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The Stacks project

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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