Definition 79.3.1. Let $S$ be a scheme contained in $\mathit{Sch}_{fppf}$. Let $F$, $G$ be presheaves on $\mathit{Sch}_{fppf}/S$. We say a morphism $a : F \to G$ is *representable by algebraic spaces* if for every $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and any $\xi : U \to G$ the fiber product $U \times _{\xi , G} F$ is an algebraic space.

## 79.3 Morphisms representable by algebraic spaces

Here we define the notion of one presheaf being relatively representable by algebraic spaces over another, and we prove some properties of this notion.

Here is a sanity check.

Lemma 79.3.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Then $f$ is representable by algebraic spaces.

**Proof.**
This is formal. It relies on the fact that the category of algebraic spaces over $S$ has fibre products, see Spaces, Lemma 64.7.3.
$\square$

Lemma 79.3.3. Let $S$ be a scheme. Let

be a fibre square of presheaves on $(\mathit{Sch}/S)_{fppf}$. If $a$ is representable by algebraic spaces so is $a'$.

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

Lemma 79.3.4. 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 representable by algebraic spaces. If $G$ is a sheaf, then so is $F$.

**Proof.**
(Same as the proof of Spaces, Lemma 64.3.5.) 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.
$\square$

Lemma 79.3.5. 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 representable by algebraic spaces. Then $\Delta _{F/G} : F \to F \times _ G F$ is representable by algebraic spaces.

**Proof.**
(Same as the proof of Spaces, Lemma 64.3.6.) 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. $\square$

The proof of Lemma 79.3.6 below is actually slightly tricky. Namely, we cannot use the argument of the proof of Spaces, Lemma 64.11.3 because we do not yet know that a composition of transformations representable by algebraic spaces is representable by algebraic spaces. In fact, we will use this lemma to prove that statement.

Lemma 79.3.6. 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 representable by algebraic spaces. If $G$ is an algebraic space, then so is $F$.

**Proof.**
We have seen in Lemma 79.3.4 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 64.10.5. $\square$

Lemma 79.3.7. Let $S$ be a scheme. Let $a : F \to G$ be a map of presheaves on $(\mathit{Sch}/S)_{fppf}$. Suppose $a : F \to G$ is representable by algebraic spaces. If $X$ is an algebraic space over $S$, and $X \to G$ is a map of presheaves then $X \times _ G F$ is an algebraic space.

**Proof.**
By Lemma 79.3.3 the transformation $X \times _ G F \to X$ is representable by algebraic spaces. Hence it is an algebraic space by Lemma 79.3.6.
$\square$

Lemma 79.3.8. Let $S$ be a scheme. Let

be maps of presheaves on $(\mathit{Sch}/S)_{fppf}$. If $a$ and $b$ are representable by algebraic spaces, so is $b \circ a$.

**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 79.3.7 we see that $T \times _ H F = (T \times _ H G) \times _ G F$ is an algebraic space as well.
$\square$

Lemma 79.3.9. Let $S$ be a scheme. 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 by algebraic spaces. Then

is a representable by algebraic spaces.

**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 79.3.8 and 79.3.3.
$\square$

Lemma 79.3.10. Let $S$ be a scheme. Let $a : F \to G$ and $b : G \to H$ be transformations of functors $(\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Assume

$\Delta : G \to G \times _ H G$ is representable by algebraic spaces, and

$b \circ a : F \to H$ is representable by algebraic spaces.

Then $a$ is representable by algebraic spaces.

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

Hence the result using Lemma 79.3.7. $\square$

Lemma 79.3.11. Let $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{fppf})$. Let $F$ be a presheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Assume

$F$ is a sheaf for the Zariski topology on $(\mathit{Sch}/S)_{fppf}$,

there exists an index set $I$ and subfunctors $F_ i \subset F$ such that

each $F_ i$ is an fppf sheaf,

each $F_ i \to F$ is representable by algebraic spaces,

$\coprod F_ i \to F$ becomes surjective after fppf sheafification.

Then $F$ is an fppf sheaf.

**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. $\square$

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