetherian}). This shows that $j$ is quasi-compact (Lemma \ref{lemma-check-quasi-compact-covering}) and that $\mathcal{X}$ is locally Noetherian. Finally, if $\mathcal{Y}$ is Noetherian, then we see from the above that $\mathcal{X}$ is quasi-compact and locally Noetherian. To finish the proof observe that $j$ is separated and hence $\mathcal{X}$ is quasi-separated because $\mathcal{Y}$ is so by Lemma \ref{lemma-separated-over-separated}.
Assume $\mathcal{X}$ is locally Noetherian. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. As $\mathcal{X}$ is locally Noetherian we see that $U$ is locally Noetherian. By Properties, Lemma \ref{properties-lemma-Noetherian-topology} this means that $|U|$ is a locally Noetherian topological space. Since $|U| \to |\mathcal{X}|$ is open and surjective we conclude that $|\mathcal{X}|$ is locally Noetherian by Topology, Lemma \ref{topology-lemma-image-Noetherian}. This proves (1). If $\mathcal{X}$ is quasi-compact and locally Noetherian, then $|\mathcal{X}|$ is quasi-compact and locally Noetherian. Hence $|\mathcal{X}|$ is Noetherian by Topology, Lemma \ref{topology-lemma-quasi-compact-locally-Noetherian-Noetherian}.
We omit the proof that (1) implies (2), and that (2) implies (3). \medskip\noindent Assume (3). Choose a surjective smooth morphism $V \to \mathcal{Y}$. We are going to show that $V \times_\mathcal{Y} \mathcal{X} \to V$ is a universally closed morphism of algebraic stacks. Let $\mathcal{Z} \to V$ be a morphism from an algebraic stack to $V$. Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W = \coprod W_i$ is a disjoint union of affine schemes. Then we have the following commutative diagram $$ \xymatrix{ \coprod_i |W_i \times_\mathcal{Y} \mathcal{X}| \ar@{=}[r] \ar[d] & |W \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \ar@{=}[r] & |\mathcal{Z} \times_V (V \times_\mathcal{Y} \mathcal{X})| \ar[ld] \\ \coprod |W_i| \ar@{=}[r] & |W| \ar[r] & |\mathcal{Z}| } $$ We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}). By assumption (3), and the fact that $W_i$ is affine we see that the left vertical arrows are closed. Hence it follows that the right vertical arrow is closed. \medskip\noindent Assume (4). We will show that $f$ is universally closed. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider the diagram $$ \xymatrix{ |(V \times_\mathcal{Y} \mathcal{Z}) \times_V (V \times_\mathcal{Y} \mathcal{X})| \ar@{=}[r] \ar[rd] & |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |Z \times_\mathcal{Y} \mathcal{X}| \ar[d] \\  & |V \times_\mathcal{Y} \mathcal{Z}| \ar[r] & |\mathcal{Z}| } $$ The south-west arrow is closed by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic stacks are surjective and smooth (see reference above). It follows that the right vertical arrow is closed.
(1) $\Rightarrow$ (2). If $f$ is universally injective, then the first projection $|\mathcal{X} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{X}|$ is injective, which implies that $|\Delta|$ is surjective. \medskip\noindent (2) $\Rightarrow$ (1). Assume $\Delta$ is surjective. Then any base change of $\Delta$ is surjective (see Properties of Stacks, Section \ref{stacks-properties-section-surjective}). Since the diagonal of a base change of $f$ is a base change of $\Delta$, we see that it suffices to show that $|\mathcal{X}| \to |\mathcal{Y}|$ is injective. If not, then by Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian} we find that the first projection $|\mathcal{X} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{X}|$ is not injective. Of course this means that $|\Delta|$ is not surjective. \medskip\noindent (3) $\Rightarrow$ (2). Let $t \in |\mathcal{X} \times_\mathcal{Y} \mathcal{X}|$. Then we can represent $t$ by a morphism $t : \Spec(k) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ with $k$ an algebraically closed field. By our construction of $2$-fibre products we can represent $t$ by $(x_1, x_2, \beta)$ where $x_1, x_2 : \Spec(k) \to \mathcal{X}$ and $\beta : f \circ x_1 \to f \circ x_2$ is a $2$-morphism. Then (3) implies that there is a $2$-morphism $\alpha : x_1 \to x_2$ mapping to $\beta$. This exactly means that $\Delta(x_1) = (x_1, x_1, \text{id})$ is isomorphic to $t$. Hence (2) holds. \medskip\noindent (2) $\Rightarrow$ (3). Let $x_1, x_2 : \Spec(k) \to \mathcal{X}$ be morphisms with $k$ an algebraically closed field. Let $\beta : f \circ x_1 \to f \circ x_2$ be a $2$-morphism. As in the previous paragraph, we obtain a morphism $t = (x_1, x_2, \beta) : \Spec(k) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$. By Lemma \ref{lemma-properties-diagonal} $$ T = \mathcal{X} \times_{\Delta, \mathcal{X} \times_\mathcal{Y} \mathcal{X}, t} \Spec(k) $$ is an algebraic space locally of finite type over $\Spec(k)$. Condition (2) implies that $T$ is nonempty. Then since $k$ is algebraically closed, there is a $k$-point in $T$. Unwinding the definitions this means there is a morphism $\alpha : x_1 \to x_2$ in $\Mor(\Spec(k), \mathcal{X})$ such that $\beta = \text{id}_f \star \alpha$.
We first remark this lemma is not a triviality, because the assumption that $y$ is in the image of $|f|$ means only that we can lift $y$ to a morphism into $\mathcal{X}$ after possibly replacing $k$ by an extension field. To prove the lemma we may base change $f$ by $y$, hence we may assume we have a nonempty algebraic stack $\mathcal{X}$ and a universally injective morphism $\mathcal{X} \to \Spec(k)$ and we want to find a $k$-valued point of $\mathcal{X}$. We may replace $\mathcal{X}$ by its reduction. We may choose a field $k'$ and a surjective, flat, locally finite type morphism $\Spec(k') \to \mathcal{X}$, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-unique-point}. Since $\mathcal{X} \to \Spec(k)$ is universally injective, we find that $$ \Spec(k') \times_\mathcal{X} \Spec(k') \to \Spec(k' \otimes_k k') $$ is surjective as the base change of the surjective morphism $\Delta : \mathcal{X} \to \mathcal{X} \times_{\Spec(k)} \mathcal{X}$ (Lemma \ref{lemma-universally-injective}). Since $k$ is algebraically closed $k' \otimes_k k'$ is a domain (Algebra, Lemma \ref{algebra-lemma-geometrically-integral-any-integral-base-change}). Let $\xi \in \Spec(k') \times_\mathcal{X} \Spec(k')$ be a point mapping to the generic point of $\Spec(k' \otimes_k k')$. Let $U$ be the reduced induced closed subscheme structure on the connected component of $\Spec(k') \times_\mathcal{X} \Spec(k')$ containing $\xi$. Then the two projections $U \to \Spec(k')$ are locally of finite type, as this was true for the projections $\Spec(k') \times_\mathcal{X} \Spec(k') \to \Spec(k')$ as base changes of the morphism $\Spec(k') \to \mathcal{X}$. Applying Varieties, Proposition \ref{varieties-proposition-unique-base-field} we find that the integral closures of the two images of $k'$ in $\Gamma(U, \mathcal{O}_U)$ are equal. Looking in $\kappa(\xi)$ means that any element of the form $\lambda \otimes 1$ is algebraically dependend on the subfield $$ 1 \otimes k' \subset  (\text{fraction field of }k' \otimes_k k') \subset \kappa(\xi). $$ Since $k$ is algebraically closed, this is only possible if $k' = k$ and the proof is complete.
The implication (1) $\Rightarrow$ (2) is immediate. Assume (2) holds. We will show that $\Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is surjective, which implies (1) by Lemma \ref{lemma-universally-injective}. Consider an affine scheme $V$ and a smooth morphism $V \to \mathcal{Y}$. Since $g : V \times_\mathcal{Y} \mathcal{X} \to V$ is universally injective by (2), we see that $\Delta_g$ is surjective. However, $\Delta_g$ is the base change of $\Delta_f$ by the smooth morphism $V \to \mathcal{Y}$. Since the collection of these morphisms $V \to \mathcal{Y}$ are jointly surjective, we conclude $\Delta_f$ is surjective.
Observe that the diagonal $\Delta_g$ of the morphism $g : W \times_\mathcal{Y} \mathcal{X} \to W$ is the base change of $\Delta_f$ by $W \to \mathcal{Y}$. Hence if $\Delta_g$ is surjective, then so is $\Delta_f$ by Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering}. Thus the lemma follows from the characterization (2) in Lemma \ref{lemma-universally-injective}.
Assume $g : W \times_\mathcal{Y} \mathcal{X} \to W$ is a universal homeomorphism. Then $g$ is universally injective, hence $f$ is universally injective by Lemma \ref{lemma-check-universally-injective-covering}. On the other hand, let $\mathcal{Z} \to \mathcal{Y}$ be a morphism with $\mathcal{Z}$ an algebraic stack. Choose a scheme $U$ and a surjective smooth morphism $U \to W \times_\mathcal{Y} \mathcal{Z}$. Consider the diagram $$ \xymatrix{ W \times_\mathcal{Y} \mathcal{X} \ar[d]^g & U \times_\mathcal{Y} \mathcal{X} \ar[d] \ar[l] \ar[r] & \mathcal{Z} \times_\mathcal{Y} \mathcal{X} \ar[d] \\ W & U \ar[l] \ar[r] & \mathcal{Z} } $$ The middle vertical arrow induces a homeomorphism on topological space by assumption on $g$. The morphism $U \to \mathcal{Z}$ and $U \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ are surjective, flat, and locally of finite presentation hence induce open maps on topological spaces. We conclude that $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is open. Surjectivity is easy to prove; we omit the proof.
Let $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} } $$ be a commutative diagram where $U$, $V$ are schemes, $V \to \mathcal{Y}$ is surjective and smooth, and $U \to V \times_\mathcal{Y} \mathcal{X}$ is surjective and smooth. Then $U \to V$ is locally of finite type. If $\mathcal{Y}$ is locally Noetherian, then $V$ is locally Noetherian. By Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian} we see that $U$ is locally Noetherian, which means that $\mathcal{X}$ is locally Noetherian.
Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times_\mathcal{Y} \mathcal{X}$. We have to show that $U \to V$ is locally of finite presentation. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times_\mathcal{Y} U$, $V' = W \times_\mathcal{Y} V$, $\mathcal{X}' = W \times_\mathcal{Y} \mathcal{X}$, and $\mathcal{Y}' = W \times_\mathcal{Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by our definition of locally finite type morphisms of algebraic stacks and the assumption that $\mathcal{X}' \to \mathcal{Y}'$ is locally of finite type, we see that $U' \to V'$ is locally of finite type. Then, since $V' \to V$ is surjective, flat, and locally of finite presentation as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ is locally of finite type by Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-locally-finite-type} and we win.
Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times_\mathcal{Z} \mathcal{Y}$. Set $U = V \times_\mathcal{Y} \mathcal{X}$ which is an algebraic space. We know that $U \to V$ is surjective, flat, and locally of finite presentation and that $U \to W$ is locally of finite type. 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-type-fppf-local-source}.
We can find a diagram $$ \xymatrix{ U \ar[r] \ar[d] & V \ar[r] \ar[d] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} } $$ where $U$, $V$, $W$ are schemes, the vertical arrow $W \to \mathcal{Z}$ is surjective and smooth, the arrow $V \to \mathcal{Y} \times_\mathcal{Z} W$ is surjective and smooth, and the arrow $U \to \mathcal{X} \times_\mathcal{Y} V$ is surjective and smooth. Then also $U \to \mathcal{X} \times_\mathcal{Z} V$ is surjective and smooth (as a composition of a surjective and smooth morphism with a base change of such). By definition we see that $U \to W$ is locally of finite type. Hence $U \to V$ is locally of finite type by Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type} which in turn means (by definition) that $\mathcal{X} \to \mathcal{Y}$ is locally of finite type.
Let $u \in U$ and $U \to \mathcal{X}$ be as in (2). Then $\Spec(\kappa(u)) \to U$ is of finite type, and $U \to \mathcal{X}$ is representable and locally of finite type (by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-locally-finite-presentation} and \ref{spaces-morphisms-lemma-finite-presentation-finite-type}). Hence we see (1) holds by Lemma \ref{lemma-composition-finite-type}. \medskip\noindent Conversely, assume $\Spec(k) \to \mathcal{X}$ is locally of finite type and represents $x$. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. By assumption $U \times_\mathcal{X} \Spec(k) \to U$ is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point $v$ of $U \times_\mathcal{X} \Spec(k)$ (there exists at least one, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-identify-finite-type-points}). By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-type-points-morphism} the image $u \in U$ of $v$ is a finite type point of $U$. Hence by Morphisms, Lemma \ref{morphisms-lemma-identify-finite-type-points} after shrinking $U$ we may assume that $u$ is a closed point of $U$, i.e., (2) holds.
Take $x \in \mathcal{X}_{\text{ft-pts}}$. Represent $x$ by a locally finite type morphism $x : \Spec(k) \to \mathcal{X}$. Then $f \circ x$ is locally of finite type by Lemma \ref{lemma-composition-finite-type}. Hence $f(x) \in \mathcal{Y}_{\text{ft-pts}}$.
We have $f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}$ by Lemma \ref{lemma-finite-type-points-morphism}. Let $y \in |\mathcal{Y}|$ be a finite type point. Represent $y$ by a morphism $\Spec(k) \to \mathcal{Y}$ which is locally of finite type. As $f$ is surjective the algebraic stack $\mathcal{X}_k = \Spec(k) \times_\mathcal{Y} \mathcal{X}$ is nonempty, therefore has a finite type point $x \in |\mathcal{X}_k|$ by Lemma \ref{lemma-identify-finite-type-points}. Now $\mathcal{X}_k \to \mathcal{X}$ is a morphism which is locally of finite type as a base change of $\Spec(k) \to \mathcal{Y}$ (Lemma \ref{lemma-base-change-finite-type}). Hence the image of $x$ in $\mathcal{X}$ is a finite type point by Lemma \ref{lemma-finite-type-points-morphism} which maps to $y$ by construction.
Let $i : \mathcal{Z} \to \mathcal{X}$ be the reduced induced substack structure on $T$, see Properties of Stacks, Remark \ref{stacks-properties-remark-stack-structure-locally-closed-subset}. An immersion is locally of finite type, see Lemma \ref{lemma-immersion-locally-finite-type}. Hence by Lemma \ref{lemma-finite-type-points-morphism} we see $\mathcal{Z}_{\text{ft-pts}} \subset \mathcal{X}_{\text{ft-pts}} \cap T$. Finally, any nonempty affine scheme $U$ with a smooth morphism towards $\mathcal{Z}$ has at least one closed point, hence $\mathcal{Z}$ has at least one finite type point by Lemma \ref{lemma-identify-finite-type-points}. The lemma follows.
Observe that (1) $\Rightarrow$ (4), (2) $\Rightarrow$ (5), and (3) $\Rightarrow$ (6) by Lemma \ref{lemma-diagonal-diagonal}. In case (4) we see that $G_x$ is a quasi-separated algebraic space and in case (5) we see that $G_x$ is a locally separated algebraic space. In both cases $G_x$ is a decent algebraic space (Decent Spaces, Section \ref{decent-spaces-section-reasonable-decent} and Lemma \ref{decent-spaces-lemma-locally-separated-decent}). Then $G_x$ is separated by More on Groupoids in Spaces, Lemma \ref{spaces-more-groupoids-lemma-group-scheme-over-field-separated} whereupon we conclude that $G_x$ is a scheme by More on Groupoids in Spaces, Proposition \ref{spaces-more-groupoids-proposition-group-space-scheme-over-field}. In case (6) we see that $G_x \to \Spec(k)$ is locally quasi-finite and hence $G_x$ is a scheme by Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-locally-quasi-finite-over-field}.
This comes down to the fact that being an isomorphism is fpqc local on the target, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-isomorphism}. Namely, suppose that $k'/k$ is an extension of fields and denote $x' : \Spec(k') \to \mathcal{X}$ the composition and set $y' = f \circ x'$. Then the morphism $G_{x'} \to G_{y'}$ is the base change of $G_x \to G_y$ by $\Spec(k') \to \Spec(k)$. Hence $G_x \to G_y$ is an isomorphism if and only if $G_{x'} \to G_{y'}$ is an isomorphism. Thus we see that the property propagates through the equivalence class if it holds for one.
Observe that the morphism $U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation by Criteria for Representability, Lemma \ref{criteria-lemma-flat-quotient-flat-presentation}. Hence the same is true for $U \times U \to \mathcal{X} \times \mathcal{X}$. We have the cartesian diagram $$ \xymatrix{ R  = U \times_\mathcal{X} U \ar[r] \ar[d] & U \times U \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times \mathcal{X} } $$ (see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}). Thus we see that $\Delta_\mathcal{X}$ has one of the properties listed in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} if and only if the morphism $R \to U \times U$ does, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering}. This explains why (1), (3), and (5) are true. The condition in (2) implies $R \to U \times U$ is separated hence (2) follows from (1). The condition in (4) implies the condition in (3) hence (4) follows from (3). The condition in (6) implies the condition in (5) by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-closed-permanence} hence (6) follows from (5).
The induced morphism $p : U \to \mathcal{X}$ is surjective, flat, and locally of finite presentation, see Criteria for Representability, Lemma \ref{criteria-lemma-flat-quotient-flat-presentation}. Hence $|U| \to |\mathcal{X}|$ is surjective by Properties of Stacks, Lemma \ref{stacks-properties-lemma-characterize-surjective}. Note that $R = U \times_\mathcal{X} U$, see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}. Hence Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian} 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.