Lemma 100.39.2. In the situation of Definition 100.39.1 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

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 100.6.1 we obtain the result we want. $\square$

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