**Proof.**
By Groupoids, Lemma 39.3.2 the morphism $j' = (t', s') : R' \to U' \times _ S U'$ defines an equivalence relation. Since $g$ is flat and locally of finite presentation we see that $g$ is universally open as well (Morphisms, Lemma 29.24.10). For the same reason $s, t$ are universally open as well. Let $W^1 = g(U') \subset U$, and let $W = t(s^{-1}(W^1))$. Then $W^1$ and $W$ are open in $U$. Moreover, as $j$ is an equivalence relation we have $t(s^{-1}(W)) = W$ (see Groupoids, Lemma 39.19.2 for example).

By Groupoids, Lemma 39.20.5 the map of sheaves $F' = U'/R' \to F = U/R$ is injective. Let $a : T \to F$ be a morphism from a scheme into $U/R$. We have to show that $T \times _ F F'$ is representable by an open subscheme of $T$.

The morphism $a$ is given by the following data: an fppf covering $\{ \varphi _ j : T_ j \to T\} _{j \in J}$ of $T$ and morphisms $a_ j : T_ j \to U$ such that the maps

\[ a_ j \times a_{j'} : T_ j \times _ T T_{j'} \longrightarrow U \times _ S U \]

factor through $j : R \to U \times _ S U$ via some (unique) maps $r_{jj'} : T_ j \times _ T T_{j'} \to R$. The system $(a_ j)$ corresponds to $a$ in the sense that the diagrams

\[ \xymatrix{ T_ j \ar[r]_{a_ j} \ar[d] & U \ar[d] \\ T \ar[r]^ a & F } \]

commute.

Consider the open subsets $W_ j = a_ j^{-1}(W) \subset T_ j$. Since $t(s^{-1}(W)) = W$ we see that

\[ W_ j \times _ T T_{j'} = r_{jj'}^{-1}(t^{-1}(W)) = r_{jj'}^{-1}(s^{-1}(W)) = T_ j \times _ T W_{j'}. \]

By Descent, Lemma 35.10.6 this means there exists an open $W_ T \subset T$ such that $\varphi _ j^{-1}(W_ T) = W_ j$ for all $j \in J$. We claim that $W_ T \to T$ represents $T \times _ F F' \to T$.

First, let us show that $W_ T \to T \to F$ is an element of $F'(W_ T)$. Since $\{ W_ j \to W_ T\} _{j \in J}$ is an fppf covering of $W_ T$, it is enough to show that each $W_ j \to U \to F$ is an element of $F'(W_ j)$ (as $F'$ is a sheaf for the fppf topology). Consider the commutative diagram

\[ \xymatrix{ W'_ j \ar[rr] \ar[dd] \ar[rd] & & U' \ar[d]^ g \\ & s^{-1}(W^1) \ar[r]_ s \ar[d]^ t & W^1 \ar[d] \\ W_ j \ar[r]^{a_ j|_{W_ j}} & W \ar[r] & F } \]

where $W'_ j = W_ j \times _ W s^{-1}(W^1) \times _{W^1} U'$. Since $t$ and $g$ are surjective, flat and locally of finite presentation, so is $W'_ j \to W_ j$. Hence the restriction of the element $W_ j \to U \to F$ to $W'_ j$ is an element of $F'$ as desired.

Suppose that $f : T' \to T$ is a morphism of schemes such that $a|_{T'} \in F'(T')$. We have to show that $f$ factors through the open $W_ T$. Since $\{ T' \times _ T T_ j \to T'\} $ is an fppf covering of $T'$ it is enough to show each $T' \times _ T T_ j \to T$ factors through $W_ T$. Hence we may assume $f$ factors as $\varphi _ j \circ f_ j : T' \to T_ j \to T$ for some $j$. In this case the condition $a|_{T'} \in F'(T')$ means that there exists some fppf covering $\{ \psi _ i : T'_ i \to T'\} _{i \in I}$ and some morphisms $b_ i : T'_ i \to U'$ such that

\[ \xymatrix{ T'_ i \ar[r]_{b_ i} \ar[d]_{f_ j \circ \psi _ i} & U' \ar[r]_ g & U \ar[d] \\ T_ j \ar[r]^{a_ j} & U \ar[r] & F } \]

is commutative. This commutativity means that there exists a morphism $r'_ i : T'_ i \to R$ such that $t \circ r'_ i = a_ j \circ f_ j \circ \psi _ i$, and $s \circ r'_ i = g \circ b_ i$. This implies that $\mathop{\mathrm{Im}}(f_ j \circ \psi _ i) \subset W_ j$ and we win.
$\square$

## Comments (2)

Comment #2964 by Yu-Liang Huang on

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