## 79.12 Algebraic spaces in the étale topology

Let $S$ be a scheme. Instead of working with sheaves over the big fppf site $(\mathit{Sch}/S)_{fppf}$ we could work with sheaves over the big étale site $(\mathit{Sch}/S)_{\acute{e}tale}$. All of the material in Algebraic Spaces, Sections 64.3 and 64.5 makes sense for sheaves over $(\mathit{Sch}/S)_{\acute{e}tale}$. Thus we get a second notion of algebraic spaces by working in the étale topology. This notion is (a priori) weaker then the notion introduced in Algebraic Spaces, Definition 64.6.1 since a sheaf in the fppf topology is certainly a sheaf in the étale topology. However, the notions are equivalent as is shown by the following lemma.

Lemma 79.12.1. Denote the common underlying category of $\mathit{Sch}_{fppf}$ and $\mathit{Sch}_{\acute{e}tale}$ by $\mathit{Sch}_\alpha $ (see Topologies, Remark 34.11.1). Let $S$ be an object of $\mathit{Sch}_\alpha $. Let

\[ F : (\mathit{Sch}_\alpha /S)^{opp} \longrightarrow \textit{Sets} \]

be a presheaf with the following properties:

$F$ is a sheaf for the étale topology,

the diagonal $\Delta : F \to F \times F$ is representable, and

there exists $U \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha /S)$ and $U \to F$ which is surjective and étale.

Then $F$ is an algebraic space in the sense of Algebraic Spaces, Definition 64.6.1.

**Proof.**
Note that properties (2) and (3) of the lemma and the corresponding properties (2) and (3) of Algebraic Spaces, Definition 64.6.1 are independent of the topology. This is true because these properties involve only the notion of a fibre product of presheaves, maps of presheaves, the notion of a representable transformation of functors, and what it means for such a transformation to be surjective and étale. Thus all we have to prove is that an étale sheaf $F$ with properties (2) and (3) is also an fppf sheaf.

To do this, let $R = U \times _ F U$. By (2) the presheaf $R$ is representable by a scheme and by (3) the projections $R \to U$ are étale. Thus $j : R \to U \times _ S U$ is an étale equivalence relation. Moreover $U \to F$ identifies $F$ as the quotient of $U$ by $R$ for the étale topology: (a) if $T \to F$ is a morphism, then $\{ T \times _ F U \to T\} $ is an étale covering, hence $U \to F$ is a surjection of sheaves for the étale topology, (b) if $a, b : T \to U$ map to the same section of $F$, then $(a, b) : T \to R$ hence $a$ and $b$ have the same image in the quotient of $U$ by $R$ for the étale topology. Next, let $U/R$ denote the quotient sheaf in the fppf topology which is an algebraic space by Spaces, Theorem 64.10.5. Thus we have morphisms (transformations of functors)

\[ U \to F \to U/R. \]

By the aforementioned Spaces, Theorem 64.10.5 the composition is representable, surjective, and étale. Hence for any scheme $T$ and morphism $T \to U/R$ the fibre product $V = T \times _{U/R} U$ is a scheme surjective and étale over $T$. In other words, $\{ V \to U\} $ is an étale covering. This proves that $U \to U/R$ is surjective as a map of sheaves in the étale topology. It follows that $F \to U/R$ is surjective as a map of sheaves in the étale topology. On the other hand, the map $F \to U/R$ is injective (as a map of presheaves) since $R = U \times _{U/R} U$ again by Spaces, Theorem 64.10.5. It follows that $F \to U/R$ is an isomorphism of étale sheaves, see Sites, Lemma 7.11.2 which concludes the proof.
$\square$

There is also an analogue of Spaces, Lemma 64.11.1.

Lemma 79.12.2. Denote the common underlying category of $\mathit{Sch}_{fppf}$ and $\mathit{Sch}_{\acute{e}tale}$ by $\mathit{Sch}_\alpha $ (see Topologies, Remark 34.11.1). Let $S$ be an object of $\mathit{Sch}_\alpha $. Let

\[ F : (\mathit{Sch}_\alpha /S)^{opp} \longrightarrow \textit{Sets} \]

be a presheaf with the following properties:

$F$ is a sheaf for the étale topology,

there exists an algebraic space $U$ over $S$ and a map $U \to F$ which is representable by algebraic spaces, surjective, and étale.

Then $F$ is an algebraic space in the sense of Algebraic Spaces, Definition 64.6.1.

**Proof.**
Set $R = U \times _ F U$. This is an algebraic space as $U \to F$ is assumed representable by algebraic spaces. The projections $s, t : R \to U$ are étale morphisms of algebraic spaces as $U \to F$ is assumed étale. The map $j = (t, s) : R \to U \times _ S U$ is a monomorphism and an equivalence relation as $R = U \times _ F U$. By Theorem 79.10.1 the fppf quotient sheaf $F' = U/R$ is an algebraic space. The morphism $U \to F'$ is surjective, flat, and locally of finite presentation by Lemma 79.11.6. The map $R \to U \times _{F'} U$ is surjective as a map of fppf sheaves by Groupoids in Spaces, Lemma 77.19.5 and since $j$ is a monomorphism it is an isomorphism. Hence the base change of $U \to F'$ by $U \to F'$ is étale, and we conclude that $U \to F'$ is étale by Descent on Spaces, Lemma 73.11.28. Thus $U \to F'$ is surjective as a map of étale sheaves. This means that $F'$ is equal to the quotient sheaf $U/R$ in the étale topology (small check omitted). Hence we obtain a canonical factorization $U \to F' \to F$ and $F' \to F$ is an injective map of sheaves. On the other hand, $U \to F$ is surjective as a map of étale sheaves and hence so is $F' \to F$. This means that $F' = F$ and the proof is complete.
$\square$

In fact, it suffices to have a smooth cover by a scheme and it suffices to assume the diagonal is representable by algebraic spaces.

Lemma 79.12.3. Denote the common underlying category of $\mathit{Sch}_{fppf}$ and $\mathit{Sch}_{\acute{e}tale}$ by $\mathit{Sch}_\alpha $ (see Topologies, Remark 34.11.1). Let $S$ be an object of $\mathit{Sch}_\alpha $.

\[ F : (\mathit{Sch}_\alpha /S)^{opp} \longrightarrow \textit{Sets} \]

be a presheaf with the following properties:

$F$ is a sheaf for the étale topology,

the diagonal $\Delta : F \to F \times F$ is representable by algebraic spaces, and

there exists $U \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha /S)$ and $U \to F$ which is surjective and smooth.

Then $F$ is an algebraic space in the sense of Algebraic Spaces, Definition 64.6.1.

**Proof.**
The proof mirrors the proof of Lemma 79.12.1. Let $R = U \times _ F U$. By (2) the presheaf $R$ is an algebraic space and by (3) the projections $R \to U$ are smooth and surjective. Denote $(U, R, s, t, c)$ the groupoid associated to the equivalence relation $j : R \to U \times _ S U$ (see Groupoids in Spaces, Lemma 77.11.3). By Theorem 79.10.1 we see that $X = U/R$ (quotient in the fppf-topology) is an algebraic space. Using that the smooth topology and the étale topology have the same sheaves (by More on Morphisms, Lemma 37.38.7) we see the map $U \to F$ identifies $F$ as the quotient of $U$ by $R$ for the smooth topology (details omitted). Thus we have morphisms (transformations of functors)

\[ U \to F \to X. \]

By Lemma 79.11.6 we see that $U \to X$ is surjective, flat and locally of finite presentation. By Groupoids in Spaces, Lemma 77.19.5 (and the fact that $j$ is a monomorphism) we have $R = U \times _ X U$. By Descent on Spaces, Lemma 73.11.26 we conclude that $U \to X$ is smooth and surjective (as the projections $R \to U$ are smooth and surjective and $\{ U \to X\} $ is an fppf covering). Hence for any scheme $T$ and morphism $T \to X$ the fibre product $T \times _ X U$ is an algebraic space surjective and smooth over $T$. Choose a scheme $V$ and a surjective étale morphism $V \to T \times _ X U$. Then $\{ V \to T\} $ is a smooth covering such that $V \to T \to X$ lifts to a morphism $V \to U$. This proves that $U \to X$ is surjective as a map of sheaves in the smooth topology. It follows that $F \to X$ is surjective as a map of sheaves in the smooth topology. On the other hand, the map $F \to X$ is injective (as a map of presheaves) since $R = U \times _ X U$. It follows that $F \to X$ is an isomorphism of smooth ($=$ étale) sheaves, see Sites, Lemma 7.11.2 which concludes the proof.
$\square$

Finally, here is the analogue of Spaces, Lemma 64.11.1 with a smooth morphism covering the space.

Lemma 79.12.4. Denote the common underlying category of $\mathit{Sch}_{fppf}$ and $\mathit{Sch}_{\acute{e}tale}$ by $\mathit{Sch}_\alpha $ (see Topologies, Remark 34.11.1). Let $S$ be an object of $\mathit{Sch}_\alpha $. Let

\[ F : (\mathit{Sch}_\alpha /S)^{opp} \longrightarrow \textit{Sets} \]

be a presheaf with the following properties:

$F$ is a sheaf for the étale topology,

there exists an algebraic space $U$ over $S$ and a map $U \to F$ which is representable by algebraic spaces, surjective, and smooth.

Then $F$ is an algebraic space in the sense of Algebraic Spaces, Definition 64.6.1.

**Proof.**
The proof is identical to the proof of Lemma 79.12.2. Set $R = U \times _ F U$. This is an algebraic space as $U \to F$ is assumed representable by algebraic spaces. The projections $s, t : R \to U$ are smooth morphisms of algebraic spaces as $U \to F$ is assumed smooth. The map $j = (t, s) : R \to U \times _ S U$ is a monomorphism and an equivalence relation as $R = U \times _ F U$. By Theorem 79.10.1 the fppf quotient sheaf $F' = U/R$ is an algebraic space. The morphism $U \to F'$ is surjective, flat, and locally of finite presentation by Lemma 79.11.6. The map $R \to U \times _{F'} U$ is surjective as a map of fppf sheaves by Groupoids in Spaces, Lemma 77.19.5 and since $j$ is a monomorphism it is an isomorphism. Hence the base change of $U \to F'$ by $U \to F'$ is smooth, and we conclude that $U \to F'$ is smooth by Descent on Spaces, Lemma 73.11.26. Thus $U \to F'$ is surjective as a map of étale sheaves (as the smooth topology is equal to the étale topology by More on Morphisms, Lemma 37.38.7). This means that $F'$ is equal to the quotient sheaf $U/R$ in the étale topology (small check omitted). Hence we obtain a canonical factorization $U \to F' \to F$ and $F' \to F$ is an injective map of sheaves. On the other hand, $U \to F$ is surjective as a map of étale sheaves (as the smooth topology is the same as the étale topology) and hence so is $F' \to F$. This means that $F' = F$ and the proof is complete.
$\square$

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