hcju/stacksqa huggingface dataset:
The functor $\text{Mod}_A \to \text{Mod}_{S^{-1}A}$ given by $M \mapsto M \otimes_A S^{-1}A$ is exact (by Algebra, Proposition \ref{algebra-proposition-localization-exact}) and maps modules in $\mathcal{T}$ to zero. Thus, by the universal property given in Homology, Lemma \ref{homology-lemma-serre-subcategory-is-kernel}, the functor descends to a functor $\text{Mod}_A/\mathcal{T} \to \text{Mod}_{S^{-1}A}$. \medskip\noindent Conversely, any $A$-module $M$ with $M \otimes_A S^{-1}A = 0$ is an object of $\mathcal{T}$, since $M \otimes_A S^{-1}A \cong S^{-1} M$ (Algebra, Lemma \ref{algebra-lemma-tensor-localization}). Thus Homology, Lemma \ref{homology-lemma-quotient-by-kernel-exact-functor} shows that the functor $\text{Mod}_A/\mathcal{T} \to \text{Mod}_{S^{-1}A}$ is faithful. \medskip\noindent Furthermore, this embedding is essentially surjective: a preimage to an $S^{-1}A$-module $N$ is $N_A$, that is $N$ regarded as an $A$-module, since the canonical map $N_A \otimes_A S^{-1}A \to N$ which maps $x \otimes a/s$ to $(a/s) \cdot x$ is an isomorphism of $S^{-1}A$-modules.
The equivalence given in Proposition \ref{proposition-quotient-by-torsion-modules} restricts along the embedding $\text{Mod}_A^{fg}/\mathcal{T}^{fg} \to \text{Mod}_A/\mathcal{T}$ to an equivalence $\text{Mod}_A^{fg}/\mathcal{T}^{fg} \to \text{Vect}_K^{fd}$. The Noetherian assumption guarantees that $\text{Mod}_A^{fg}$ is an abelian category (see More on Algebra, Section \ref{more-algebra-section-abelian-categories-modules}) and that the canonical functor $\text{Mod}_A^{fg}/\mathcal{T}^{fg} \to \text{Mod}_A/\mathcal{T}$ is full (else torsion submodules of finitely generated modules might not be objects of $\mathcal{T}^{fg}$).
We will use the remarks directly below Definition \ref{definition-property-transformation} without further mention. In the situation of the theorem, let $U \to X$ be a surjective \'etale morphism from a scheme towards $X$. By Lemma \ref{lemma-composition-transformation} $U \to F$ is surjective and \'etale also. Hence the theorem boils down to proving that $\Delta_F$ is representable. This follows immediately from Lemma \ref{lemma-bootstrap-diagonal}. On the other hand we can circumvent this lemma and show directly $F$ is an algebraic space as in the next paragraph. \medskip\noindent Let $U$ be a scheme, and let $U \to F$ be surjective and \'etale. Set $R = U \times_F U$, which is an algebraic space (see Lemma \ref{lemma-representable-diagonal}). The morphism of algebraic spaces $R \to U \times_S U$ is a monomorphism, hence separated (as the diagonal of a monomorphism is an isomorphism). Moreover, since $U \to F$ is \'etale, we see that $R \to U$ is \'etale, by Lemma \ref{lemma-base-change-transformation-property}. In particular, we see that $R \to U$ is locally quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-etale-locally-quasi-finite}. We conclude that also $R \to U \times_S U$ is locally quasi-finite by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-quasi-finite}. Hence Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme} applies and $R$ is a scheme. Hence $F = U/R$ is an algebraic space according to Spaces, Theorem \ref{spaces-theorem-presentation}.
(Same as the proof of Spaces, Lemma \ref{spaces-lemma-representable-transformation-to-sheaf}.) Let $\{\varphi_i : T_i \to T\}$ be a covering of the site $(\Sch/S)_{fppf}$. Let $s_i \in F(T_i)$ which satisfy the sheaf condition. Then $\sigma_i = a(s_i) \in G(T_i)$ satisfy the sheaf condition also. Hence there exists a unique $\sigma \in G(T)$ such that $\sigma_i = \sigma|_{T_i}$. By assumption $F' = h_T \times_{\sigma, G, a} F$ is a sheaf. Note that $(\varphi_i, s_i) \in F'(T_i)$ satisfy the sheaf condition also, and hence come from some unique $(\text{id}_T, s) \in F'(T)$. Clearly $s$ is the section of $F$ we are looking for.
(Same as the proof of Spaces, Lemma \ref{spaces-lemma-representable-transformation-diagonal}.) Let $U$ be a scheme. Let $\xi = (\xi_1, \xi_2) \in (F \times_G F)(U)$. Set $\xi' = a(\xi_1) = a(\xi_2) \in G(U)$. By assumption there exist an algebraic space $V$ and a morphism $V \to U$ representing the fibre product $U \times_{\xi', G} F$. In particular, the elements $\xi_1, \xi_2$ give morphisms $f_1, f_2 : U \to V$ over $U$. Because $V$ represents the fibre product $U \times_{\xi', G} F$ and because $\xi' = a \circ \xi_1 = a \circ \xi_2$ we see that if $g : U' \to U$ is a morphism then $$ g^*\xi_1 = g^*\xi_2 \Leftrightarrow f_1 \circ g = f_2 \circ g. $$ In other words, we see that $U \times_{\xi, F \times_G F} F$ is represented by $V \times_{\Delta, V \times V, (f_1, f_2)} U$ which is an algebraic space.
We have seen in Lemma \ref{lemma-representable-by-spaces-transformation-to-sheaf} that $F$ is a sheaf. \medskip\noindent Let $U$ be a scheme and let $U \to G$ be a surjective \'etale morphism. In this case $U \times_G F$ is an algebraic space. Let $W$ be a scheme and let $W \to U \times_G F$ be a surjective \'etale morphism. \medskip\noindent First we claim that $W \to F$ is representable. To see this let $X$ be a scheme and let $X \to F$ be a morphism. Then $$ W \times_F X = W \times_{U \times_G F} U \times_G F \times_F X = W \times_{U \times_G F} (U \times_G X) $$ Since both $U \times_G F$ and $G$ are algebraic spaces we see that this is a scheme. \medskip\noindent Next, we claim that $W \to F$ is surjective and \'etale (this makes sense now that we know it is representable). This follows from the formula above since both $W \to U \times_G F$ and $U \to G$ are \'etale and surjective, hence $W \times_{U \times_G F} (U \times_G X) \to U \times_G X$ and $U \times_G X \to X$ are surjective and \'etale, and the composition of surjective \'etale morphisms is surjective and \'etale. \medskip\noindent Set $R = W \times_F W$. By the above $R$ is a scheme and the projections $t, s : R \to W$ are \'etale. It is clear that $R$ is an equivalence relation, and $W \to F$ is a surjection of sheaves. Hence $R$ is an \'etale equivalence relation and $F = W/R$. Hence $F$ is an algebraic space by Spaces, Theorem \ref{spaces-theorem-presentation}.
Write $a_1 \times a_2$ as the composition $F_1 \times F_2 \to G_1 \times F_2 \to G_1 \times G_2$. The first arrow is the base change of $a_1$ by the map $G_1 \times F_2 \to G_1$, and the second arrow is the base change of $a_2$ by the map $G_1 \times G_2 \to G_2$. Hence this lemma is a formal consequence of Lemmas \ref{lemma-composition-transformation} and \ref{lemma-base-change-transformation}.
Let $U$ be a scheme over $S$ and let $\xi \in G(U)$. Then $$ U \times_{\xi, G, a} F = (U \times_{b(\xi), H, b \circ a} F) \times_{(\xi, a), (G \times_H G), \Delta} G $$ Hence the result using Lemma \ref{lemma-representable-by-spaces}.
Let $T \in \Ob((\Sch/S)_{fppf})$ and let $s \in F(T)$. By (2)(c) there exists an fppf covering $\{T_j \to T\}$ such that $s|_{T_j}$ is a section of $F_{\alpha(j)}$ for some $\alpha(j) \in I$. Let $W_j \subset T$ be the image of $T_j \to T$ which is an open subscheme Morphisms, Lemma \ref{morphisms-lemma-fppf-open}. By (2)(b) we see $F_{\alpha(j)} \times_{F, s|_{W_j}} W_j \to W_j$ is a monomorphism of algebraic spaces through which $T_j$ factors. Since $\{T_j \to W_j\}$ is an fppf covering, we conclude that $F_{\alpha(j)} \times_{F, s|_{W_j}} W_j = W_j$, in other words $s|_{W_j} \in F_{\alpha(j)}(W_j)$. Hence we conclude that $\coprod F_i \to F$ is surjective for the Zariski topology. \medskip\noindent Let $\{T_j \to T\}$ be an fppf covering in $(\Sch/S)_{fppf}$. Let $s, s' \in F(T)$ with $s|_{T_j} = s'|_{T_j}$ for all $j$. We want to show that $s, s'$ are equal. As $F$ is a Zariski sheaf by (1) we may work Zariski locally on $T$. By the result of the previous paragraph we may assume there exist $i$ such that $s \in F_i(T)$. Then we see that $s'|_{T_j}$ is a section of $F_i$. By (2)(b) we see $F_{i} \times_{F, s'} T \to T$ is a monomorphism of algebraic spaces through which all of the $T_j$ factor. Hence we conclude that $s' \in F_i(T)$. Since $F_i$ is a sheaf for the fppf topology we conclude that $s = s'$. \medskip\noindent Let $\{T_j \to T\}$ be an fppf covering in $(\Sch/S)_{fppf}$ and let $s_j \in F(T_j)$ such that $s_j|_{T_j \times_T T_{j'}} = s_{j'}|_{T_j \times_T T_{j'}}$. By assumption (2)(b) we may refine the covering and assume that $s_j \in F_{\alpha(j)}(T_j)$ for some $\alpha(j) \in I$. Let $W_j \subset T$ be the image of $T_j \to T$ which is an open subscheme Morphisms, Lemma \ref{morphisms-lemma-fppf-open}. Then $\{T_j \to W_j\}$ is an fppf covering. Since $F_{\alpha(j)}$ is a sub presheaf of $F$ we see that the two restrictions of $s_j$ to $T_j \times_{W_j} T_j$ agree as elements of $F_{\alpha(j)}(T_j \times_{W_j} T_j)$. Hence, the sheaf condition for $F_{\alpha(j)}$ implies there exists a $s'_j \in F_{\alpha(j)}(W_j)$ whose restriction to $T_j$ is $s_j$. For a pair of indices $j$ and $j'$ the sections $s'_j|_{W_j \cap W_{j'}}$ and $s'_{j'}|_{W_j \cap W_{j'}}$ of $F$ agree by the result of the previous paragraph. This finishes the proof by the fact that $F$ is a Zariski sheaf.
Let $T$ be a scheme over $S$ and let $g : T \to G$ be a $T$-valued point of $G$. By assumption $T' = F \times_G T$ is an algebraic space and the morphism $T' \to T$ is a flat, locally of finite presentation, and surjective morphism of algebraic spaces. Let $U \to T'$ be a surjective \'etale morphism, where $U$ is a scheme. Then by the definition of flat morphisms of algebraic spaces the morphism of schemes $U \to T$ is flat. Similarly for ``locally of finite presentation''. The morphism $U \to T$ is surjective also, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-surjective-local}. Hence we see that $\{U \to T\}$ is an fppf covering such that $g|_U \in G(U)$ comes from an element of $F(U)$, namely the map $U \to T' \to F$. This proves the map is surjective as a map of sheaves, see Sites, Definition \ref{sites-definition-sheaves-injective-surjective}.
Assume (1). Let $X \to F$ be as in (3). Let $T$ be a scheme, and let $T \to F$ be a morphism. Then we have $$ T \times_F X = (T \times_S X) \times_{F \times F, \Delta} F $$ which is an algebraic space by Lemma \ref{lemma-representable-by-spaces} and (1). Hence $X \to F$ is representable, i.e., (3) holds. The implication (3) $\Rightarrow$ (2) is trivial. Assume (2). Let $T$ be a scheme, and let $(a, b) : T \to F \times F$ be a morphism. Then $$ F \times_{\Delta_F, F \times F} T = (T \times_{a, F, b} T) \times_{T \times T, \Delta_T} T $$ which is an algebraic space by assumption. Hence $\Delta_F$ is representable by algebraic spaces, i.e., (1) holds.
Let $T$ be a scheme, and let $T \to G$ be a morphism. We have to show that $T \times_G H$ is a scheme, and that the morphism $T \times_G H \to T$ is separated and locally quasi-finite. Thus we may base change the whole diagram to $T$ and assume that $G$ is a scheme. In this case $F$ is an algebraic space. Let $U$ be a scheme, and let $U \to F$ be a surjective \'etale morphism. Then $U \to F$ is representable, surjective, flat and locally of finite presentation by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-flat} and \ref{spaces-morphisms-lemma-etale-locally-finite-presentation}. By Lemma \ref{lemma-composition-transformation} $U \to G$ is surjective, flat and locally of finite presentation also. Note that the base change $E \times_F U \to U$ of $a$ is still separated and locally quasi-finite (by Lemma \ref{lemma-base-change-transformation-property}). Hence we may replace the upper part of the diagram of the lemma by $E \times_F U \to U$. In other words, we may assume that $F \to G$ is a surjective, flat morphism of schemes which is locally of finite presentation. In particular, $\{F \to G\}$ is an fppf covering of schemes. By Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme} we conclude that $E$ is a scheme also. By Descent, Lemma \ref{descent-lemma-descent-data-sheaves} the fact that $E = H \times_G F$ means that we get a descent datum on $E$ relative to the fppf covering $\{F \to G\}$. By More on Morphisms, Lemma \ref{more-morphisms-lemma-separated-locally-quasi-finite-morphisms-fppf-descend} this descent datum is effective. By Descent, Lemma \ref{descent-lemma-descent-data-sheaves} again this implies that $H$ is a scheme. By Descent, Lemmas \ref{descent-lemma-descending-property-separated} and \ref{descent-lemma-descending-property-quasi-finite} it now follows that $b$ is separated and locally quasi-finite.
Let $U \to X$ be a surjective \'etale morphism from a scheme towards $X$. Then $U \to X$ is representable, surjective, flat and locally of finite presentation by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-flat} and \ref{spaces-morphisms-lemma-etale-locally-finite-presentation}. By Lemma \ref{lemma-composition-transformation-property} the composition $U \to F$ is representable by algebraic spaces, surjective, flat and locally of finite presentation also. Thus we see that $R = U \times_F U$ is an algebraic space, see Lemma \ref{lemma-representable-by-spaces}. The morphism of algebraic spaces $R \to U \times_S U$ is a monomorphism, hence separated (as the diagonal of a monomorphism is an isomorphism, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism}). Since $U \to F$ is locally of finite presentation, both morphisms $R \to U$ are locally of finite presentation, see Lemma \ref{lemma-base-change-transformation-property}. Hence $R \to U \times_S U$ is locally of finite type (use Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-finite-presentation-finite-type} and \ref{spaces-morphisms-lemma-permanence-finite-type}). Altogether this means that $R \to U \times_S U$ is a monomorphism which is locally of finite type, hence a separated and locally quasi-finite morphism, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite}. \medskip\noindent Now we are ready to prove that $\Delta_F$ is representable. Let $T$ be a scheme, and let $(a, b) : T \to F \times F$ be a morphism. Set $$ T' = (U \times_S U) \times_{F \times F} T. $$ Note that $U \times_S U \to F \times F$ is representable by algebraic spaces, surjective, flat and locally of finite presentation by Lemma \ref{lemma-product-transformations-property}. Hence $T'$ is an algebraic space, and the projection morphism $T' \to T$ is surjective, flat, and locally of finite presentation. Consider $Z = T \times_{F \times F} F$ (this is a sheaf) and $$ Z' = T' \times_{U \times_S U} R = T' \times_T Z. $$ We see that $Z'$ is an algebraic space, and $Z' \to T'$ is separated and locally quasi-finite by the discussion in the first paragraph of the proof which showed that $R$ is an algebraic space and that the morphism $R \to U \times_S U$ has those properties. Hence we may apply Lemma \ref{lemma-after-fppf-sep-lqf} to the diagram $$ \xymatrix{ Z' \ar[r] \ar[d] & T' \ar[d] \\ Z \ar[r] & T } $$ and we conclude.
Let $T$ be a scheme and let $T \to F$ be a morphism. We have to show that the algebraic space $X \times_F T$ is representable by a scheme. Consider the morphism $$ X \times_F T  \longrightarrow X \times_{\Spec(\mathbf{Z})} T $$ Since $X \times_F T \to T$ is locally quasi-finite, so is the displayed arrow (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-quasi-finite}). On the other hand, the displayed arrow is a monomorphism and hence separated (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-separated}). Thus $X \times_F T$ is a scheme by Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme}.
Consider the morphism $\coprod F_i \to F$. This is the base change of $\coprod S_i \to S$ via $F \to S$. Hence it is representable, locally of finite presentation, flat and surjective by our definition of an fppf covering and Lemma \ref{lemma-base-change-transformation-property}. Thus Theorem \ref{theorem-final-bootstrap} applies to show that $F$ is an algebraic space.
We will use Lemma \ref{lemma-locally-algebraic-space} above. To do this we will show that the assumption that $F_i$ is of finite type over $S_i$ to prove that the set theoretic condition in the lemma is satisfied (after perhaps refining the given covering of $S$ a bit). We suggest the reader skip the rest of the proof. \medskip\noindent If $S'_i \to S_i$ is a morphism of schemes then $$ h_{S'_i} \times F = h_{S'_i} \times_{h_{S_i}} h_{S_i} \times F = h_{S'_i} \times_{h_{S_i}} F_i $$ is an algebraic space of finite type over $S'_i$, see Spaces, Lemma \ref{spaces-lemma-fibre-product-spaces} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type}. Thus we may refine the given covering. After doing this we may assume: (a) each $S_i$ is affine, and (b) the cardinality of $I$ is at most the cardinality of the set of points of $S$. (Since to cover all of $S$ it is enough that each point is in the image of $S_i \to S$ for some $i$.) \medskip\noindent Since each $S_i$ is affine and each $F_i$ of finite type over $S_i$ we conclude that $F_i$ is quasi-compact. Hence by Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover} we can find an affine $U_i \in \Ob((\Sch/S)_{fppf})$ and a surjective \'etale morphism $U_i \to F_i$. The fact that $F_i \to S_i$ is locally of finite type then implies that $U_i \to S_i$ is locally of finite type, and in particular $U_i \to S$ is locally of finite type. By Sets, Lemma \ref{sets-lemma-bound-finite-type} we conclude that $\text{size}(U_i) \leq \text{size}(S)$. Since also $|I| \leq \text{size}(S)$ we conclude that $\coprod_{i \in I} U_i$ is isomorphic to an object of $(\Sch/S)_{fppf}$ by Sets, Lemma \ref{sets-lemma-bound-size} and the construction of $\Sch$. This implies that $\coprod F_i$ is an algebraic space by Spaces, Lemma \ref{spaces-lemma-coproduct-algebraic-spaces} and we win.
Proof of (1). By Descent on Spaces, Lemma \ref{spaces-descent-lemma-descent-data-sheaves} this translates into the statement that an fppf sheaf $F$ endowed with a map $F \to X$ is an algebraic space provided that each $F \times_X X_i$ is an algebraic space. The restriction on the cardinality of $I$ implies that coproducts of algebraic spaces indexed by $I$ are algebraic spaces, see Spaces, Lemma \ref{spaces-lemma-coproduct-algebraic-spaces} and Sets, Lemma \ref{sets-lemma-what-is-in-it}. The morphism $$ \coprod F \times_X X_i \longrightarrow F $$ is representable by algebraic spaces (as the base change of $\coprod X_i \to X$, see Lemma \ref{lemma-base-change-transformation}), and surjective, flat, and locally of finite presentation (as the base change of $\coprod X_i \to X$, see Lemma \ref{lemma-base-change-transformation-property}). Hence part (1) follows from Theorem \ref{theorem-final-bootstrap}. \medskip\noindent Proof of (2). First we apply Descent on Spaces, Lemma \ref{spaces-descent-lemma-descent-data-sheaves} to obtain an fppf sheaf $F$ endowed with a map $F \to X$ such that $F \times_X X_i = Y_i$ for all $i \in I$. Our goal is to show that $F$ is an algebraic space. Choose a scheme $U$ and a surjective \'etale morphism $U \to X$. Then $F' = U \times_X F \to F$ is representable, surjective, and \'etale as the base change of $U \to X$. By Theorem \ref{theorem-final-bootstrap} it suffices to show that $F' = U \times_X F$ is an algebraic space. We may choose an fppf covering $\{U_j \to U\}_{j \in J}$ where $U_j$ is a scheme refining the fppf covering $\{X_i \times_X U \to U\}_{i \in I}$, see Topologies on Spaces, Lemma \ref{spaces-topologies-lemma-refine-fppf-schemes}. Thus we get a map $a : J \to I$ and for each $j$ a morphism $U_j \to X_{a(j)}$ over $X$. Then we see that $U_j \times_U F' = U_j \times_{X_{a(j)}} Y_{a(j)}$ is of finite type over $U_j$. Hence $F'$ is an algebraic space by Lemma \ref{lemma-locally-algebraic-space-finite-type}.
Let $U$ be a scheme over $S$ and let $\xi \in H(U)$. We have to show that $U \times_{\xi, H} G$ is an algebraic space. On the other hand, we know that $U \times_{\xi, H} F$ is an algebraic space and that $U \times_{\xi, H} F \to U \times_{\xi, H} G$ is representable by algebraic spaces, flat, locally of finite presentation, and surjective as a base change of the morphism $a$ (see Lemma \ref{lemma-base-change-transformation-property}). Thus the result follows from Theorem \ref{theorem-final-bootstrap}.
By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-isom} there exists an fppf covering $\{T_i \to T\}_{i \in I}$ such that $\mathit{Isom}(x, y)|_{(\Sch/T_i)_{fppf}}$ is an algebraic space for each $i$. By Spaces, Lemma \ref{spaces-lemma-rephrase} this means that each $F_i = h_{S_i} \times \mathit{Isom}(x, y)$ is an algebraic space. Thus to prove the lemma we only have to verify the set theoretic condition that $\coprod F_i$ is an algebraic space of Lemma \ref{lemma-locally-algebraic-space} above to conclude. To do this we use Spaces, Lemma \ref{spaces-lemma-coproduct-algebraic-spaces} which requires showing that $I$ and the $F_i$ are not ``too large''. We suggest the reader skip the rest of the proof. \medskip\noindent Choose $U' \in \Ob(\Sch/S)_{fppf}$ and a surjective \'etale morphism $U' \to U$. Let $R'$ be the restriction of $R$ to $U'$. Since $[U/R] = [U'/R']$ we may, after replacing $U$ by $U'$, assume that $U$ is a scheme. (This step is here so that the fibre products below are over a scheme.) \medskip\noindent Note that if we refine the covering $\{T_i \to T\}$ then it remains true that each $F_i$ is an algebraic space. Hence we may assume that each $T_i$ is affine. Since $T_i \to T$ is locally of finite presentation, this then implies that $\text{size}(T_i) \leq \text{size}(T)$, see Sets, Lemma \ref{sets-lemma-bound-finite-type}. We may also assume that the cardinality of the index set $I$ is at most the cardinality of the set of points of $T$ since to get a covering it suffices to check that each point of $T$ is in the image. Hence $|I| \leq \text{size}(T)$. Choose $W \in \Ob((\Sch/S)_{fppf})$ and a surjective \'etale morphism $W \to R$. Note that in the proof of Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-isom} we showed that $F_i$ is representable by $T_i \times_{(y_i, x_i), U \times_B U} R$ for some $x_i, y_i : T_i \to U$. Hence now we see that $V_i = T_i \times_{(y_i, x_i), U \times_B U} W$ is a scheme which comes with an \'etale surjection $V_i \to F_i$. By Sets, Lemma \ref{sets-lemma-bound-size-fibre-product} we see that $$ \text{size}(V_i) \leq \max\{\text{size}(T_i), \text{size}(W)\} \leq \max\{\text{size}(T), \text{size}(W)\} $$ Hence, by Sets, Lemma \ref{sets-lemma-bound-size} we conclude that $$ \text{size}(\coprod\nolimits_{i \in I} V_i) \leq \max\{|I|, \text{size}(T), \text{size}(W)\}. $$ Hence we conclude by our construction of $\Sch$ that $\coprod_{i \in I} V_i$ is isomorphic to an object $V$ of $(\Sch/S)_{fppf}$. This verifies the hypothesis of Spaces, Lemma \ref{spaces-lemma-coproduct-algebraic-spaces} and we win.
This is almost but not quite a triviality. Namely, by Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-pre-equivalence} and the fact that $j$ is a monomorphism we see that $R = U \times_F U$. Choose a scheme $W$ and a surjective \'etale morphism $W \to F$. As $U \to F$ is a surjection of sheaves we can find an fppf covering $\{W_i \to W\}$ and maps $W_i \to U$ lifting the morphisms $W_i \to F$. Then we see that $$ W_i \times_F U = W_i \times_U U \times_F U = W_i \times_{U, t} R $$ and the projection $W_i \times_F U \to W_i$ is the base change of $t : R \to U$ hence flat and locally of finite presentation, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-base-change-flat} and \ref{spaces-morphisms-lemma-base-change-finite-presentation}. Hence by Descent on Spaces, Lemmas \ref{spaces-descent-lemma-descending-property-flat} and \ref{spaces-descent-lemma-descending-property-locally-finite-presentation} we see that $U \to F$ is flat and locally of finite presentation. It is surjective by Spaces, Remark \ref{spaces-remark-warning}.