## 62.5 Properties of representable morphisms of presheaves

Here is the definition that makes this work.

Definition 62.5.1. With $S$, and $a : F \to G$ representable as above. Let $\mathcal{P}$ be a property of morphisms of schemes which

is preserved under any base change, see Schemes, Definition 26.18.3, and

is fppf local on the base, see Descent, Definition 35.19.1.

In this case we say that $a$ has *property $\mathcal{P}$* if for every $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $\xi \in G(U)$ the resulting morphism of schemes $V_\xi \to U$ has property $\mathcal{P}$.

It is important to note that we will only use this definition for properties of morphisms that are stable under base change, and local in the fppf topology on the base. This is not because the definition doesn't make sense otherwise; rather it is because we may want to give a different definition which is better suited to the property we have in mind.

Here is a sanity check.

Lemma 62.5.3. Let $S$, $X$, $Y$ be objects of $\mathit{Sch}_{fppf}$. Let $f : X \to Y$ be a morphism of schemes. Let $\mathcal{P}$ be as in Definition 62.5.1. Then $h_ X \longrightarrow h_ Y$ has property $\mathcal{P}$ if and only if $f$ has property $\mathcal{P}$.

**Proof.**
Note that the lemma makes sense by Lemma 62.3.1. Proof omitted.
$\square$

Lemma 62.5.4. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G, H : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $\mathcal{P}$ be a property as in Definition 62.5.1 which is stable under composition. Let $a : F \to G$, $b : G \to H$ be representable transformations of functors. If $a$ and $b$ have property $\mathcal{P}$ so does $b \circ a : F \longrightarrow H$.

**Proof.**
Note that the lemma makes sense by Lemma 62.3.2. Proof omitted.
$\square$

Lemma 62.5.5. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G, H : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $\mathcal{P}$ be a property as in Definition 62.5.1. Let $a : F \to G$ be a representable transformations of functors. Let $b : H \to G$ be any transformation of functors. Consider the fibre product diagram

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

If $a$ has property $\mathcal{P}$ then also the base change $a'$ has property $\mathcal{P}$.

**Proof.**
Note that the lemma makes sense by Lemma 62.3.3. Proof omitted.
$\square$

Lemma 62.5.6. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F, G, H : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Let $\mathcal{P}$ be a property as in Definition 62.5.1. Let $a : F \to G$ be a representable transformations of functors. Let $b : H \to G$ be any transformation of functors. Consider the fibre product diagram

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

Assume that $b$ induces a surjective map of fppf sheaves $H^\# \to G^\# $. In this case, if $a'$ has property $\mathcal{P}$, then also $a$ has property $\mathcal{P}$.

**Proof.**
First we remark that by Lemma 62.3.3 the transformation $a'$ is representable. Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, 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.
$\square$

Lemma 62.5.7. 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. Let $\mathcal{P}$ be a property as in Definition 62.5.1 which is stable under composition. If $a_1$ and $a_2$ have property $\mathcal{P}$ so does $a_1 \times a_2 : F_1 \times F_2 \longrightarrow G_1 \times G_2$.

**Proof.**
Note that the lemma makes sense by Lemma 62.3.4. Proof omitted.
$\square$

Lemma 62.5.8. 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. Let $\mathcal{P}$, $\mathcal{P}'$ be properties as in Definition 62.5.1. Suppose that for any morphism of schemes $f : X \to Y$ we have $\mathcal{P}(f) \Rightarrow \mathcal{P}'(f)$. If $a$ has property $\mathcal{P}$ then $a$ has property $\mathcal{P}'$.

**Proof.**
Formal.
$\square$

Lemma 62.5.9. Let $S$ be a scheme. Let $F, G : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be sheaves. Let $a : F \to G$ be representable, flat, locally of finite presentation, and surjective. Then $a : F \to G$ is surjective as a map of sheaves.

**Proof.**
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 7.11.1.
$\square$

Here is a characterization of those functors for which the diagonal is representable.

Lemma 62.5.10. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$ be a presheaf of sets on $(\mathit{Sch}/S)_{fppf}$. The following are equivalent:

the diagonal $F \to F \times F$ is representable,

for $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $a \in F(U)$ the map $a : h_ U \to F$ is representable,

for every pair $U, V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $a \in F(U)$, $b \in F(V)$ the fibre product $h_ U \times _{a, F, b} h_ V$ is representable.

**Proof.**
This is completely formal, see Categories, Lemma 4.8.4. It depends only on the fact that the category $(\mathit{Sch}/S)_{fppf}$ has products of pairs of objects and fibre products, see Topologies, Lemma 34.7.10.
$\square$

In the situation of the lemma, for any morphism $\xi : h_ U \to F$ as in the lemma, it makes sense to say that $\xi $ has property $\mathcal{P}$, for any property as in Definition 62.5.1. In particular this holds for $\mathcal{P} = $ “surjective” and $\mathcal{P} = $ “étale”, see Remark 62.4.3 above. We will use this remark in the definition of algebraic spaces below.

Lemma 62.5.11. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$ be a presheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Let $\mathcal{P}$ be a property as in Definition 62.5.1. If for every $U, V \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and $a \in F(U)$, $b \in F(V)$ we have

$h_ U \times _{a, F, b} h_ V$ is representable, say by the scheme $W$, and

the morphism $W \to U \times _ S V$ corresponding to $h_ U \times _{a, F, b} h_ V \to h_ U \times h_ V$ has property $\mathcal{P}$,

then $\Delta : F \to F \times F$ is representable and has property $\mathcal{P}$.

**Proof.**
Observe that $\Delta $ is representable by Lemma 62.5.10. 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 62.5.3. Consider $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ 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.
$\square$

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