Lemma 77.4.4. Let $B \to S$ as in Section 77.3. Let $j : R \to U \times _ B U$ be a pre-relation of algebraic spaces over $B$. Consider the relation on $|U|$ defined by the rule

\[ x \sim y \Leftrightarrow \exists \ r \in |R| : t(r) = x, s(r) = y. \]

If $j$ is a pre-equivalence relation then this is an equivalence relation.

**Proof.**
Suppose that $x \sim y$ and $y \sim z$. Pick $r \in |R|$ with $t(r) = x$, $s(r) = y$ and pick $r' \in |R|$ with $t(r') = y$, $s(r') = z$. We may pick a field $K$ such that $r$ and $r'$ can be represented by morphisms $r, r' : \mathop{\mathrm{Spec}}(K) \to R$ with $s \circ r = t \circ r'$. Denote $x = t \circ r$, $y = s \circ r = t \circ r'$, and $z = s \circ r'$, so $x, y, z : \mathop{\mathrm{Spec}}(K) \to U$. By construction $(x, y) \in j(R(K))$ and $(y, z) \in j(R(K))$. Since $j$ is a pre-equivalence relation we see that also $(x, z) \in j(R(K))$. This clearly implies that $x \sim z$.

The proof that $\sim $ is reflexive and symmetric is omitted.
$\square$

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