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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