Lemma 79.7.1. Let $B \to S$ be as in Section 79.2. Let $(U, R, s, t, c)$ be a groupoid in algebraic spaces over $B$. Let $K$ be a field and let $r, r' : \mathop{\mathrm{Spec}}(K) \to R$ be morphisms such that $t \circ r = t \circ r' : \mathop{\mathrm{Spec}}(K) \to U$. Set $u = s \circ r$, $u' = s \circ r'$ and denote $F_ u = \mathop{\mathrm{Spec}}(K) \times _{u, U, s} R$ and $F_{u'} = \mathop{\mathrm{Spec}}(K) \times _{u', U, s} R$ the fibre products. Then $F_ u \cong F_{u'}$ as algebraic spaces over $K$.
Proof. We use the properties and the existence of Diagram (79.3.0.1). There exists a morphism $\xi : \mathop{\mathrm{Spec}}(K) \to R \times _{s, U, t} R$ with $\text{pr}_0 \circ \xi = r$ and $c \circ \xi = r'$. Let $\tilde r = \text{pr}_1 \circ \xi : \mathop{\mathrm{Spec}}(K) \to R$. Then looking at the bottom two squares of Diagram (79.3.0.1) we see that both $F_ u$ and $F_{u'}$ are identified with the algebraic space $\mathop{\mathrm{Spec}}(K) \times _{\tilde r, R, \text{pr}_1} (R \times _{s, U, t} R)$. $\square$
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