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 $K$.
79.7 Comparing fibres
This section is the analogue of More on Groupoids, Section 40.7. The reader is strongly encouraged to read that section first.
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 $\mathop{\mathrm{Spec}}(K) \times _{\tilde r, R, \text{pr}_1} (R \times _{s, U, t} R)$. $\square$
Actually, in the situation of the lemma the morphisms of pairs $s : (R, r) \to (U, u)$ and $s : (R, r') \to (U, u')$ are locally isomorphic in the $\tau $-topology, provided $\{ s: R \to U\} $ is a $\tau $-covering. We will insert a precise statement here if needed.
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)