The Stacks project

Proof. This is formal. It relies on the fact that the category of algebraic spaces over $S$ has fibre products, see Spaces, Lemma 65.7.3. $\square$

Proof. Omitted. Hint: This is formal. $\square$

Proof. (Same as the proof of Spaces, Lemma 65.3.5.) Let $\{ \varphi _ i : T_ i \to T\} $ be a covering of the site $(\mathit{Sch}/S)_{fppf}$. 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. $\square$

Proof. (Same as the proof of Spaces, Lemma 65.3.6.) 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

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. $\square$

Proof. We have seen in Lemma 80.3.4 that $F$ is a sheaf.

Let $U$ be a scheme and let $U \to G$ be a surjective étale 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 étale morphism.

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

Since both $U \times _ G F$ and $G$ are algebraic spaces we see that this is a scheme.

Next, we claim that $W \to F$ is surjective and étale (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 étale 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 étale, and the composition of surjective étale morphisms is surjective and étale.

Set $R = W \times _ F W$. By the above $R$ is a scheme and the projections $t, s : R \to W$ are étale. It is clear that $R$ is an equivalence relation, and $W \to F$ is a surjection of sheaves. Hence $R$ is an étale equivalence relation and $F = W/R$. Hence $F$ is an algebraic space by Spaces, Theorem 65.10.5. $\square$

Proof. By Lemma 80.3.3 the transformation $X \times _ G F \to X$ is representable by algebraic spaces. Hence it is an algebraic space by Lemma 80.3.6. $\square$

Proof. Let $T$ be a scheme over $S$, and let $T \to H$ be a morphism. By assumption $T \times _ H G$ is an algebraic space. Hence by Lemma 80.3.7 we see that $T \times _ H F = (T \times _ H G) \times _ G F$ is an algebraic space as well. $\square$

Proof. 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 80.3.8 and 80.3.3. $\square$

Proof. Let $U$ be a scheme over $S$ and let $\xi \in G(U)$. Then

Hence the result using Lemma 80.3.7. $\square$

Proof. Let $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ 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 29.25.10. 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.

Let $\{ T_ j \to T\} $ be an fppf covering in $(\mathit{Sch}/S)_{fppf}$. 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 $s = s'$.

Let $\{ T_ j \to T\} $ be an fppf covering in $(\mathit{Sch}/S)_{fppf}$ 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 29.25.10. 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. $\square$