Lemma 39.20.4. Let $\tau \in \{ Zariski, {\acute{e}tale}, fppf, smooth, syntomic\}$. Let $S$ be a scheme. Let $j : R \to U \times _ S U$ be a pre-equivalence relation over $S$. Assume $U, R, S$ are objects of a $\tau$-site $\mathit{Sch}_\tau$. For $T \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_\tau )$ and $a, b \in U(T)$ the following are equivalent:

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

2. there exists a $\tau$-covering $\{ f_ i : T_ i \to T\}$ of $T$ and morphisms $r_ i : T_ i \to R$ 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 $\tau$-sheaves

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

is surjective.

Proof. Omitted. Hint: The reason this works is that the presheaf (39.20.0.1) 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 39.3.1. $\square$

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