The Stacks project

Lemma 64.15.2. 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 64.15.1 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 64.9.1 and Theorem 64.10.5 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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