Lemma 77.19.5. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-equivalence relation over $B$. For a scheme $S'$ over $S$ and $a, b \in U(S')$ the following are equivalent:

1. $a$ and $b$ map to the same element of $(U/R)(S')$, and

2. there exists an fppf covering $\{ f_ i : S_ i \to S'\}$ of $S'$ and morphisms $r_ i : S_ i \to R$ such that $a \circ f_ i = s \circ r_ i$ and $b \circ f_ i = t \circ r_ i$.

In other words, in this case the map of sheaves

$R \longrightarrow U \times _{U/R} U$

is surjective.

Proof. Omitted. Hint: The reason this works is that the presheaf (77.19.0.1) in this case is really given by $T \mapsto U(T)/j(R(T))$ as $j(R(T)) \subset U(T) \times U(T)$ is an equivalence relation, see Definition 77.4.1. $\square$

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