Lemma 39.3.4. Let $j : R \to U \times _ S U$ be a pre-relation. Consider the relation on points of the scheme $U$ 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 $t(r') = y$, $s(r') = z$. Pick a field $K$ fitting into the following commutative diagram

$\xymatrix{ \kappa (r) \ar[r] & K \\ \kappa (y) \ar[u] \ar[r] & \kappa (r') \ar[u] }$

Denote $x_ K, y_ K, z_ K : \mathop{\mathrm{Spec}}(K) \to U$ the morphisms

$\begin{matrix} \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(\kappa (r)) \to \mathop{\mathrm{Spec}}(\kappa (x)) \to U \\ \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(\kappa (r)) \to \mathop{\mathrm{Spec}}(\kappa (y)) \to U \\ \mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(\kappa (r')) \to \mathop{\mathrm{Spec}}(\kappa (z)) \to U \end{matrix}$

By construction $(x_ K, y_ K) \in j(R(K))$ and $(y_ K, z_ K) \in j(R(K))$. Since $j$ is a pre-equivalence relation we see that also $(x_ K, z_ K) \in j(R(K))$. This clearly implies that $x \sim z$.

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

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