hcju/stacksqa huggingface dataset:
The equivalence of (1) and (2) follows from Stacks, Lemma \ref{stacks-lemma-isom-as-2-fibre-product} and the definitions. Let us prove the equivalence of (1) and (3). Write $\mathcal{C} = (\Sch/S)_{fppf}$ for the base category. We will use some of the observations of the proof of the similar Categories, Lemma \ref{categories-lemma-representable-diagonal-groupoids}. We will use the symbol $\cong$ to mean ``equivalence of categories fibred in groupoids over $\mathcal{C} = (\Sch/S)_{fppf}$''. Assume (1). Suppose given $U$ and $x$ as in (3). For any scheme $V$ and $y \in \Ob(\mathcal{X}_V)$ we see (compare reference above) that $$ \mathcal{C}/U \times_{x, \mathcal{X}, y} \mathcal{C}/V \cong (\mathcal{C}/U \times_S V) \times_{(x, y), \mathcal{X} \times \mathcal{X}, \Delta} \mathcal{X} $$ which is representable by an algebraic space by assumption. Conversely, assume (3). Consider any scheme $U$ over $S$ and a pair $(x, x')$ of objects of $\mathcal{X}$ over $U$. We have to show that $\mathcal{X} \times_{\Delta, \mathcal{X} \times \mathcal{X}, (x, x')} U$ is representable by an algebraic space. This is clear because (compare reference above) $$ \mathcal{X} \times_{\Delta, \mathcal{X} \times \mathcal{X}, (x, x')} \mathcal{C}/U \cong (\mathcal{C}/U \times_{x, \mathcal{X}, x'} \mathcal{C}/U) \times_{\mathcal{C}/U \times_S U, \Delta} \mathcal{C}/U $$ and the right hand side is representable by an algebraic space by assumption and the fact that the category of algebraic spaces over $S$ has fibre products and contains $U$ and $S$. \medskip\noindent The equivalences (3) $\Leftrightarrow$ (4), (5) $\Leftrightarrow$ (6), and (7) $\Leftrightarrow$ (8) are formal. The equivalences (3) $\Leftrightarrow$ (5) and (4) $\Leftrightarrow$ (6) follow from Lemma \ref{lemma-representable-by-spaces-morphism-equivalent}. Assume (3), and let $\mathcal{U} \to \mathcal{X}$ be as in (7). To prove (7) we have to show that for every scheme $V$ and $1$-morphism $y : (\Sch/V)_{fppf} \to \mathcal{X}$ the $2$-fibre product $(\Sch/V)_{fppf} \times_{y, \mathcal{X}} \mathcal{U}$ is representable by an algebraic space. Property (3) tells us that $y$ is representable by algebraic spaces hence Lemma \ref{lemma-base-change-by-space-representable-by-space} implies what we want. Finally, (7) directly implies (3).
Assume $\mathcal{X}$ is an algebraic stack (resp.\ a Deligne-Mumford stack). By Stacks, Lemma \ref{stacks-lemma-stack-in-groupoids-equivalent} this implies that $\mathcal{Y}$ is a stack in groupoids over $\Sch_{fppf}$. Choose an equivalence $f : \mathcal{X} \to \mathcal{Y}$ over $\Sch_{fppf}$. This gives a $2$-commutative diagram $$ \xymatrix{ \mathcal{X} \ar[r]_f \ar[d]_{\Delta_\mathcal{X}} & \mathcal{Y} \ar[d]^{\Delta_\mathcal{Y}} \\ \mathcal{X} \times \mathcal{X} \ar[r]^{f \times f} & \mathcal{Y} \times \mathcal{Y} } $$ whose horizontal arrows are equivalences. This implies that $\Delta_\mathcal{Y}$ is representable by algebraic spaces according to Lemma \ref{lemma-representable-by-spaces-morphism-equivalent}. Finally, let $U$ be a scheme over $S$, and let $x : (\Sch/U)_{fppf} \to \mathcal{X}$ be a $1$-morphism which is surjective and smooth (resp.\ \'etale). Considering the diagram $$ \xymatrix{ (\Sch/U)_{fppf} \ar[r]_{\text{id}} \ar[d]_x & (\Sch/U)_{fppf} \ar[d]^{f \circ x} \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } $$ and applying Lemma \ref{lemma-property-morphism-equivalent} we conclude that $f \circ x$ is surjective and smooth (resp.\ \'etale) as desired.
It is clear that (2) implies (3). Parts (1) and (2) are equivalent by Lemma \ref{lemma-equivalent}. Hence it suffices to prove (2). First, we note that $\mathcal{S}_F$ is stack in sets since $F$ is a sheaf (Stacks, Lemma \ref{stacks-lemma-stack-in-setoids-characterize}). A fortiori it is a stack in groupoids. Second the diagonal morphism $\mathcal{S}_F \to \mathcal{S}_F  \times \mathcal{S}_F$ is the same as the morphism $\mathcal{S}_F \to \mathcal{S}_{F \times F}$ which comes from the diagonal of $F$. Hence this is representable by algebraic spaces according to Lemma \ref{lemma-morphism-spaces-gives-representable-by-spaces}. Actually it is even representable (by schemes), as the diagonal of an algebraic space is representable, but we do not need this. Let $U$ be a scheme and let $h_U \to F$ be a surjective \'etale morphism. We may think of this as a surjective \'etale morphism of algebraic spaces. Hence by Lemma \ref{lemma-map-presheaves-representable-by-spaces-transformation-property} the corresponding $1$-morphism $(\Sch/U)_{fppf} \to \mathcal{S}_F$ is surjective and \'etale.
The equivalence of (1) and (2) follows from Stacks, Lemma \ref{stacks-lemma-characterize-stack-in-setoids}. The implication (3) $\Rightarrow$ (1) follows from Lemma \ref{lemma-representable-algebraic}. Finally, assume (1). By Stacks, Lemma \ref{stacks-lemma-stack-in-setoids-characterize} there exists a sheaf $F$ on $(\Sch/S)_{fppf}$ and an equivalence $j : \mathcal{X} \to \mathcal{S}_F$. By Lemma \ref{lemma-map-presheaves-representable-by-algebraic-spaces} the fact that $\Delta_\mathcal{X}$ is representable by algebraic spaces, means that $\Delta_F : F \to F \times F$ is representable by algebraic spaces. Let $U$ be a scheme, and let $x : (\Sch/U)_{fppf} \to \mathcal{X}$ be a surjective \'etale morphism. The composition $j \circ x : (\Sch/U)_{fppf} \to \mathcal{S}_F$ corresponds to a morphism $h_U \to F$ of sheaves. By Bootstrap, Lemma \ref{bootstrap-lemma-representable-diagonal} this morphism is representable by algebraic spaces. Hence by Lemma \ref{lemma-map-fibred-setoids-property} we conclude that $h_U \to F$ is surjective and \'etale. Finally, we apply Bootstrap, Theorem \ref{bootstrap-theorem-bootstrap} to see that $F$ is an algebraic space.
An object of $\mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y}$ over $T$ is just a pair $(x, y)$ where $x$ is an object of $\mathcal{X}_T$ and $y$ is an object of $\mathcal{Y}_T$. Hence it is immediate from the definitions that $\mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y}$ is a stack in groupoids. If $(x, y)$ and $(x', y')$ are two objects of $\mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y}$ over $T$, then $$ \mathit{Isom}((x, y), (x', y')) = \mathit{Isom}(x, x') \times \mathit{Isom}(y, y'). $$ Hence it follows from the equivalences in Lemma \ref{lemma-representable-diagonal} and the fact that the category of algebraic spaces has products that the diagonal of $\mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y}$ is representable by algebraic spaces. Finally, suppose that $U, V \in \Ob((\Sch/S)_{fppf})$, and let $x, y$ be surjective smooth morphisms $x : (\Sch/U)_{fppf} \to \mathcal{X}$, $y : (\Sch/V)_{fppf} \to \mathcal{Y}$. Note that $$ (\Sch/U \times_S V)_{fppf} = (\Sch/U)_{fppf} \times_{(\Sch/S)_{fppf}} (\Sch/V)_{fppf}. $$ The object $(\text{pr}_U^*x, \text{pr}_V^*y)$ of $\mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y}$ over $(\Sch/U \times_S V)_{fppf}$ thus defines a $1$-morphism $$ (\Sch/U \times_S V)_{fppf} \longrightarrow \mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y} $$ which is the composition of base changes of $x$ and $y$, hence is surjective and smooth, see Lemmas \ref{lemma-base-change-representable-transformations-property} and \ref{lemma-composition-representable-transformations-property}. We conclude that $\mathcal{X} \times_{(\Sch/S)_{fppf}} \mathcal{Y}$ is indeed an algebraic stack. We omit the verification that it really is a product.
We have to check conditions (1), (2), and (3) of Definition \ref{definition-algebraic-stack}. The first condition follows from Stacks, Lemma \ref{stacks-lemma-2-product-stacks-in-groupoids}. \medskip\noindent The second condition we have to check is that the $\mathit{Isom}$-sheaves are representable by algebraic spaces. To do this, suppose that $T$ is a scheme over $S$, and $u, v$ are objects of $(\mathcal{X} \times_{f, \mathcal{Z}, g} \mathcal{Y})_T$. By our construction of $2$-fibre products (which goes all the way back to Categories, Lemma \ref{categories-lemma-2-product-categories-over-C}) we may write $u = (x, y, \alpha)$ and $v = (x', y', \alpha')$. Here $\alpha : f(x) \to g(y)$ and similarly for $\alpha'$. Then it is clear that $$ \xymatrix{ \mathit{Isom}(u, v) \ar[d] \ar[rr] & & \mathit{Isom}(y, y') \ar[d]^{\phi \mapsto g(\phi) \circ \alpha} \\ \mathit{Isom}(x, x') \ar[rr]^-{\psi \mapsto \alpha' \circ f(\psi)} & & \mathit{Isom}(f(x), g(y')) } $$ is a cartesian diagram of sheaves on $(\Sch/T)_{fppf}$. Since by assumption the sheaves $\mathit{Isom}(y, y')$, $\mathit{Isom}(x, x')$, $\mathit{Isom}(f(x), g(y'))$ are algebraic spaces (see Lemma \ref{lemma-representable-diagonal}) we see that $\mathit{Isom}(u, v)$ is an algebraic space. \medskip\noindent Let $U, V \in \Ob((\Sch/S)_{fppf})$, and let $x, y$ be surjective smooth morphisms $x : (\Sch/U)_{fppf} \to \mathcal{X}$, $y : (\Sch/V)_{fppf} \to \mathcal{Y}$. Consider the morphism $$ (\Sch/U)_{fppf} \times_{f \circ x, \mathcal{Z}, g \circ y} (\Sch/V)_{fppf} \longrightarrow \mathcal{X} \times_{f, \mathcal{Z}, g} \mathcal{Y}. $$ As the diagonal of $\mathcal{Z}$ is representable by algebraic spaces the source of this arrow is representable by an algebraic space $F$, see Lemma \ref{lemma-representable-diagonal}. Moreover, the morphism is the composition of base changes of $x$ and $y$, hence surjective and smooth, see Lemmas \ref{lemma-base-change-representable-transformations-property} and \ref{lemma-composition-representable-transformations-property}. Choosing a scheme $W$ and a surjective \'etale morphism $W \to F$ we see that the composition of the displayed $1$-morphism with the corresponding $1$-morphism $$ (\Sch/W)_{fppf} \longrightarrow (\Sch/U)_{fppf} \times_{f \circ x, \mathcal{Z}, g \circ y} (\Sch/V)_{fppf} $$ is surjective and smooth which proves the last condition.
The fact that $\mathcal{X} \times_{f, \mathcal{Z}, g} \mathcal{Y}$ is an algebraic stack follows from the stronger Lemma \ref{lemma-2-fibre-product-general}. The fact that $\mathcal{X} \times_{f, \mathcal{Z}, g} \mathcal{Y}$ is a $2$-fibre product in the $2$-category of algebraic stacks over $S$ follows formally from the fact that the $2$-category of algebraic stacks over $S$ is a full sub $2$-category of the $2$-category of stacks in groupoids over $(\Sch/S)_{fppf}$.
First choose $W \in \Ob((\Sch/S)_{fppf})$ and a surjective smooth $1$-morphism $z : (\Sch/W)_{fppf} \to \mathcal{X}$. As $\mathcal{Y}$ is an algebraic stack we may choose an equivalence $$ j : \mathcal{S}_F \longrightarrow (\Sch/W)_{fppf} \times_{f \circ z, \mathcal{Y}, y} (\Sch/V)_{fppf} $$ where $F$ is an algebraic space. By Lemma \ref{lemma-base-change-representable-transformations-property} the morphism $\mathcal{S}_F \to (\Sch/W)_{fppf}$ is surjective and smooth as a base change of $y$. Hence by Lemma \ref{lemma-composition-representable-transformations-property} we see that $\mathcal{S}_F \to \mathcal{X}$ is surjective and smooth. Choose an object $U \in \Ob((\Sch/S)_{fppf})$ and a surjective \'etale morphism $U \to F$. Then applying Lemma \ref{lemma-composition-representable-transformations-property} once more we obtain the desired properties.
Parts (1) and (2) are equivalent by general properties of $1$-morphisms of categories fibred in groupoids, see Categories, Lemma \ref{categories-lemma-equivalence-fibred-categories}. We see that (3) implies (2) by Lemma \ref{lemma-criterion-map-representable-spaces-fibred-in-groupoids}. Finally, assume (2). Let $U$ be a scheme. Let $y \in \Ob(\mathcal{Y}_U)$. We have to prove that $$ \mathcal{W} = (\Sch/U)_{fppf} \times_{y, \mathcal{Y}} \mathcal{X} $$ is representable by an algebraic space over $U$. Since $(\Sch/U)_{fppf}$ is an algebraic stack we see from Lemma \ref{lemma-2-fibre-product} that $\mathcal{W}$ is an algebraic stack. On the other hand the explicit description of objects of $\mathcal{W}$ as triples $(V, x, \alpha : y(V) \to f(x))$ and the fact that $f$ is faithful, shows that the fibre categories of $\mathcal{W}$ are setoids. Hence Proposition \ref{proposition-algebraic-stack-no-automorphisms} guarantees that $\mathcal{W}$ is representable by an algebraic space.
We have to show that $\Delta : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is representable by algebraic spaces, see Definition \ref{definition-algebraic-stack}. Given two schemes $T_1$, $T_2$ over $S$ denote $\mathcal{T}_i = (\Sch/T_i)_{fppf}$ the associated representable fibre categories. Suppose given $1$-morphisms $f_i : \mathcal{T}_i \to \mathcal{X}$. According to Lemma \ref{lemma-representable-diagonal} it suffices to prove that the $2$-fibered product $\mathcal{T}_1 \times_\mathcal{X} \mathcal{T}_2$ is representable by an algebraic space. By Stacks, Lemma \ref{stacks-lemma-2-fibre-product-stacks-in-setoids-over-stack-in-groupoids} this is in any case a stack in setoids. Thus $\mathcal{T}_1 \times_\mathcal{X} \mathcal{T}_2$ corresponds to some sheaf $F$ on $(\Sch/S)_{fppf}$, see Stacks, Lemma \ref{stacks-lemma-stack-in-setoids-characterize}. Let $U$ be the algebraic space which represents $\mathcal{U}$. By assumption $$ \mathcal{T}_i' = \mathcal{U} \times_{u, \mathcal{X}, f_i} \mathcal{T}_i $$ is representable by an algebraic space $T'_i$ over $S$. Hence $\mathcal{T}_1' \times_\mathcal{U} \mathcal{T}_2'$ is representable by the algebraic space $T'_1 \times_U T'_2$. Consider the commutative diagram $$ \xymatrix{ & \mathcal{T}_1 \times_{\mathcal X} \mathcal{T}_2 \ar[rr]\ar'[d][dd] & & \mathcal{T}_1 \ar[dd] \\ \mathcal{T}_1' \times_\mathcal{U} \mathcal{T}_2' \ar[ur]\ar[rr]\ar[dd] & & \mathcal{T}_1' \ar[ur]\ar[dd] \\ & \mathcal{T}_2 \ar'[r][rr] & & \mathcal X \\ \mathcal{T}_2' \ar[rr]\ar[ur] & & \mathcal{U} \ar[ur] } $$ In this diagram the bottom square, the right square, the back square, and the front square are $2$-fibre products. A formal argument then shows that $\mathcal{T}_1' \times_\mathcal{U} \mathcal{T}_2' \to \mathcal{T}_1 \times_{\mathcal X} \mathcal{T}_2$ is the ``base change'' of $\mathcal{U} \to \mathcal{X}$, more precisely the diagram $$ \xymatrix{ \mathcal{T}_1' \times_\mathcal{U} \mathcal{T}_2' \ar[d] \ar[r] & \mathcal{U} \ar[d] \\ \mathcal{T}_1 \times_{\mathcal X} \mathcal{T}_2 \ar[r] & \mathcal{X} } $$ is a $2$-fibre square. Hence $T'_1 \times_U T'_2 \to F$ is representable by algebraic spaces, smooth, and surjective, see Lemmas \ref{lemma-map-fibred-setoids-representable-algebraic-spaces}, \ref{lemma-base-change-representable-by-spaces}, \ref{lemma-map-fibred-setoids-property}, and \ref{lemma-base-change-representable-transformations-property}. Therefore $F$ is an algebraic space by Bootstrap, Theorem \ref{bootstrap-theorem-final-bootstrap} and we win.
Let $\mathcal{V} \to \mathcal{Y}$ be a surjective smooth $1$-morphism from a representable stack in groupoids to $\mathcal{Y}$. This exists by Definition \ref{definition-algebraic-stack}. Then the $2$-fibre product $\mathcal{U} = \mathcal{V} \times_{\mathcal Y} \mathcal X$ is representable by an algebraic space by Lemma \ref{lemma-base-change-by-space-representable-by-space}. The $1$-morphism $\mathcal{U} \to \mathcal X$ is representable by algebraic spaces, smooth, and surjective, see Lemmas \ref{lemma-base-change-representable-by-spaces} and \ref{lemma-base-change-representable-transformations-property}. By Lemma \ref{lemma-smooth-surjective-morphism-implies-algebraic} we conclude that $\mathcal{X}$ is an algebraic stack.
The statement on algebraic stacks will follow from the statement on stacks in groupoids by Lemma \ref{lemma-representable-morphism-to-algebraic}. If $j$ is representable by algebraic spaces, then $j$ is faithful on fibre categories and for each $U$ and each $y \in \Ob(\mathcal{Y}_U)$ the presheaf $$ (h : V \to U) \longmapsto \{(x, \phi) \mid x \in \Ob(\mathcal{X}_V), \phi : h^*y \to f(x)\}/\cong $$ is an algebraic space over $U$. See Lemma \ref{lemma-criterion-map-representable-spaces-fibred-in-groupoids}. In particular this presheaf is a sheaf and the conclusion follows from Stacks, Lemma \ref{stacks-lemma-relative-sheaf-over-stack-is-stack}.
The morphisms $s, t$ are smooth by Lemmas \ref{lemma-property-morphism-equivalent} and \ref{lemma-map-presheaves-representable-by-spaces-transformation-property}. As the $1$-morphism $f$ is smooth and surjective it is clear that given any scheme $T$ and any object $a \in \Ob(\mathcal{X}_T)$ there exists a smooth and surjective morphism $T' \to T$ such that $a|_T'$ comes from an object of $[U/R]_{T'}$. Since $f_{can} : [U/R] \to \mathcal{X}$ is fully faithful, we deduce that $[U/R] \to \mathcal{X}$ is essentially surjective as descent data on objects are effective on both sides, see Stacks, Lemma \ref{stacks-lemma-characterize-essentially-surjective-when-ff}.
Let $T$ be a scheme and let $x : (\Sch/T)_{fppf} \to [U/R]$ be a $1$-morphism. We have to show that the projection $$ \mathcal{S}_U \times_{[U/R]} (\Sch/T)_{fppf} \longrightarrow (\Sch/T)_{fppf} $$ is surjective and smooth. We already know that the left hand side is representable by an algebraic space $F$, see Lemmas \ref{lemma-diagonal-quotient-stack} and \ref{lemma-representable-diagonal}. Hence we have to show the corresponding morphism $F \to T$ of algebraic spaces is surjective and smooth. Since we are working with properties of morphisms of algebraic spaces which are local on the target in the fppf topology we may check this fppf locally on $T$. By construction, there exists an fppf covering $\{T_i \to T\}$ of $T$ such that $x|_{(\Sch/T_i)_{fppf}}$ comes from a morphism $x_i : T_i \to U$. (Note that $F \times_T T_i$ represents the $2$-fibre product $\mathcal{S}_U \times_{[U/R]} (\Sch/T_i)_{fppf}$ so everything is compatible with the base change via $T_i \to T$.) Hence we may assume that $x$ comes from $x : T \to U$. In this case we see that $$ \mathcal{S}_U \times_{[U/R]} (\Sch/T)_{fppf} = (\mathcal{S}_U \times_{[U/R]} \mathcal{S}_U) \times_{\mathcal{S}_U} (\Sch/T)_{fppf} = \mathcal{S}_R \times_{\mathcal{S}_U} (\Sch/T)_{fppf} $$ The first equality by Categories, Lemma \ref{categories-lemma-2-fibre-product-erase-factor} and the second equality by Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}. Clearly the last $2$-fibre product is represented by the algebraic space $F = R \times_{s, U, x} T$ and the projection $R \times_{s, U, x} T \to T$ is smooth as the base change of the smooth morphism of algebraic spaces $s : R \to U$. It is also surjective as $s$ has a section (namely the identity $e : U \to R$ of the groupoid). This proves the lemma.
Let us prove (3). By Lemma \ref{lemma-stack-presentation} we may write $\mathcal{X} = [U/R]$ for some smooth groupoid in algebraic spaces $(U, R, s, t, c)$. By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-change-big-site} we see that $f^{-1}[U/R] = [f^{-1}U/f^{-1}R]$. Of course $(f^{-1}U, f^{-1}R, f^{-1}s, f^{-1}t, f^{-1}c)$ is a smooth groupoid in algebraic spaces too. Hence (3) is proved. \medskip\noindent Now the other cases (1), (2), (4) each mean that $\mathcal{X}$ has a presentation $[U/R]$ of a particular kind, and hence translate into the same kind of presentation for $f^{-1}\mathcal{X} = [f^{-1}U/f^{-1}R]$. Whence the lemma is proved.
The statement makes sense as the functor $j : (\Sch/S)_{fppf} \to (\Sch/S')_{fppf}$ is the localization functor associated to the object $S/S'$ of $(\Sch/S')_{fppf}$. By Stacks, Lemma \ref{stacks-lemma-localize-stacks} the only thing to show is that the constructions A and B preserve the subcategories of algebraic stacks. For example, if $\mathcal{X} = [U/R]$ then construction A applied to $\mathcal{X}$ just produces $\mathcal{X}' = \mathcal{X}$. Conversely, if $\mathcal{X}' = [U'/R']$ the morphism $p$ induces morphisms of algebraic spaces $U' \to S$ and $R' \to S$, and then $\mathcal{X} = [U'/R']$ but now viewed as a stack over $S$. Hence the lemma is clear.
The equivalence of (1) and (2) follows from Stacks, Lemma \ref{stacks-lemma-characterize-stack-in-setoids}. The implication (3) $\Rightarrow$ (1) follows from Lemma \ref{lemma-algebraic-stack-no-automorphisms}. Finally, assume (1). By Stacks, Lemma \ref{stacks-lemma-stack-in-setoids-characterize} there exists an equivalence $j : \mathcal{X} \to \mathcal{S}_F$ where $F$ is a sheaf on $(\Sch/S)_{fppf}$.  By Lemma \ref{lemma-map-presheaves-representable-by-algebraic-spaces} the fact that $\Delta_\mathcal{X}$ is representable by algebraic spaces, means that $\Delta_F : F \to F \times F$ is representable by algebraic spaces. Let $U$ be a scheme and let $x : (\Sch/U)_{fppf} \to \mathcal{X}$ be a surjective smooth morphism. The composition $j \circ x : (\Sch/U)_{fppf} \to \mathcal{S}_F$ corresponds to a morphism $h_U \to F$ of sheaves. By Bootstrap, Lemma \ref{bootstrap-lemma-representable-diagonal} this morphism is representable by algebraic spaces. Hence by Lemma \ref{lemma-map-fibred-setoids-property} we conclude that $h_U \to F$ is surjective and smooth. In particular it is surjective, flat and locally of finite presentation (by Lemma \ref{lemma-representable-transformations-property-implication} and the fact that a smooth morphism of algebraic spaces is flat and locally of finite presentation, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation} and \ref{spaces-morphisms-lemma-smooth-flat}). Finally, we apply Bootstrap, Theorem \ref{bootstrap-theorem-final-bootstrap} to see that $F$ is an algebraic space.
Proof of (\ref{item-sep}). Suppose that $s, s' \in \mathcal{F}^+(U)$ and suppose that there exists some covering $\{U_i \to U\}$ such that $s|_{U_i} = s'|_{U_i}$ for all $i$. We now have three coverings of $U$: the covering $\{U_i \to U\}$ above, a covering $\mathcal{U}$ for $s$ as in Lemma \ref{lemma-plus-surjective}, and a similar covering $\mathcal{U}'$ for $s'$. By Lemma \ref{lemma-common-refinement}, we can find a common refinement, say $\{W_j \to U\}$. This means we have $s_j, s'_j \in \mathcal{F}(W_j)$ such that $s|_{W_j} = \theta(s_j)$, similarly for $s'|_{W_j}$, and such that $\theta(s_j) = \theta(s'_j)$. This last equality means that there exists some covering $\{W_{jk} \to W_j\}$ such that $s_j|_{W_{jk}} = s'_j|_{W_{jk}}$. Then since $\{W_{jk} \to U\}$ is a covering we see that $s, s'$ map to the same element of $H^0(\{W_{jk} \to U\}, \mathcal{F})$ as desired. \medskip\noindent Proof of (\ref{item-sheaf}). It is clear that $\mathcal{F} \to \mathcal{F}^+$ is injective because all the maps $\mathcal{F}(U) \to H^0(\mathcal{U}, \mathcal{F})$ are injective. It is also clear that, if $\mathcal{U} \to \mathcal{U}'$ is a refinement, then $H^0(\mathcal{U}', \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F})$ is injective. Now, suppose that $\{U_i \to U\}$ is a covering, and let $(s_i)$ be a family of elements of $\mathcal{F}^+(U_i)$ satisfying the sheaf condition $s_i|_{U_i \times_U U_j} = s_j|_{U_i \times_U U_j}$ for all $i, j \in I$. Choose coverings (as in Lemma \ref{lemma-plus-surjective}) $\{U_{ij} \to U_i\}$ such that $s_i|_{U_{ij}}$ is the image of the (unique) element $s_{ij} \in \mathcal{F}(U_{ij})$. The sheaf condition implies that $s_{ij}$ and $s_{i'j'}$ agree over $U_{ij} \times_U U_{i'j'}$ because it maps to $U_i \times_U U_{i'}$ and we have the equality there. Hence $(s_{ij}) \in H^0(\{U_{ij} \to U\}, \mathcal{F})$ gives rise to an element $s \in \mathcal{F}^+(U)$. We leave it to the reader to verify that $s|_{U_i} = s_i$. \medskip\noindent Proof of (\ref{item-plus-iso}). This is immediate from the definitions because the sheaf property says exactly that every map $\mathcal{F} \to H^0(\mathcal{U}, \mathcal{F})$ is bijective (for every covering $\mathcal{U}$ of $U$). \medskip\noindent Statement (\ref{item-plusplus}) is now obvious.
It is a tautology that if $J = J'$ then the notions of sheaves are the same. Conversely, Lemma \ref{lemma-sieve-sheafification} characterizes covering sieves in terms of the sheafification functor. But the sheafification functor $\textit{PSh}(\mathcal{C}) \to \Sh(\mathcal{C}, J)$ is the right adjoint of the inclusion functor $\Sh(\mathcal{C}, J) \to \textit{PSh}(\mathcal{C})$. Hence if the subcategories $\Sh(\mathcal{C}, J)$ and $\Sh(\mathcal{C}, J')$ are the same, then the sheafification functors are the same and hence the collections of covering sieves are the same.