Lemma 39.3.5 (0DT7)—The Stacks project

Table of contents
Part 2: Schemes
Chapter 39: Groupoid Schemes
Section 39.3: Equivalence relations
Lemma 39.3.5 (cite)

previous lemma

Lemma 39.3.5. Let $j : R \to U \times _ S U$ be a pre-relation. Assume

$s, t$ are unramified,
for any algebraically closed field $k$ over $S$ the map $R(k) \to U(k) \times U(k)$ is an equivalence relation,
there are morphisms $e : U \to R$, $i : R \to R$, $c : R \times _{s, U, t} R \to R$ such that

\[ \xymatrix{ U \ar[r]_ e \ar[d]_\Delta & R \ar[d]_ j & R \ar[d]^ j \ar[r]_ i & R \ar[d]^ j & R \times _{s, U, t} R \ar[d]^{j \times j} \ar[r]_ c & R \ar[d]^ j \\ U \times _ S U \ar[r] & U \times _ S U & U \times _ S U \ar[r]^{flip} & U \times _ S U & U \times _ S U \times _ S U \ar[r]^{\text{pr}_{02}} & U \times _ S U } \]

are commutative.

Then $j$ is an equivalence relation.

Proof. By condition (1) and Morphisms, Lemma 29.35.16 we see that $j$ is a unramified. Then $\Delta _ j : R \to R \times _{U \times _ S U} R$ is an open immersion by Morphisms, Lemma 29.35.13. However, then condition (2) says $\Delta _ j$ is bijective on $k$-valued points, hence $\Delta _ j$ is an isomorphism, hence $j$ is a monomorphism. Then it easily follows from the commutative diagrams that $R(T) \subset U(T) \times U(T)$ is an equivalence relation for all schemes $T$ over $S$. $\square$

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