Lemma 101.39.2. In the situation of Definition 101.39.1 the category of dotted arrows is a groupoid. If $\Delta _ f$ is separated, then it is a setoid.

Proof. Since $2$-arrows are invertible it is clear that the category of dotted arrows is a groupoid. Given a dotted arrow $(a, \alpha , \beta )$ an automorphism of $(a, \alpha , \beta )$ is a $2$-morphism $\theta : a \to a$ satisfying two conditions. The first condition $\beta = (\text{id}_ f \star \theta ) \circ \beta$ signifies that $\theta$ defines a morphism $(a, \theta ) : \mathop{\mathrm{Spec}}(A) \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. The second condition $\alpha = \alpha \circ (\theta \star \text{id}_ j)$ implies that the restriction of $(a, \theta )$ to $\mathop{\mathrm{Spec}}(K)$ is the identity. Picture

$\xymatrix{ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] & & \mathop{\mathrm{Spec}}(K) \ar[d]^ j \ar[ll]_{(a \circ j, \text{id})} \\ \mathcal{X} & & \mathop{\mathrm{Spec}}(A) \ar[ll]_ a \ar[llu]_{(a, \theta )} }$

In other words, if $G \to \mathop{\mathrm{Spec}}(A)$ is the group algebraic space we get by pulling back the relative inertia by $a$, then $\theta$ defines a point $\theta \in G(A)$ whose image in $G(K)$ is trivial. Certainly, if the identity $e : \mathop{\mathrm{Spec}}(A) \to G$ is a closed immersion, then this can happen only if $\theta$ is the identity. Looking at Lemma 101.6.1 we obtain the result we want. $\square$

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