## 63.3 Representable morphisms of presheaves

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

$\xymatrix{ V_\xi \ar@{~>}[r] \ar[d]_{a_\xi } & h_{V_\xi } \ar[d] \ar[r] & F \ar[d]^ a \\ U \ar@{~>}[r] & h_ U \ar[r]^\xi & G }$

where the squiggly arrows represent the Yoneda embedding. Here are some lemmas about this notion that work in great generality.

Lemma 63.3.1. Let $S$, $X$, $Y$ be objects of $\mathit{Sch}_{fppf}$. Let $f : X \to Y$ be a morphism of schemes. Then

$h_ f : h_ X \longrightarrow h_ Y$

is a representable transformation of functors.

Proof. This is formal and relies only on the fact that the category $(\mathit{Sch}/S)_{fppf}$ has fibre products. $\square$

Lemma 63.3.2. 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

$b \circ a : F \longrightarrow H$

is a representable transformation of functors.

Proof. This is entirely formal and works in any category. $\square$

Lemma 63.3.3. 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

$\xymatrix{ H \times _{b, G, a} F \ar[r]_-{b'} \ar[d]_{a'} & F \ar[d]^ a \\ H \ar[r]^ b & G }$

Then the base change $a'$ is a representable transformation of functors.

Proof. This is entirely formal and works in any category. $\square$

Lemma 63.3.4. 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

$a_1 \times a_2 : F_1 \times F_2 \longrightarrow G_1 \times G_2$

is a representable transformation of functors.

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 63.3.2 and 63.3.3. $\square$

Lemma 63.3.5. 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 63.2) 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 63.3.6. 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

$g^*\xi _1 = g^*\xi _2 \Leftrightarrow f_1 \circ g = f_2 \circ g.$

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. $\square$

Comment #5432 by P. Licht on

There is a typo in Lemma 02W9: X,Y,f should live in the category of fppf schemes over S.

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