hcju/stacksqa huggingface dataset:
By Lemma \ref{lemma-when-it-works-it-works} it suffices to prove that $U/R$ is an algebraic space. Let $U' \to U$ be a surjective, \'etale morphism. Then $\{U' \to U\}$ is in particular an fppf covering. Let $R'$ be the restriction of $R$ to $U'$, see Groupoids, Definition \ref{groupoids-definition-restrict-relation}. According to Groupoids, Lemma \ref{groupoids-lemma-quotient-groupoid-restrict} we see that $U/R \cong U'/R'$. By Lemma \ref{lemma-pullback-etale-equivalence-relation} $R'$ is an \'etale equivalence relation on $U'$. Thus we may replace $U$ by $U'$. \medskip\noindent We apply the previous remark to $U' = \coprod U_i$, where $U = \bigcup U_i$ is an affine open covering of $S$. Hence we may and do assume that $U = \coprod U_i$ where each $U_i$ is an affine scheme. \medskip\noindent Consider the restriction $R_i$ of $R$ to $U_i$. By Lemma \ref{lemma-pullback-etale-equivalence-relation} this is an \'etale equivalence relation. Set $F_i = U_i/R_i$ and $F = U/R$. It is clear that $\coprod F_i \to F$ is surjective. By Lemma \ref{lemma-finding-opens} each $F_i \to F$ is representable, and an open immersion. By Lemma \ref{lemma-presentation-quasi-compact} applied to $(U_i, R_i)$ we see that $F_i$ is an algebraic space. Then by Lemma \ref{lemma-when-it-works-it-works} we see that $U_i \to F_i$ is \'etale and surjective. From Lemma \ref{lemma-coproduct-algebraic-spaces} it follows that $\coprod F_i$ is an algebraic space. Finally, we have verified all hypotheses of Lemma \ref{lemma-glueing-algebraic-spaces} and it follows that $F = U/R$ is an algebraic space.
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-representable-transformations} and \ref{lemma-base-change-representable-transformations}.
First we remark that by Lemma \ref{lemma-base-change-representable-transformations} the transformation $a'$ is representable. Let $U \in \Ob((\Sch/S)_{fppf})$, and let $\xi \in G(U)$. By assumption there exists an fppf covering $\{U_i \to U\}_{i \in I}$ and elements $\xi_i \in H(U_i)$ mapping to $\xi|_U$ via $b$. From general category theory it follows that for each $i$ we have a fibre product diagram $$ \xymatrix{ U_i \times_{\xi_i, H, a'} (H \times_{b, G, a} F) \ar[r] \ar[d] & U \times_{\xi, G, a} F \ar[d] \\ U_i \ar[r] & U } $$ By assumption the left vertical arrow is a morphism of schemes which has property $\mathcal{P}$. Since $\mathcal{P}$ is local in the fppf topology this implies that also the right vertical arrow has property $\mathcal{P}$ as desired.
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 (representable by) a scheme and the morphism $T' \to T$ is a flat, locally of finite presentation, and surjective. Hence $\{T' \to T\}$ is an fppf covering such that $g|_{T'} \in G(T')$ comes from an element of $F(T')$, namely the map $T' \to F$. This proves the map is surjective as a map of sheaves, see Sites, Definition \ref{sites-definition-sheaves-injective-surjective}.
Observe that $\Delta$ is representable by Lemma \ref{lemma-representable-diagonal}. We can formulate condition (2) as saying that the transformation $h_U \times_{a, F, b} h_V \to h_{U \times_S V}$ has property $\mathcal{P}$, see Lemma \ref{lemma-morphism-schemes-gives-representable-transformation-property}. Consider $T \in \Ob((\Sch/S)_{fppf})$ and $(a, b) \in (F \times F)(T)$. Observe that we have the commutative diagram $$ \xymatrix{ F \times_{\Delta, F \times F, (a, b)} h_T \ar[d] \ar[r] & h_T \ar[d]^{\Delta_{T/S}} \\ h_T \times_{a, F, b} h_T \ar[r] \ar[d] & h_{T \times_S T} \ar[d]^{(a, b)} \\ F \ar[r]^\Delta & F \times F } $$ both of whose squares are cartesian. In this way we see that the morphism $F \times_{F \times F} h_T \to h_T$ is the base change of a morphism having property $\mathcal{P}$ by $\Delta_{T/S}$. Since $\mathcal{P}$ is preserved under base change this finishes the proof.
It is clear that $H = F \times G$ is a sheaf. The diagonal of $H$ is simply the product of the diagonals of $F$ and $G$. Hence it is representable by Lemma \ref{lemma-product-representable-transformations}. Finally, if $U \to F$ and $V \to G$ are surjective \'etale morphisms, with $U, V \in \Ob((\Sch/S)_{fppf})$, then $U \times V \to F \times G$ is surjective \'etale by Lemma \ref{lemma-product-representable-transformations-property}.
We check the 3 conditions of Definition \ref{definition-algebraic-space}. A fibre product of sheaves is a sheaf, hence $F \times_H G$ is a sheaf. The diagonal of $F \times_H G$ is the left vertical arrow in $$ \xymatrix{ F \times_H G \ar[r] \ar[d]_\Delta & F \times G \ar[d]^{\Delta_F \times \Delta_G} \\ (F \times F) \times_{(H \times H)} (G \times G) \ar[r] & (F \times F) \times (G \times G) } $$ which is cartesian. Hence $\Delta$ is representable as the base change of the morphism on the right which is representable, see Lemmas \ref{lemma-product-representable-transformations} and \ref{lemma-base-change-representable-transformations}. Finally, let $U, V \in \Ob((\Sch/S)_{fppf})$ and $a : U \to F$, $b : V \to G$ be surjective and \'etale. As $\Delta_H$ is representable, we see that $U \times_H V$ is a scheme. The morphism $$ U \times_H V \longrightarrow F \times_H G $$ is surjective and \'etale as a composition of the base changes $U \times_H V \to U \times_H G$ and $U \times_H G \to F \times_H G$ of the \'etale surjective morphisms $U \to F$ and $V \to G$, see Lemmas \ref{lemma-composition-representable-transformations} and \ref{lemma-base-change-representable-transformations}. This proves the last condition of Definition \ref{definition-algebraic-space} holds and we conclude that $F \times_H G$ is an algebraic space.
It follows from the stronger Lemma \ref{lemma-fibre-product-spaces-over-sheaf-with-representable-diagonal} that $F \times_H G$ is an algebraic space. It is clear that $F \times_H G$ is a fibre product in the category of algebraic spaces over $S$ since that is a full subcategory of the category of (pre)sheaves of sets on $(\Sch/S)_{fppf}$.
Let $U$ be a scheme and let $\xi \in (F \amalg G)(\xi)$. Recall the coproduct in the category of sheaves is the sheafification of the coproduct presheaf (Sites, Lemma \ref{sites-lemma-colimit-sheaves}). Thus there exists an fppf covering $\{g_i : U_i \to U\}_{i \in I}$ and a disjoint union decomposition $I = I' \amalg I''$ such that $U_i \to U \to F \amalg G$ factors through $F$, resp.\ $G$ if and only if $i \in I'$, resp.\ $i \in I''$. Since $F$ and $G$ have empty intersection in $F \amalg G$ we conclude that $U_i \times_U U_j$ is empty if $i \in I'$ and $j \in I''$. Hence $U' = \bigcup_{i \in I'} g_i(U_i)$ and $U'' = \bigcup_{i \in I''} g_i(U_i)$ are disjoint open (Morphisms, Lemma \ref{morphisms-lemma-fppf-open}) subschemes of $U$ with $U = U' \amalg U''$. We omit the verification that $U' = U \times_{F \amalg G} F$.
By Lemma \ref{lemma-coproduct-sheaves-open-and-closed} the map $F_i \to U$ is representable by open and closed immersions. Hence $F_i$ is representable by an open and closed subscheme $U_i$ of $U$. We have $U = \coprod U_i$ because we have $U \cong \coprod F_i$ as sheaves and we can test the equality on points.
The representability of $F \to F \times F$ implies that each diagonal morphism $F_i \to F_i \times F_i$ is representable (immediate from the definitions and the fact that $F \times_{(F \times F)} (F_i \times F_i) = F_i$). Choose a scheme $U$ in $(\Sch/S)_{fppf}$ and a surjective \'etale morphism $U \to F$ (this exist by hypothesis). The base change $U \times_F F_i \to F_i$ is surjective and \'etale by Lemma \ref{lemma-base-change-representable-transformations-property}. On the other hand, $U \times_F F_i$ is a scheme by Lemma \ref{lemma-coproduct-sheaves-open-and-closed}. Thus we have verified all the conditions in Definition \ref{definition-algebraic-space} and $F_i$ is an algebraic space.
Let $T$ be an object of $(\Sch/S)_{fppf}$. Let $T \to F$ be a morphism. By assumption (2)(b) and (2)(c) the fibre product $F_i \times_F T$ is representable by an open subscheme $V_i \subset T$. It follows that $(\coprod F_i) \times_F T$ is represented by the scheme $\coprod V_i$ over $T$. By assumption (2)(d) there exists an fppf covering $\{T_j \to T\}_{j \in J}$ such that $T_j \to T \to F$ factors through $F_i$, $i = i(j)$. Hence $T_j \to T$ factors through the open subscheme $V_{i(j)} \subset T$. Since $\{T_j \to T\}$ is jointly surjective, it follows that $T = \bigcup V_i$ is an open covering. In particular, the transformation of functors $\coprod F_i \to F$ is representable and surjective in the sense of Definition \ref{definition-relative-representable-property} (see Remark \ref{remark-warning} for a discussion). \medskip\noindent Next, let $T' \to F$ be a second morphism from an object in $(\Sch/S)_{fppf}$. Write as above $T' = \bigcup V'_i$ with $V'_i = T' \times_F F_i$. To show that the diagonal $F \to F \times F$ is representable we have to show that $G = T \times_F T'$ is representable, see Lemma \ref{lemma-representable-diagonal}. Consider the subfunctors $G_i = G \times_F F_i$. Note that $G_i = V_i \times_{F_i} V'_i$, and hence is representable as $F_i$ is an algebraic space. By the above the $G_i$ form a Zariski covering of $G$. Hence by Schemes, Lemma \ref{schemes-lemma-glue-functors} we see $G$ is representable. \medskip\noindent Choose a scheme $U \in \Ob((\Sch/S)_{fppf})$ and a surjective \'etale morphism $U \to \coprod F_i$ (this exists by hypothesis). We may write $U = \coprod U_i$ with $U_i$ the inverse image of $F_i$, see Lemma \ref{lemma-representable-sheaf-coproduct-sheaves}. We claim that $U \to F$ is surjective and \'etale. Surjectivity follows as $\coprod F_i \to F$ is surjective (see first paragraph of the proof) by applying Lemma \ref{lemma-composition-representable-transformations-property}. Consider the fibre product $U \times_F T$ where $T \to F$ is as above. We have to show that $U \times_F T \to T$ is \'etale. Since $U \times_F T = \coprod U_i \times_F T$ it suffices to show each $U_i \times_F T \to T$ is \'etale. Since $U_i \times_F T = U_i \times_{F_i} V_i$ this follows from the fact that $U_i \to F_i$ is \'etale and $V_i \to T$ is an open immersion (and Morphisms, Lemmas \ref{morphisms-lemma-open-immersion-etale} and \ref{morphisms-lemma-composition-etale}).
Let $T/S$ be an object of $(\Sch/S)_{fppf}$. Then $R(T) = \{(a, b) \in U(T) \times U(T) \mid f \circ a = f \circ b\}$ which is clearly defines an equivalence relation on $U(T)$. The morphisms $s, t : R \to U$ are \'etale because the morphism $U \to F$ is \'etale. \medskip\noindent To prove (3) we first show that $U \to F$ is a surjection of sheaves, see Sites, Definition \ref{sites-definition-sheaves-injective-surjective}. Let $\xi \in F(T)$ with $T$ as above. Let $V = T \times_{\xi, F, f}U$. By assumption $V$ is a scheme and $V \to T$ is surjective \'etale. Hence $\{V \to T\}$ is a covering for the fppf topology. Since $\xi|_V$ factors through $U$ by construction we conclude $U \to F$ is surjective. Surjectivity implies that $F$ is the coequalizer of the diagram by Sites, Lemma \ref{sites-lemma-coequalizer-surjection}.
It is clear from the description of $s', t'$ in Groupoids, Lemma \ref{groupoids-lemma-restrict-groupoid} that $s' , t' : R' \to U'$ are \'etale as compositions of base changes of \'etale morphisms (see Morphisms, Lemma \ref{morphisms-lemma-base-change-etale} and \ref{morphisms-lemma-composition-etale}).
Denote $c : U \to U/R$ the morphism in question. Let $T$ be a scheme and let $a : T \to U/R$ be a morphism. We have to show that the morphism (of schemes) $\pi : T \times_{a, U/R, c} U \to T$ is \'etale and surjective. The morphism $a$ corresponds to an fppf covering $\{\varphi_i : T_i \to T\}$ and morphisms $a_i : T_i \to U$ such that $a_i \times a_{i'} : T_i \times_T T_{i'} \to U \times_S U$ factors through $R$, and such that $c \circ a_i = a \circ \varphi_i$. Hence $$ T_i \times_{\varphi_i, T} T \times_{a, U/R, c} U = T_i \times_{c \circ a_i, U/R, c} U = T_i \times_{a_i, U} U \times_{c, U/R, c} U = T_i \times_{a_i, U, t} R. $$ Since $t$ is \'etale and surjective we conclude that the base change of $\pi$ to $T_i$ is surjective and \'etale. Since the property of being surjective and \'etale is local on the base in the fpqc topology (see Remark \ref{remark-list-properties-fpqc-local-base}) we win.
Set $R = U \times_F U$. This is a scheme as $U \to F$ is assumed representable. The projections $s, t : R \to U$ are \'etale as $U \to F$ is assumed \'etale. The map $j = (t, s) : R \to U \times_S U$ is a monomorphism and an equivalence relation as $R = U \times_F U$. By Theorem \ref{theorem-presentation} the quotient sheaf $F' = U/R$ is an algebraic space and $U \to F'$ is surjective and \'etale. Again since $R = U \times_F U$ we obtain a canonical factorization $U \to F' \to F$ and $F' \to F$ is an injective map of sheaves. On the other hand, $U \to F$ is surjective as a map of sheaves by Lemma \ref{lemma-surjective-flat-locally-finite-presentation}. Thus $F' \to F$ is also surjective and we conclude $F' = F$ is an algebraic space.
Pick a scheme $U$ and a surjective \'etale morphism $U \to G$. Since $G$ is an algebraic space $U \to G$ is representable. Hence the composition $U \to G \to F$ is representable, surjective, and \'etale. See Lemmas \ref{lemma-composition-representable-transformations} and \ref{lemma-composition-representable-transformations-property}. Thus $F$ is an algebraic space by Lemma \ref{lemma-etale-locally-representable-gives-space}.
By Lemma \ref{lemma-representable-transformation-to-sheaf} we see that $G$ is a sheaf. The diagram $$ \xymatrix{ G \times_F G \ar[r] \ar[d] & F \ar[d]^{\Delta_F} \\ G \times G \ar[r] & F \times F } $$ is cartesian. Hence we see that $G \times_F G \to G \times G$ is representable by Lemma \ref{lemma-base-change-representable-transformations}. By Lemma \ref{lemma-representable-transformation-diagonal} we see that $G \to G \times_F G$ is representable. Hence $\Delta_G : G \to G \times G$ is representable as a composition of representable transformations, see Lemma \ref{lemma-composition-representable-transformations}. Finally, let $U$ be an object of $(\Sch/S)_{fppf}$ and let $U \to F$ be surjective and \'etale. By assumption $U \times_F G$ is representable by a scheme $U'$. By Lemma \ref{lemma-base-change-representable-transformations-property} the morphism $U' \to G$ is surjective and \'etale. This verifies the final condition of Definition \ref{definition-algebraic-space} and we win.
(This lemma follows from Lemmas \ref{lemma-base-change-representable-transformations-property} and \ref{lemma-descent-representable-transformations-property}, but we give a direct proof here also.) It is clear from the definitions that if $G \to F$ has property $\mathcal{P}$, then $V \to U$ has property $\mathcal{P}$. Conversely, assume $V \to U$ has property $\mathcal{P}$. Let $T \to F$ be a morphism from a scheme to $F$. Let $T' = T \times_F G$ which is a scheme since $G \to F$ is representable. We have to show that $T' \to T$ has property $\mathcal{P}$. Consider the commutative diagram of schemes $$ \xymatrix{ V \ar[d] & T \times_F V \ar[d] \ar[l] \ar[r] & T \times_F G \ar[d] \ar@{=}[r] & T' \\ U & T \times_F U \ar[l] \ar[r] & T } $$ where both squares are fibre product squares. Hence we conclude the middle arrow has property $\mathcal{P}$ as a base change of $V \to U$. Finally, $\{T \times_F U \to T\}$ is a fppf covering as it is surjective \'etale, and hence we conclude that $T' \to T$ has property $\mathcal{P}$ as it is local on the base in the fppf topology.
Let $R = U \times_F U$, and denote $t, s : R \to U$ the projection morphisms as usual. Let $T$ be a scheme and let $T \to F$ be a morphism. Then $U \times_F T \to T$ is surjective \'etale, hence $\{U \times_F T \to T\}$ is a covering for the \'etale topology. Consider $$ W = G \times_F (U \times_F T) = V \times_F T = V \times_U (U \times_F T). $$ It is a scheme since $F$ is an algebraic space. The morphism $W \to U \times_F T$ has property $\mathcal{P}$ since it is a base change of $V \to U$. There is an isomorphism \begin{align*} W \times_T (U \times_F T) & = (G \times_F (U \times_F T)) \times_T (U \times_F T) \\ & = (U \times_F T) \times_T (G \times_F (U \times_F T)) \\ & = (U \times_F T) \times_T W \end{align*} over $(U \times_F T) \times_T (U \times_F T)$. The middle equality maps $((g, (u_1, t)), (u_2, t))$ to $((u_1, t), (g, (u_2, t)))$. This defines a descent datum for $W/U \times_F T/T$, see Descent, Definition \ref{descent-definition-descent-datum}. This follows from Descent, Lemma \ref{descent-lemma-descent-data-sheaves}. Namely we have a sheaf $G \times_F T$, whose base change to $U \times_F T$ is represented by $W$ and the isomorphism above is the one from the proof of Descent, Lemma \ref{descent-lemma-descent-data-sheaves}. By assumption on $\mathcal{P}$ the descent datum above is representable. Hence by the last statement of Descent, Lemma \ref{descent-lemma-descent-data-sheaves} we see that $G \times_F T$ is representable. This proves that $G \to F$ is a representable transformation of functors. \medskip\noindent As $G \to F$ is representable, we see that $G$ is an algebraic space by Lemma \ref{lemma-representable-over-space}. The fact that $G \to F$ has property $\mathcal{P}$ now follows from Lemma \ref{lemma-representable-morphisms-spaces-property}.
First choose $W \in \Ob((\Sch/S)_{fppf})$ with surjective \'etale morphism $W \to F$. Next, put $U = W \times_G V$. Since $G$ is an algebraic space we see that $U$ is isomorphic to an object of $(\Sch/S)_{fppf}$. As $q$ is surjective \'etale, we see that $U \to W$ is surjective \'etale (see Lemma \ref{lemma-base-change-representable-transformations-property}). Thus $U \to F$ is surjective \'etale as a composition of surjective \'etale morphisms (see Lemma \ref{lemma-composition-representable-transformations-property}).
Let $F = U/R$ be a presentation of $F$. As in the proof of Lemma \ref{lemma-presentation-quasi-compact} the diagram $$ \xymatrix{ R \ar[r] \ar[d]_j & F \ar[d]^\Delta \\ U \times_S U \ar[r] & F \times F } $$ is cartesian. Hence according to Lemma \ref{lemma-representable-morphisms-spaces-property} it suffices to show that $j$ has the properties listed in the lemma. (Note that each of the properties (1) -- (4) occur in the lists of Remarks \ref{remark-list-properties-stable-base-change} and \ref{remark-list-properties-fpqc-local-base}.) Since $j$ is an equivalence relation it is a monomorphism. Hence it is separated by Schemes, Lemma \ref{schemes-lemma-monomorphism-separated}. As $R$ is an \'etale equivalence relation we see that $s, t : R \to U$ are \'etale. Hence $s, t$ are locally of finite type. Then it follows from Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type} that $j$ is locally of finite type. Finally, as it is a monomorphism its fibres are finite. Thus we conclude that it is locally quasi-finite by Morphisms, Lemma \ref{morphisms-lemma-finite-fibre}.
In the statement of the lemma the symbol $\text{Aut}_S(U)$ denotes the group of automorphisms of $U$ over $S$. Assume $(*)$ holds. Let us show that $$ j : R = \coprod\nolimits_{g \in G} U \longrightarrow U \times_S U, \quad (g, x) \longmapsto (g(x), x) $$ is a monomorphism. This signifies that if $T$ is a nonempty scheme, and $h : T \to U$ is a $T$-valued point such that $g \circ h = g' \circ h$ then $g = g'$. Suppose $T \not = \emptyset$, $h : T \to U$ and $g \circ h = g' \circ h$. Let $t \in T$. Consider the composition $\Spec(\kappa(t)) \to \Spec(\kappa(h(t))) \to U$. Then we conclude that $g^{-1} \circ g'$ fixes $u = h(t)$ and acts as the identity on its residue field. Hence $g = g'$ by $(*)$. \medskip\noindent Thus if $(*)$ holds we see that $j$ is a relation (see Groupoids, Definition \ref{groupoids-definition-equivalence-relation}). Moreover, it is an equivalence relation since on $T$-valued points for a connected scheme $T$ we see that $R(T) = G \times U(T) \to U(T) \times U(T)$ (recall that we always work over $S$). Moreover, the morphisms $s, t : R \to U$ are \'etale since $R$ is a disjoint product of copies of $U$. This proves that $j : R \to U \times_S U$ is an \'etale equivalence relation.
In the proof of Lemma \ref{lemma-properties-diagonal} we saw that it suffices to prove the corresponding properties for the morphism $j : R \to U \times_S U$. If $U \to S$ is quasi-separated, then for every affine open $V \subset U$ which maps into an affine of $S$ the opens $g(V) \cap V$ are quasi-compact. It follows that $j$ is quasi-compact. If $U \to S$ is separated, the diagonal $\Delta_{U/S}$ is a closed immersion. Hence $j : R \to U \times_S U$ is a finite coproduct of closed immersions with disjoint images. Hence $j$ is a closed immersion.
Consider the fibre product $V = \Spec(k) \times_{U/G} U$. Here is a diagram $$ \xymatrix{ V \ar[r] \ar[d] & U \ar[d] \\ \Spec(k) \ar[r] & U/G } $$ Then $V$ is a nonempty scheme \'etale over $\Spec(k)$ and hence is a disjoint union $V = \coprod_{i \in I} \Spec(k_i)$ of spectra of fields $k_i$ finite separable over $k$ (Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}). We have \begin{align*} V \times_{\Spec(k)} V & = (\Spec(k) \times_{U/G} U) \times_{\Spec(k)}(\Spec(k) \times_{U/G} U) \\ & =  \Spec(k) \times_{U/G} U \times_{U/G} U \\ & = \Spec(k) \times_{U/G} U \times G \\ & = V \times G \end{align*} The action of $G$ on $U$ induces an action of $a : G \times V \to V$. The displayed equality means that $G \times V \to V \times_{\Spec(k)} V$, $(g, v) \mapsto (a(g, v), v)$ is an isomorphism. In particular we see that for every $i$ we have an isomorphism $H_i \times \Spec(k_i) \to \Spec(k_i \otimes_k k_i)$ where $H_i \subset G$ is the subgroup of elements fixing $i \in I$. Thus $H_i$ is finite and is the Galois group of $k_i/k$. We omit the converse construction.
Let $X \in \Ob((\Sch/S)_{fppf})$. Let us write $h_X$ for the representable sheaf on $(\Sch/S)_{fppf}$ associated to $X$, and $h'_X$ for the representable sheaf on $(\Sch'/S)_{fppf}$ associated to $X$. By the description of $f^{-1}$ in Topologies, Section \ref{topologies-section-change-alpha} we see that $f^{-1}h_X = h'_X$. This proves (1). \medskip\noindent Next, suppose that $F$ is an algebraic space over $S$. By Lemma \ref{lemma-space-presentation} this means that $F = h_U/h_R$ for some \'etale equivalence relation $R \to U \times_S U$ in $(\Sch/S)_{fppf}$. Since $f^{-1}$ is an exact functor we conclude that $f^{-1}F = h'_U/h'_R$. Hence $f^{-1}F$ is an algebraic space over $S$ by Theorem \ref{theorem-presentation}.