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65.15 Change of big site

In this section we briefly discuss what happens when we change big sites. The upshot is that we can always enlarge the big site at will, hence we may assume any set of schemes we want to consider is contained in the big fppf site over which we consider our algebraic space. Here is a precise statement of the result.

Lemma 65.15.1. Suppose given big sites $\mathit{Sch}_{fppf}$ and $\mathit{Sch}'_{fppf}$. Assume that $\mathit{Sch}_{fppf}$ is contained in $\mathit{Sch}'_{fppf}$, see Topologies, Section 34.12. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Let

\begin{align*} g : \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf}) \longrightarrow \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}'/S)_{fppf}), \\ f : \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}'/S)_{fppf}) \longrightarrow \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf}) \end{align*}

be the morphisms of topoi of Topologies, Lemma 34.12.2. Let $F$ be a sheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Then

  1. if $F$ is representable by a scheme $X \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ over $S$, then $f^{-1}F$ is representable too, in fact it is representable by the same scheme $X$, now viewed as an object of $(\mathit{Sch}'/S)_{fppf}$, and

  2. if $F$ is an algebraic space over $S$, then $f^{-1}F$ is an algebraic space over $S$ also.

Proof. Let $X \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Let us write $h_ X$ for the representable sheaf on $(\mathit{Sch}/S)_{fppf}$ associated to $X$, and $h'_ X$ for the representable sheaf on $(\mathit{Sch}'/S)_{fppf}$ associated to $X$. By the description of $f^{-1}$ in Topologies, Section 34.12 we see that $f^{-1}h_ X = h'_ X$. This proves (1).

Next, suppose that $F$ is an algebraic space over $S$. By Lemma 65.9.1 this means that $F = h_ U/h_ R$ for some étale equivalence relation $R \to U \times _ S U$ in $(\mathit{Sch}/S)_{fppf}$. Since $f^{-1}$ is an exact functor we conclude that $f^{-1}F = h'_ U/h'_ R$. Hence $f^{-1}F$ is an algebraic space over $S$ by Theorem 65.10.5. $\square$

Note that this lemma is purely set theoretical and has virtually no content. Moreover, it is not true (in general) that the restriction of an algebraic space over the bigger site is an algebraic space over the smaller site (simply by reasons of cardinality). Hence we can only ever use a simple lemma of this kind to enlarge the base category and never to shrink it.

Lemma 65.15.2. Suppose $\mathit{Sch}_{fppf}$ is contained in $\mathit{Sch}'_{fppf}$. Let $S$ be an object of $\mathit{Sch}_{fppf}$. Denote $\textit{Spaces}/S$ the category of algebraic spaces over $S$ defined using $\mathit{Sch}_{fppf}$. Similarly, denote $\textit{Spaces}'/S$ the category of algebraic spaces over $S$ defined using $\mathit{Sch}'_{fppf}$. The construction of Lemma 65.15.1 defines a fully faithful functor

\[ \textit{Spaces}/S \longrightarrow \textit{Spaces}'/S \]

whose essential image consists of those $X' \in \mathop{\mathrm{Ob}}\nolimits (\textit{Spaces}'/S)$ such that there exist $U, R \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$1 and morphisms

\[ U \longrightarrow X' \quad \text{and}\quad R \longrightarrow U \times _{X'} U \]

in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}'/S)_{fppf})$ which are surjective as maps of sheaves (for example if the displayed morphisms are surjective and étale).

Proof. In Sites, Lemma 7.21.8 we have seen that the functor $f^{-1} : \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf}) \to \mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}'/S)_{fppf})$ is fully faithful (see discussion in Topologies, Section 34.12). Hence we see that the displayed functor of the lemma is fully faithful.

Suppose that $X' \in \mathop{\mathrm{Ob}}\nolimits (\textit{Spaces}'/S)$ such that there exists $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a map $U \to X'$ in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}'/S)_{fppf})$ which is surjective as a map of sheaves. Let $U' \to X'$ be a surjective étale morphism with $U' \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}'/S)_{fppf})$. Let $\kappa = \text{size}(U)$, see Sets, Section 3.9. Then $U$ has an affine open covering $U = \bigcup _{i \in I} U_ i$ with $|I| \leq \kappa $. Observe that $U' \times _{X'} U \to U$ is étale and surjective. For each $i$ we can pick a quasi-compact open $U'_ i \subset U'$ such that $U'_ i \times _{X'} U_ i \to U_ i$ is surjective (because the scheme $U' \times _{X'} U_ i$ is the union of the Zariski opens $W \times _{X'} U_ i$ for $W \subset U'$ affine and because $U' \times _{X'} U_ i \to U_ i$ is étale hence open). Then $\coprod _{i \in I} U'_ i \to X$ is surjective étale because of our assumption that $U \to X$ and hence $\coprod U_ i \to X$ is a surjection of sheaves (details omitted). Because $U'_ i \times _{X'} U \to U'_ i$ is a surjection of sheaves and because $U'_ i$ is quasi-compact, we can find a quasi-compact open $W_ i \subset U'_ i \times _{X'} U$ such that $W_ i \to U'_ i$ is surjective as a map of sheaves (details omitted). Then $W_ i \to U$ is étale and we conclude that $\text{size}(W_ i) \leq \text{size}(U)$, see Sets, Lemma 3.9.7. By Sets, Lemma 3.9.11 we conclude that $\text{size}(U'_ i) \leq \text{size}(U)$. Hence $\coprod _{i \in I} U'_ i$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$ by Sets, Lemma 3.9.5.

Now let $X'$, $U \to X'$ and $R \to U \times _{X'} U$ be as in the statement of the lemma. In the previous paragraph we have seen that we can find $U' \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $U' \to X'$ in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}'/S)_{fppf})$. Then $U' \times _{X'} U \to U'$ is a surjection of sheaves, i.e., we can find an fppf covering $\{ U'_ i \to U'\} $ such that $U'_ i \to U'$ factors through $U' \times _{X'} U \to U'$. By Sets, Lemma 3.9.12 we can find $\tilde U \to U'$ which is surjective, flat, and locally of finite presentation, with $\text{size}(\tilde U) \leq \text{size}(U')$, such that $\tilde U \to U'$ factors through $U' \times _{X'} U \to U'$. Then we consider

\[ \xymatrix{ U' \times _{X'} U' \ar[d] & \tilde U \times _{X'} \tilde U \ar[l] \ar[d] \ar[r] & U \times _{X'} U \ar[d] \\ U' \times _ S U' & \tilde U \times _ S \tilde U \ar[l] \ar[r] & U \times _ S U } \]

The squares are cartesian. We know the objects of the bottom row are represented by objects of $(\mathit{Sch}/S)_{fppf}$. By the result of the argument of the previous paragraph, the same is true for $U \times _{X'} U$ (as we have the surjection of sheaves $R \to U \times _{X'} U$ by assumption). Since $(\mathit{Sch}/S)_{fppf}$ is closed under fibre products (by construction), we see that $\tilde U \times _{X'} \tilde U$ is represented by an object of $(\mathit{Sch}/S)_{fppf}$. Finally, the map $\tilde U \times _{X'} \tilde U \to U' \times _{X'} U'$ is a surjection of fppf sheaves as $\tilde U \to U'$ is so. Thus we can once more apply the result of the previous paragraph to conclude that $R' = U' \times _{X'} U'$ is represented by an object of $(\mathit{Sch}/S)_{fppf}$. At this point Lemma 65.9.1 and Theorem 65.10.5 imply that $X = h_{U'}/h_{R'}$ is an object of $\textit{Spaces}/S$ such that $f^{-1}X \cong X'$ as desired. $\square$

[1] Requiring the existence of $R$ is necessary because of our choice of the function $Bound$ in Sets, Equation ( The size of the fibre product $U \times _{X'} U$ can grow faster than $Bound$ in terms of the size of $U$. We can illustrate this by setting $S = \mathop{\mathrm{Spec}}(A)$, $U = \mathop{\mathrm{Spec}}(A[x_ i, i \in I])$ and $R = \coprod _{(\lambda _ i) \in A^ I} \mathop{\mathrm{Spec}}(A[x_ i, y_ i]/(x_ i - \lambda _ i y_ i))$. In this case the size of $R$ grows like $\kappa ^\kappa $ where $\kappa $ is the size of $U$.

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