Lemma 94.6.1. Let $f : X \to Y$ be a morphism of $(\mathit{Sch}/S)_{fppf}$. Then the $1$-morphism induced by $f$
\[ (\mathit{Sch}/X)_{fppf} \longrightarrow (\mathit{Sch}/Y)_{fppf} \]
is a representable $1$-morphism.
94.6 Representable morphisms of categories fibred in groupoids
Let $\mathcal{X}$, $\mathcal{Y}$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a representable $1$-morphism, see Categories, Definition 4.42.3. This means that for every $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ the $2$-fibre product $(\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}$ is representable. Choose a representing object $V_ y$ and an equivalence
\[ (\mathit{Sch}/V_ y)_{fppf} \longrightarrow (\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}. \]
The projection $(\mathit{Sch}/V_ y)_{fppf} \to (\mathit{Sch}/U)_{fppf} \times _\mathcal {Y} \mathcal{Y} \to (\mathit{Sch}/U)_{fppf}$ comes from a morphism of schemes $f_ y : V_ y \to U$, see Section 94.4. We represent this by the diagram
94.6.0.1
\begin{equation} \label{algebraic-equation-representable} \vcenter { \xymatrix{ V_ y \ar@{~>}[r] \ar[d]_{f_ y} & (\mathit{Sch}/V_ y)_{fppf} \ar[d] \ar[r] & \mathcal{X} \ar[d]^ f \\ U \ar@{~>}[r] & (\mathit{Sch}/U)_{fppf} \ar[r]^-y & \mathcal{Y} } } \end{equation}
where the squiggly arrows represent the $2$-Yoneda embedding. Here are some lemmas about this notion that work in great generality (namely, they work for categories fibred in groupoids over any base category which has fibre products).
Proof. This is formal and relies only on the fact that the category $(\mathit{Sch}/S)_{fppf}$ has fibre products. $\square$
Lemma 94.6.2. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Consider a $2$-commutative diagram of $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume the horizontal arrows are equivalences. Then $f$ is representable if and only if $f'$ is representable.
\[ \xymatrix{ \mathcal{X}' \ar[r] \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Y}' \ar[r] & \mathcal{Y} } \]
Proof. Omitted. $\square$
Lemma 94.6.3. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be representable $1$-morphisms. Then is a representable $1$-morphism.
\[ g \circ f : \mathcal{X} \longrightarrow \mathcal{Z} \]
Proof. This is entirely formal and works in any category. $\square$
Lemma 94.6.4. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$ Let $f : \mathcal{X} \to \mathcal{Y}$ be a representable $1$-morphism. Let $g : \mathcal{Z} \to \mathcal{Y}$ be any $1$-morphism. Consider the fibre product diagram Then the base change $f'$ is a representable $1$-morphism.
\[ \xymatrix{ \mathcal{Z} \times _{g, \mathcal{Y}, f} \mathcal{X} \ar[r]_-{g'} \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Z} \ar[r]^ g & \mathcal{Y} } \]
Proof. This is entirely formal and works in any category. $\square$
Lemma 94.6.5. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $\mathcal{X}_ i, \mathcal{Y}_ i$ be categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$, $i = 1, 2$. Let $f_ i : \mathcal{X}_ i \to \mathcal{Y}_ i$, $i = 1, 2$ be representable $1$-morphisms. Then is a representable $1$-morphism.
\[ f_1 \times f_2 : \mathcal{X}_1 \times \mathcal{X}_2 \longrightarrow \mathcal{Y}_1 \times \mathcal{Y}_2 \]
Proof. Write $f_1 \times f_2$ as the composition $\mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{Y}_2$. The first arrow is the base change of $f_1$ by the map $\mathcal{Y}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1$, and the second arrow is the base change of $f_2$ by the map $\mathcal{Y}_1 \times \mathcal{Y}_2 \to \mathcal{Y}_2$. Hence this lemma is a formal consequence of Lemmas 94.6.3 and 94.6.4. $\square$
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