Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $a : F \to G$ be a representable transformation of functors, see Categories, Definition 4.8.2. This means that for every $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $\xi \in G(U)$ the fiber product $h_ U \times _{\xi , G} F$ is representable. Choose a representing object $V_\xi $ and an isomorphism $h_{V_\xi } \to h_ U \times _ G F$. By the Yoneda lemma, see Categories, Lemma 4.3.5, the projection $h_{V_\xi } \to h_ U \times _ G F \to h_ U$ comes from a unique morphism of schemes $a_\xi : V_\xi \to U$. Suggestively we could represent this by the diagram
where the squiggly arrows represent the Yoneda embedding. Here are some lemmas about this notion that work in great generality.
Lemma 65.3.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$ and let $X$, $Y$ be objects of $(\mathit{Sch}/S)_{fppf}$. Let $f : X \to Y$ be a morphism of schemes. Then is a representable transformation of functors.
\[ h_ f : h_ X \longrightarrow h_ Y \]
Proof. This is formal and relies only on the fact that the category $(\mathit{Sch}/S)_{fppf}$ has fibre products. $\square$
Lemma 65.3.2. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G, H : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $a : F \to G$, $b : G \to H$ be representable transformations of functors. Then is a representable transformation of functors.
\[ b \circ a : F \longrightarrow H \]
Proof. This is entirely formal and works in any category. $\square$
Lemma 65.3.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G, H : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $a : F \to G$ be a representable transformation of functors. Let $b : H \to G$ be any transformation of functors. Consider the fibre product diagram Then the base change $a'$ is a representable transformation of functors.
\[ \xymatrix{ H \times _{b, G, a} F \ar[r]_-{b'} \ar[d]_{a'} & F \ar[d]^ a \\ H \ar[r]^ b & G } \]
Proof. This is entirely formal and works in any category. $\square$
Lemma 65.3.4. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F_ i, G_ i : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$, $i = 1, 2$. Let $a_ i : F_ i \to G_ i$, $i = 1, 2$ be representable transformations of functors. Then is a representable transformation of functors.
\[ a_1 \times a_2 : F_1 \times F_2 \longrightarrow G_1 \times G_2 \]
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 65.3.2 and 65.3.3. $\square$
Lemma 65.3.5. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $a : F \to G$ be a representable transformation of functors. If $G$ is a sheaf, then so is $F$.
Proof. 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 representable presheaf and hence (see remarks in Section 65.2) 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. $\square$
Lemma 65.3.6. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $a : F \to G$ be a representable transformation of functors. Then $\Delta _{F/G} : F \to F \times _ G F$ is representable.
Proof. Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. 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 a scheme $V$ and a morphism $V \to U$ representing the fibre product $h_ U \times _{\xi ', G} F$. 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 $h_ U \times _{\xi ', G} F$ 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 $h_ U \times _{\xi , F \times _ G F} F$ is represented by $V \times _{\Delta , V \times V, (f_1, f_2)} U$ which is a scheme. $\square$
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