hcju/stacksqa huggingface dataset:
Let $U \to \mathcal{X}$ surjective and smooth with $U$ an algebraic space. Let $V \to \mathcal{X}$ be a smooth morphism with $V$ an algebraic space. Choose schemes $U'$ and $V'$ and surjective \'etale morphisms $U' \to U$ and $V' \to V$. Finally, choose a scheme $W$ and a surjective \'etale morphism $W \to V' \times_\mathcal{X} U'$. Then $W \to V'$ and $W \to U'$ are smooth morphisms of schemes as compositions of \'etale and smooth morphisms of algebraic spaces, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-smooth} and \ref{spaces-morphisms-lemma-composition-smooth}. Moreover, $W \to V'$ is surjective as $U' \to \mathcal{X}$ is surjective. Hence, we have $$ \mathcal{P}(U) \Leftrightarrow \mathcal{P}(U') \Rightarrow \mathcal{P}(W) \Rightarrow \mathcal{P}(V') \Leftrightarrow \mathcal{P}(V) $$ where the equivalences are by definition of property $\mathcal{P}$ for algebraic spaces, and the two implications come from Descent, Definition \ref{descent-definition-property-local}. This proves (3) $\Rightarrow$ (4). \medskip\noindent The implications (2) $\Rightarrow$ (1), (1) $\Rightarrow$ (3), and (4) $\Rightarrow$ (2) are immediate.
Let $a : U \to \mathcal{X}$ and $u \in |U|$ as in (3). Let $b : V \to \mathcal{X}$ be another smooth morphism with $V$ an algebraic space and $v \in |V|$ with $b(v) = x$ also. Choose a scheme $U'$, an \'etale morphism $U' \to U$ and $u' \in U'$ mapping to $u$. Choose a scheme $V'$, an \'etale morphism $V' \to V$ and $v' \in V'$ mapping to $v$. By Lemma \ref{lemma-points-cartesian} there exists a point $\overline{w} \in |V' \times_\mathcal{X} U'|$ mapping to $u'$ and $v'$. Choose a scheme $W$ and a surjective \'etale morphism $W \to V' \times_\mathcal{X} U'$. We may choose a $w \in |W|$ mapping to $\overline{w}$ (see Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-surjective}). Then $W \to V'$ and $W \to U'$ are smooth morphisms of schemes as compositions of \'etale and smooth morphisms of algebraic spaces, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-smooth} and \ref{spaces-morphisms-lemma-composition-smooth}. Hence $$ \mathcal{P}(U, u) \Leftrightarrow \mathcal{P}(U', u') \Leftrightarrow \mathcal{P}(W, w) \Leftrightarrow \mathcal{P}(V', v') \Leftrightarrow \mathcal{P}(V, v) $$ The outer two equivalences by Properties of Spaces, Definition \ref{spaces-properties-definition-property-at-point} and the other two by what it means to be a smooth local property of germs of schemes. This proves (4) $\Rightarrow$ (3). \medskip\noindent The implications (1) $\Rightarrow$ (2), (2) $\Rightarrow$ (4), and (3) $\Rightarrow$ (1) are immediate.
The equivalence of (1) and (4) follows from the general discussion in Section \ref{section-properties-morphisms} and in particular Lemmas \ref{lemma-check-representable-covering} and \ref{lemma-check-property-covering}. \medskip\noindent The equivalence of (2) and (3) is Categories, Lemma \ref{categories-lemma-fully-faithful-diagonal-equivalence}. \medskip\noindent Assume the equivalent conditions (2) and (3). Then $f$ is representable by algebraic spaces according to Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}. Moreover, the $2$-Yoneda lemma combined with the fully faithfulness implies that for every scheme $T$ the functor $$ \Mor(T, \mathcal{X}) \longrightarrow \Mor(T, \mathcal{Y}) $$ is fully faithful. Hence given a morphism $y : T \to \mathcal{Y}$ there exists up to unique $2$-isomorphism at most one morphism $x : T \to \mathcal{X}$ such that $y \cong f \circ x$. In particular, given a morphism of schemes $h : T' \to T$ there exists at most one lift $\tilde h : T' \to T \times_\mathcal{Y} \mathcal{X}$ of $h$. Thus $T \times_\mathcal{Y} \mathcal{X} \to T$ is a monomorphism of algebraic spaces, which proves that (1) holds. \medskip\noindent Finally, assume that (1) holds. Then for any scheme $T$ and morphism $y : T \to \mathcal{Y}$ the fibre product $T \times_\mathcal{Y} \mathcal{X}$ is an algebraic space, and $T \times_\mathcal{Y} \mathcal{X} \to T$ is a monomorphism. Hence there exists up to unique isomorphism exactly one pair $(x, \alpha)$ where $x : T \to \mathcal{X}$ is a morphism and $\alpha : f \circ x \to y$ is a $2$-morphism. Applying the $2$-Yoneda lemma this says exactly that $f$ is fully faithful, i.e., that (2) holds.
Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Suppose that $x_i : \Spec(K_i) \to \mathcal{X}$ be morphisms such that $f \circ x_1$ and $f \circ x_2$ define the same element of $|\mathcal{Y}|$. Applying the definition we find a common extension $\Omega$ with corresponding morphisms $c_i : \Spec(\Omega) \to \Spec(K_i)$ and a $2$-isomorphism $\beta : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$. As $f$ is fully faithful, see Lemma \ref{lemma-monomorphism}, we can lift $\beta$ to an isomorphism $\alpha : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$. Hence $x_1$ and $x_2$ define the same point of $|\mathcal{X}|$ as desired.
By Lemma \ref{lemma-representable-in-terms-presentations} we get a commutative diagram $$ \xymatrix{ [U'/R'] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld] \\ & [U/R] } $$ where $U' = \mathcal{Z} \times_{[U/R]} U$ and $R' = \mathcal{Z} \times_{[U/R]} R$. Since $\mathcal{Z} \to [U/R]$ is an immersion we see that $U' \to U$ is an immersion of algebraic spaces. Let $Z \subset U$ be the locally closed subspace such that $U' \to U$ factors through $Z$ and induces an isomorphism $U' \to Z$. It is clear from the construction of $R'$ that $R' = U' \times_{U, t} R = R \times_{s, U} U'$. This implies that $Z \cong U'$ is $R$-invariant and that the image of $R' \to R$ identifies $R'$ with the restriction $R_Z = s^{-1}(Z) = t^{-1}(Z)$ of $R$ to $Z$. Hence the lemma holds.
Recall that by Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-invariant-open} (see also discussion following the definition) we have $R_Z = s^{-1}(Z) = t^{-1}(Z)$ as locally closed subspaces of $R$. Hence the two morphisms $R_Z \to Z$ are smooth as base changes of $s$ and $t$. Hence $(Z, R_Z, s|_{R_Z}, t|_{R_Z}, c|_{R_Z \times_{s, Z, t} R_Z})$ is a smooth groupoid in algebraic spaces, and we see that $[Z/R_Z]$ is an algebraic stack, see Algebraic Stacks, Theorem \ref{algebraic-theorem-smooth-groupoid-gives-algebraic-stack}. The assumptions of Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-criterion-fibre-product} are all satisfied and it follows that we have a $2$-fibre square $$ \xymatrix{ Z \ar[d] \ar[r] & [Z/R_Z] \ar[d] \\ U \ar[r] & [U/R] } $$ It follows from this and Lemma \ref{lemma-check-representable-covering} that $[Z/R_Z] \to [U/R]$ is representable by algebraic spaces, whereupon it follows from Lemma \ref{lemma-check-property-covering} that the right vertical arrow is an immersion (resp.\ closed immersion, resp.\ open immersion) if and only if the left vertical arrow is.
By Lemmas \ref{lemma-immersion-into-presentation} and \ref{lemma-immersion-presentation} we find that the map is a bijection. If $\mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ then of course the base change $\mathcal{Y} \times_\mathcal{X} U \to U$ factors through $Z$. Converse, suppose that $\mathcal{Y} \to \mathcal{X}$ is a morphism such that $\mathcal{Y} \times_\mathcal{X} U \to U$ factors through $Z$. We will show that for every scheme $T$ and morphism $T \to \mathcal{Y}$, given by an object $y$ of the fibre category of $\mathcal{Y}$ over $T$, the object $y$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the fibre product $T \times_\mathcal{X} U$ is an algebraic space and $T \times_\mathcal{X} U \to T$ is a surjective smooth morphism. Hence there is an fppf covering $\{T_i \to T\}$ such that $T_i \to T$ factors through $T \times_\mathcal{X} U \to T$ for all $i$. Then $T_i \to \mathcal{X}$ factors through $\mathcal{Y} \times_\mathcal{X} U$ and hence through $Z \subset U$. Thus $y|_{T_i}$ is an object of $\mathcal{Z}$ (as $Z$ is the fibre product of $U$ with $\mathcal{Z}$ over $\mathcal{X}$). Since $\mathcal{Z}$ is a strictly full substack, we conclude that $y$ is an object of $\mathcal{Z}$ as desired.
Choose a presentation $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation}. By Lemma \ref{lemma-substacks-presentation} we see that open substacks correspond to $R$-invariant open subschemes of $U$. On the other hand Lemmas \ref{lemma-points-presentation} and \ref{lemma-topology-points} guarantee these correspond bijectively to open subsets of $|\mathcal{X}|$.
We define a category fibred in groupoids $\mathcal Y$ by letting the fiber category $\mathcal{Y}_T$ over an object $T$ of $(\Sch/S)_{fppf}$ be the full subcategory of $\mathcal{X}_T$ consisting of all $y \in \Ob(\mathcal{X}_T)$ such that the projection morphism $V \times_{\mathcal X, y} T \to T$ surjective. Now for any morphism $x : T \to \mathcal X$, the $2$-fibred product $T \times_{x, \mathcal X} \mathcal Y$ has fiber category over $T'$ consisting of triples $(f : T' \to T, y \in \mathcal{X}_{T'}, f^*x \simeq y)$ such that $V \times_{\mathcal X, y} T' \to T'$ is surjective. Note that $T \times_{x, \mathcal X} \mathcal Y$ is fibered in setoids since $\mathcal Y \to \mathcal X$ is faithful (see Stacks, Lemma \ref{stacks-lemma-2-fibre-product-gives-stack-in-setoids}). Now the isomorphism $f^*x \simeq y$ gives the diagram $$ \xymatrix{ V \times_{\mathcal X, y} T' \ar[d] \ar[r] & V \times_{\mathcal X, x} T \ar[r] \ar[d] & V \ar[d] \\ T' \ar[r]^f & T \ar[r]^x & \mathcal X } $$ where both squares are cartesian. The morphism $V \times_{\mathcal X, x} T \to T$ is smooth by base change, and hence open. Let $T_0 \subset T$ be its image. From the cartesian squares we deduce that $V \times_{\mathcal X, y} T' \to T'$ is surjective if and only if $f$ lands in $T_0$.  Therefore $T \times_{x, \mathcal X} \mathcal Y$ is representable by $T_0$, so the inclusion $\mathcal Y \to \mathcal X$ is an open immersion. By Algebraic Stacks, Lemma \ref{algebraic-lemma-open-fibred-category-is-algebraic} we conclude that $\mathcal{Y}$ is an algebraic stack. Lastly if we denote the morphism $V \to \mathcal X$ by $g$, we have $V \times_{\mathcal X} V \to V$ is surjective (the diagonal gives a section). Hence $g$ is in the image of $\mathcal{Y}_V \to \mathcal{X}_V$, i.e., we obtain a morphism $g' : V \to \mathcal{Y}$ fitting into the commutative diagram $$ \xymatrix{ V \ar[r] \ar[d]^{g'} & U \ar[d] \\ \mathcal{Y} \ar[r] & \mathcal{X} } $$ Since $V \times_{g, \mathcal X} \mathcal Y \to V$ is a monomorphism, it is in fact an isomorphism since $(1, g')$ defines a section. Therefore $g' : V \to \mathcal Y$ is a smooth morphism, as it is the base change of the smooth morphism $g : V \to \mathcal{X}$. It is surjective by our construction of $\mathcal{Y}$ which finishes the proof of the lemma.
We define a fibred subcategory $\mathcal{X}' = \bigcup_{i \in I} \mathcal{X}_i$ by letting the fiber category over an object $T$ of $(\Sch/S)_{fppf}$ be the full subcategory of $\mathcal{X}_T$ consisting of all $x \in \Ob(\mathcal{X}_T)$ such that the morphism $\coprod_{i \in I} (\mathcal{X}_i \times_{\mathcal X} T) \to T$ is surjective. Let $x_i \in \Ob((\mathcal{X}_i)_T)$. Then $(x_i, 1)$ gives a section of $\mathcal{X}_i \times_{\mathcal X} T \to T$, so we have an isomorphism. Thus $\mathcal{X}_i \subset \mathcal{X}'$ is a full subcategory. Now let $x \in \Ob(\mathcal{X}_T)$. Then $\mathcal{X}_i \times_{\mathcal X} T$ is representable by an open subscheme $T_i \subset T$. The $2$-fibred product $\mathcal{X}' \times_{\mathcal X} T$ has fiber over $T'$ consisting of $(y \in \mathcal{X}_{T'}, f : T' \to T, f^*x \simeq y)$ such that $\coprod (\mathcal{X}_i \times_{\mathcal X, y} T') \to T'$ is surjective. The isomorphism $f^*x \simeq y$ induces an isomorphism $\mathcal{X}_i \times_{\mathcal X, y} T' \simeq T_i \times_T T'$. Then the $T_i \times_T T'$ cover $T'$ if and only if $f$ lands in $\bigcup T_i$. Therefore we have a diagram $$ \xymatrix{ T_i \ar[r] \ar[d] & \bigcup T_i \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X}_i \ar[r] & \mathcal{X}' \ar[r] & \mathcal{X} } $$ with both squares cartesian. By Algebraic Stacks, Lemma \ref{algebraic-lemma-open-fibred-category-is-algebraic} we conclude that $\mathcal{X'} \subset \mathcal{X}$ is algebraic and an open substack. It is also clear from the cartesian squares above that the morphism $\coprod_{i \in I} \mathcal{X}_i \to \mathcal{X}'$ which finishes the proof of the lemma.
Since $\mathcal X$ is algebraic, there exists a scheme $U$ with a surjective smooth morphism $U \to \mathcal X$. Let $U_i \subset U$ be the open subscheme representing $\mathcal{X}_i \times_{\mathcal X} U$ and $U' \subset U$ the open subscheme representing $\mathcal{X}' \times_{\mathcal X} U$. By hypothesis, $U'\subset \bigcup_{i\in I} U_i$. From the proof of Lemma \ref{lemma-quasi-compact-stack}, there is a quasi-compact open $V \subset U'$ such that $V \to \mathcal{X}'$ is a surjective smooth morphism. Therefore there exists a finite subset $I' \subset I$ such that $V \subset \bigcup_{i \in I'} U_i$. We claim that $\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i$. Take $x \in \Ob(\mathcal{X}'_T)$ for $T \in \Ob((\Sch/S)_{fppf})$. Since $\mathcal{X}' \to \mathcal{X}$ is a monomorphism, we have cartesian squares $$ \xymatrix{ V \times_\mathcal{X} T \ar[r] \ar[d] & T \ar[d]^x \ar@{=}[r] & T \ar[d]^x \\ V \ar[r] & \mathcal{X}' \ar[r] & \mathcal X } $$ By base change, $V \times_{\mathcal X} T \to T$ is surjective. Therefore $\bigcup_{i \in I'} U_i \times_{\mathcal X} T \to T$ is also surjective. Let $T_i \subset T$ be the open subscheme representing $\mathcal{X}_i \times_{\mathcal X} T$. By a formal argument, we have a Cartesian square $$ \xymatrix{ U_i \times_{\mathcal{X}_i} T_i \ar[r] \ar[d] & U \times_{\mathcal X} T \ar[d] \\ T_i \ar[r] & T } $$ where the vertical arrows are surjective by base change. Since $U_i \times_{\mathcal{X}_i} T_i \simeq U_i \times_{\mathcal X} T$, we find that $\bigcup_{i \in I'} T_i = T$. Hence $x$ is an object of $(\bigcup_{i\in I'} \mathcal{X}_i)_T$ by definition of the union. Observe that the inclusion $\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i$ is automatically an open substack.
Apply Stacks, Lemma \ref{stacks-lemma-stack-in-setoids-descent} to the morphism $\coprod_{i \in I} \mathcal{X}_i \to \mathcal{X}$ and the morphism $\text{id} : \mathcal{X} \to \mathcal{X}$ to see that $\mathcal{X}$ is a stack in setoids. Hence $\mathcal{X}$ is an algebraic space, see Algebraic Stacks, Proposition \ref{algebraic-proposition-algebraic-stack-no-automorphisms}.
Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = V \times_\mathcal{Y} \mathcal{X}$ and let $f' : U \to V$ be the base change of $f$. The morphism of algebraic spaces $f' : U \to V$ has property $\mathcal{Q}$. Thus we obtain the open $W(\mathcal{P}, f') \subset U$ by assumption (3). Note that $U \times_\mathcal{X} U = (V \times_\mathcal{Y} V) \times_\mathcal{Y} \mathcal{X}$ hence the morphism $f'' : U \times_\mathcal{X} U \to V \times_\mathcal{Y} V$ is the base change of $f$ via either projection $V \times_\mathcal{Y} V \to V$. By our choice of $V$ these projections are smooth, hence have property $\mathcal{R}$ by (2). Thus by (4) we see that the inverse images of $W(\mathcal{P}, f')$ under the two projections $\text{pr}_i : U \times_\mathcal{X} U \to U$ agree. In other words, $W(\mathcal{P}, f')$ is an $R$-invariant subspace of $U$ (where $R = U \times_\mathcal{X} U$). Let $\mathcal{X}'$ be the open substack of $\mathcal{X}$ corresponding to $W(\mathcal{P}, f)$ via Lemma \ref{lemma-immersion-into-presentation}. By construction $W(\mathcal{P}, f') = \mathcal{X}' \times_\mathcal{Y} V$ hence $f|_{\mathcal{X}'}$ has property $\mathcal{P}$ by Lemma \ref{lemma-check-property-covering}. Also, $\mathcal{X}'$ is the largest open substack such that $f|_{\mathcal{X}'}$ has $\mathcal{P}$ as the same maximality holds for $W(\mathcal{P}, f)$. This proves (A). \medskip\noindent Finally, if $\mathcal{Z} \to \mathcal{Y}$ is a morphism of algebraic stacks representable by algebraic spaces which has $\mathcal{R}$ then we set $T = V \times_\mathcal{Y} \mathcal{Z}$ and we see that $T \to V$ is a morphism of algebraic spaces having property $\mathcal{R}$. Set $f'_T : T \times_V U \to T$ the base change of $f'$. By (4) again we see that $W(\mathcal{P}, f'_T)$ is the inverse image of $W(\mathcal{P}, f)$ in $T \times_V U$. This implies (B); some details omitted.
Let $U \to \mathcal{X}$ be a surjective smooth morphism, where $U$ is an algebraic space. Set $R = U \times_\mathcal{X} U$, so that there is a presentation $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation}. As usual we denote $s, t : R \to U$ the two smooth projection morphisms. By Lemma \ref{lemma-points-presentation} we see that $T$ corresponds to a closed subset $T' \subset |U|$ such that $|s|^{-1}(T') = |t|^{-1}(T')$. Let $Z \subset U$ be the reduced induced algebraic space structure on $T'$, see Properties of Spaces, Definition \ref{spaces-properties-definition-reduced-induced-space}. The fibre products $Z \times_{U, t} R$ and $R \times_{s, U} Z$ are closed subspaces of $R$ (Spaces, Lemma \ref{spaces-lemma-base-change-immersions}). The projections $Z \times_{U, t} R \to Z$ and $R \times_{s, U} Z \to Z$ are smooth by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-smooth}. Thus as $Z$ is reduced, it follows that $Z \times_{U, t} R$ and $R \times_{s, U} Z$ are reduced, see Remark \ref{remark-list-properties-local-smooth-topology}. Since $$ |Z \times_{U, t} R| = |t|^{-1}(T') = |s|^{-1}(T') = R \times_{s, U} Z $$ we conclude from the uniqueness in Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace} that $Z \times_{U, t} R = R \times_{s, U} Z$. Hence $Z$ is an $R$-invariant closed subspace of $U$. By the correspondence of Lemma \ref{lemma-substacks-presentation} we obtain a closed substack $\mathcal{Z} \subset \mathcal{X}$ with $Z = \mathcal{Z} \times_\mathcal{X} U$. Then $[Z/R_Z] \to \mathcal{Z}$ is a presentation (Lemma \ref{lemma-immersion-into-presentation}). Then $|\mathcal{Z}| = |Z|/|R_Z| = |T'|/\sim$ is the given closed subset $T$. We omit the proof of unicity.
Choose a presentation $[U/R] \to \mathcal{X}$ with $U$ a scheme. As $\mathcal{X}$ is reduced, we see that $U$ is reduced (by definition of reduced algebraic stacks). By Lemma \ref{lemma-substacks-presentation} $\mathcal{X}'$ corresponds to an $R$-invariant closed subscheme $Z \subset U$. But now $|Z| \subset |U|$ is the inverse image of $|\mathcal{X}'|$, and hence $|Z| = |U|$. Hence $Z$ is a closed subscheme of $U$ whose underlying sets of points agree. By Schemes, Lemma \ref{schemes-lemma-map-into-reduction} the map $\text{id}_U : U \to U$ factors through $Z \to U$, and hence $Z = U$, i.e., $\mathcal{X}' = \mathcal{X}$.
Assume $f(|\mathcal{Y}|) \subset |\mathcal{Z}|$. Consider $\mathcal{Y} \times_\mathcal{X} \mathcal{Z} \to \mathcal{Y}$. There is an equivalence $\mathcal{Y} \times_\mathcal{X} \mathcal{Z} \to \mathcal{Y}'$ where $\mathcal{Y}'$ is a closed substack of $\mathcal{Y}$, see Lemmas \ref{lemma-base-change-immersion} and \ref{lemma-substack-image}. Using Lemmas \ref{lemma-points-cartesian}, \ref{lemma-monomorphism-injective-points}, and \ref{lemma-immersion-monomorphism} we see that $|\mathcal{Y}'| = |\mathcal{Y}|$. Hence we have reduced the lemma to Lemma \ref{lemma-reduced-stack-determined-by-points}.
Consider the fibre square $$ \xymatrix{ T \ar[d] \ar[r] & \Spec(k) \ar[d] \\ \Spec(k') \ar[r] & \mathcal{Z} } $$ Note that $T \to \Spec(k')$ is flat and surjective hence $T$ is not empty. On the other hand $T \to \Spec(k)$ is flat as $k$ is a field. Hence $T \to \mathcal{Z}$ is flat and surjective. It follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-permanence} (via the discussion in Section \ref{section-properties-morphisms}) that $\Spec(k') \to \mathcal{Z}$ is flat. It is clear that it is surjective as by assumption $|\mathcal{Z}|$ is a singleton.
Assume (1). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Then $W$ is a reduced scheme. Let $\eta \in W$ be a generic point of an irreducible component of $W$. Since $W$ is reduced we have $\mathcal{O}_{W, \eta} = \kappa(\eta)$. It follows that the canonical morphism $\eta = \Spec(\kappa(\eta)) \to W$ is flat. We see that the composition $\eta \to \mathcal{Z}$ is flat (see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-flat}). It is also surjective as $|\mathcal{Z}|$ is a singleton. In other words (2) holds. \medskip\noindent Assume (2). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a surjective flat morphism $\Spec(k) \to \mathcal{Z}$. Then $W \times_\mathcal{Z} \Spec(k)$ is an algebraic space smooth over $k$, hence regular (see Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular}) and in particular reduced. Since $W \times_\mathcal{Z} \Spec(k) \to W$ is surjective and flat we conclude that $W$ is reduced (Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-reduced}). In other words (1) holds. \medskip\noindent It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. The composition $$ \Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z} $$ is locally of finite type by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-type} and \ref{spaces-morphisms-lemma-smooth-locally-finite-type}. It is also flat and surjective by Lemma \ref{lemma-flat-cover-by-field}. Hence (3) holds.
Assume (2) holds. By Lemma \ref{lemma-unique-point} we see that $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton. Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a locally finitely presented, surjective, flat morphism $\Spec(k) \to \mathcal{Z}$. Then $W \times_\mathcal{Z} \Spec(k)$ is an algebraic space smooth over $k$, hence locally Noetherian (see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-locally-noetherian}). Since $W \times_\mathcal{Z} \Spec(k) \to W$ is flat, surjective, and locally of finite presentation, we see that $\{W \times_\mathcal{Z} \Spec(k) \to W\}$ is an fppf covering and we conclude that $W$ is locally Noetherian (Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-locally-Noetherian}). In other words (1) holds. \medskip\noindent Assume (1). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. Because $W$ is locally Noetherian the morphism $w : \Spec(k) \to W$ is of finite presentation, see Morphisms, Lemma \ref{morphisms-lemma-closed-immersion-finite-presentation}. Hence the composition $$ \Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z} $$ is locally of finite presentation by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-presentation} and \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation}. It is also flat and surjective by Lemma \ref{lemma-flat-cover-by-field}. Hence (2) holds.
We may assume that $\mathcal{Z}'$ is nonempty. In this case the fibre product $T = \mathcal{Z}' \times_\mathcal{Z} \Spec(k)$ is nonempty, see Lemma \ref{lemma-points-cartesian}. Now $T$ is an algebraic space and the projection $T \to \Spec(k)$ is a monomorphism. Hence $T = \Spec(k)$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-toward-field}. We conclude that $\Spec(k) \to \mathcal{Z}$ factors through $\mathcal{Z}'$. Suppose the morphism $z : \Spec(k) \to \mathcal{Z}$ is given by the object $\xi$ over $\Spec(k)$. We have just seen that $\xi$ is isomorphic to an object $\xi'$ of $\mathcal{Z}'$ over $\Spec(k)$. Since $z$ is surjective, flat, and locally of finite presentation we see that every object of $\mathcal{Z}$ over any scheme is fppf locally isomorphic to a pullback of $\xi$, hence also to a pullback of $\xi'$. By descent of objects for stacks in groupoids this implies that $\mathcal{Z}' \to \mathcal{Z}$ is essentially surjective (as well as fully faithful, see Lemma \ref{lemma-monomorphism}). Hence we win.