## 34.5 The smooth topology

In this section we define the smooth topology. This is a bit pointless as it will turn out later (see More on Morphisms, Section 37.38) that this topology defines the same topos as the étale topology. But still it makes sense and it is used occasionally.

Definition 34.5.1. Let $T$ be a scheme. A smooth covering of $T$ is a family of morphisms $\{ f_ i : T_ i \to T\} _{i \in I}$ of schemes such that each $f_ i$ is smooth and such that $T = \bigcup f_ i(T_ i)$.

Lemma 34.5.2. Any étale covering is a smooth covering, and a fortiori, any Zariski covering is a smooth covering.

Proof. This is clear from the definitions, the fact that an étale morphism is smooth see Morphisms, Definition 29.36.1 and Lemma 34.4.2. $\square$

Next, we show that this notion satisfies the conditions of Sites, Definition 7.6.2.

Lemma 34.5.3. Let $T$ be a scheme.

1. If $T' \to T$ is an isomorphism then $\{ T' \to T\}$ is a smooth covering of $T$.

2. If $\{ T_ i \to T\} _{i\in I}$ is a smooth covering and for each $i$ we have a smooth covering $\{ T_{ij} \to T_ i\} _{j\in J_ i}$, then $\{ T_{ij} \to T\} _{i \in I, j\in J_ i}$ is a smooth covering.

3. If $\{ T_ i \to T\} _{i\in I}$ is a smooth covering and $T' \to T$ is a morphism of schemes then $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is a smooth covering.

Proof. Omitted. $\square$

Lemma 34.5.4. Let $T$ be an affine scheme. Let $\{ T_ i \to T\} _{i \in I}$ be a smooth covering of $T$. Then there exists a smooth covering $\{ U_ j \to T\} _{j = 1, \ldots , m}$ which is a refinement of $\{ T_ i \to T\} _{i \in I}$ such that each $U_ j$ is an affine scheme, and such that each morphism $U_ j \to T$ is standard smooth, see Morphisms, Definition 29.34.1. Moreover, we may choose each $U_ j$ to be open affine in one of the $T_ i$.

Proof. Omitted, but see Algebra, Lemma 10.137.10. $\square$

Thus we define the corresponding standard coverings of affines as follows.

Definition 34.5.5. Let $T$ be an affine scheme. A standard smooth covering of $T$ is a family $\{ f_ j : U_ j \to T\} _{j = 1, \ldots , m}$ with each $U_ j$ is affine, $U_ j \to T$ standard smooth and $T = \bigcup f_ j(U_ j)$.

Definition 34.5.6. A big smooth site is any site $\mathit{Sch}_{smooth}$ as in Sites, Definition 7.6.2 constructed as follows:

1. Choose any set of schemes $S_0$, and any set of smooth coverings $\text{Cov}_0$ among these schemes.

2. As underlying category take any category $\mathit{Sch}_\alpha$ constructed as in Sets, Lemma 3.9.2 starting with the set $S_0$.

3. Choose any set of coverings as in Sets, Lemma 3.11.1 starting with the category $\mathit{Sch}_\alpha$ and the class of smooth coverings, and the set $\text{Cov}_0$ chosen above.

See the remarks following Definition 34.3.5 for motivation and explanation regarding the definition of big sites.

Before we continue with the introduction of the big smooth site of a scheme $S$, let us point out that the topology on a big smooth site $\mathit{Sch}_{smooth}$ is in some sense induced from the smooth topology on the category of all schemes.

Lemma 34.5.7. Let $\mathit{Sch}_{smooth}$ be a big smooth site as in Definition 34.5.6. Let $T \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{smooth})$. Let $\{ T_ i \to T\} _{i \in I}$ be an arbitrary smooth covering of $T$.

1. There exists a covering $\{ U_ j \to T\} _{j \in J}$ of $T$ in the site $\mathit{Sch}_{smooth}$ which refines $\{ T_ i \to T\} _{i \in I}$.

2. If $\{ T_ i \to T\} _{i \in I}$ is a standard smooth covering, then it is tautologically equivalent to a covering of $\mathit{Sch}_{smooth}$.

3. If $\{ T_ i \to T\} _{i \in I}$ is a Zariski covering, then it is tautologically equivalent to a covering of $\mathit{Sch}_{smooth}$.

Proof. For each $i$ choose an affine open covering $T_ i = \bigcup _{j \in J_ i} T_{ij}$ such that each $T_{ij}$ maps into an affine open subscheme of $T$. By Lemma 34.5.3 the refinement $\{ T_{ij} \to T\} _{i \in I, j \in J_ i}$ is a smooth covering of $T$ as well. Hence we may assume each $T_ i$ is affine, and maps into an affine open $W_ i$ of $T$. Applying Sets, Lemma 3.9.9 we see that $W_ i$ is isomorphic to an object of $\mathit{Sch}_{smooth}$. But then $T_ i$ as a finite type scheme over $W_ i$ is isomorphic to an object $V_ i$ of $\mathit{Sch}_{smooth}$ by a second application of Sets, Lemma 3.9.9. The covering $\{ V_ i \to T\} _{i \in I}$ refines $\{ T_ i \to T\} _{i \in I}$ (because they are isomorphic). Moreover, $\{ V_ i \to T\} _{i \in I}$ is combinatorially equivalent to a covering $\{ U_ j \to T\} _{j \in J}$ of $T$ in the site $\mathit{Sch}_{smooth}$ by Sets, Lemma 3.9.9. The covering $\{ U_ j \to T\} _{j \in J}$ is a refinement as in (1). In the situation of (2), (3) each of the schemes $T_ i$ is isomorphic to an object of $\mathit{Sch}_{smooth}$ by Sets, Lemma 3.9.9, and another application of Sets, Lemma 3.11.1 gives what we want. $\square$

Definition 34.5.8. Let $S$ be a scheme. Let $\mathit{Sch}_{smooth}$ be a big smooth site containing $S$.

1. The big smooth site of $S$, denoted $(\mathit{Sch}/S)_{smooth}$, is the site $\mathit{Sch}_{smooth}/S$ introduced in Sites, Section 7.25.

2. The big affine smooth site of $S$, denoted $(\textit{Aff}/S)_{smooth}$, is the full subcategory of $(\mathit{Sch}/S)_{smooth}$ whose objects are affine $U/S$. A covering of $(\textit{Aff}/S)_{smooth}$ is any covering $\{ U_ i \to U\}$ of $(\mathit{Sch}/S)_{smooth}$ which is a standard smooth covering.

Next, we check that the big affine site defines the same topos as the big site.

Lemma 34.5.9. Let $S$ be a scheme. Let $\mathit{Sch}_{smooth}$ be a big smooth site containing $S$. The functor $(\textit{Aff}/S)_{smooth} \to (\mathit{Sch}/S)_{smooth}$ is special cocontinuous and induces an equivalence of topoi from $\mathop{\mathit{Sh}}\nolimits ((\textit{Aff}/S)_{smooth})$ to $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{smooth})$.

Proof. The notion of a special cocontinuous functor is introduced in Sites, Definition 7.29.2. Thus we have to verify assumptions (1) – (5) of Sites, Lemma 7.29.1. Denote the inclusion functor $u : (\textit{Aff}/S)_{smooth} \to (\mathit{Sch}/S)_{smooth}$. Being cocontinuous just means that any smooth covering of $T/S$, $T$ affine, can be refined by a standard smooth covering of $T$. This is the content of Lemma 34.5.4. Hence (1) holds. We see $u$ is continuous simply because a standard smooth covering is a smooth covering. Hence (2) holds. Parts (3) and (4) follow immediately from the fact that $u$ is fully faithful. And finally condition (5) follows from the fact that every scheme has an affine open covering. $\square$

To be continued...

Lemma 34.5.10. Let $\mathit{Sch}_{smooth}$ be a big smooth site. Let $f : T \to S$ be a morphism in $\mathit{Sch}_{smooth}$. The functor

$u : (\mathit{Sch}/T)_{smooth} \longrightarrow (\mathit{Sch}/S)_{smooth}, \quad V/T \longmapsto V/S$

is cocontinuous, and has a continuous right adjoint

$v : (\mathit{Sch}/S)_{smooth} \longrightarrow (\mathit{Sch}/T)_{smooth}, \quad (U \to S) \longmapsto (U \times _ S T \to T).$

They induce the same morphism of topoi

$f_{big} : \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/T)_{smooth}) \longrightarrow \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{smooth})$

We have $f_{big}^{-1}(\mathcal{G})(U/T) = \mathcal{G}(U/S)$. We have $f_{big, *}(\mathcal{F})(U/S) = \mathcal{F}(U \times _ S T/T)$. Also, $f_{big}^{-1}$ has a left adjoint $f_{big!}$ which commutes with fibre products and equalizers.

Proof. The functor $u$ is cocontinuous, continuous, and commutes with fibre products and equalizers. Hence Sites, Lemmas 7.21.5 and 7.21.6 apply and we deduce the formula for $f_{big}^{-1}$ and the existence of $f_{big!}$. Moreover, the functor $v$ is a right adjoint because given $U/T$ and $V/S$ we have $\mathop{\mathrm{Mor}}\nolimits _ S(u(U), V) = \mathop{\mathrm{Mor}}\nolimits _ T(U, V \times _ S T)$ as desired. Thus we may apply Sites, Lemmas 7.22.1 and 7.22.2 to get the formula for $f_{big, *}$. $\square$

Comment #264 by Ray Hoobler on

The two proofs of Lemma 33.28.6 are displayed incorrectly. They both begin "\begin{proof}". Also the second proof should begin "Pick a point". Thanks for all the work involved in maintaining the Stacks Project!

Comment #266 by on

I have fixed the incorrect displaying, see here. In a short while you'll see this change on the website too. Thanks for noticing this!

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