Lemma 40.10.6. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. Assume $U = \mathop{\mathrm{Spec}}(k)$ with $k$ a field. If $r \in R$ is a point such that $s, t$ induce isomorphisms $k \to \kappa (r)$, then the map
(see proof for precise notation) is an automorphism $R \to R$ which maps $e$ to $r$.
Proof. This is completely obvious if you think about groupoids in a functorial way. But we will also spell it out completely. Denote $a : U \to R$ the morphism with image $r$ such that $s \circ a = \text{id}_ U$ which exists by the hypothesis that $s : k \to \kappa (r)$ is an isomorphism. Similarly, denote $b : U \to R$ the morphism with image $r$ such that $t \circ b = \text{id}_ U$. Note that $b = a \circ (t \circ a)^{-1}$, in particular $a \circ s \circ b = b$.
Consider the morphism $\Psi : R \to R$ given on $T$-valued points by
To see this is defined we have to check that $s \circ a \circ t \circ f = t \circ f$ which is obvious as $s \circ a = 1$. Note that $\Phi (e) = a$, so that in order to prove the lemma it suffices to show that $\Phi $ is an automorphism of $R$. Let $\Phi : R \to R$ be the morphism given on $T$-valued points by
This is defined because $s \circ i \circ b \circ t \circ g = t \circ b \circ t \circ g = t \circ g$. We claim that $\Phi $ and $\Psi $ are inverse to each other. To see this we compute
where we have used the relation $a \circ s \circ b = b$ shown above. In the other direction we have
The lemma is proved. $\square$
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