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Tag 0CLC

Chapter 87: Morphisms of Algebraic Stacks > Section 87.38: Valuative criteria

Lemma 87.38.2. In the situation of Definition 87.38.1 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 $(a, \theta) : \mathop{\rm Spec}(A) \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. The second condition $\alpha = \alpha \circ (\theta \star \text{id}_j)$ implies that the restriction of $(a, \theta)$ to $\mathop{\rm Spec}(K)$ is the identity. Picture $$ \xymatrix{ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] & & \mathop{\rm Spec}(K) \ar[d]^j \ar[ll]_{(a \circ j, \text{id})} \\ \mathcal{X} & & \mathop{\rm Spec}(A) \ar[ll]_a \ar[llu]_{(a, \theta)} } $$ In other words, if $G \to \mathop{\rm Spec}(A)$ 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 $e : \mathop{\rm Spec}(A) \to G$ is a closed immersion, then this can happen only if $\theta$ is the identity. Looking at Lemma 87.6.1 we obtain the result we want. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 8593–8598 (see updates for more information).

    \begin{lemma}
    \label{lemma-cat-dotted-arrows}
    In the situation of Definition \ref{definition-fill-in-diagram}
    the category of dotted arrows is a groupoid. If $\Delta_f$
    is separated, then it is a setoid.
    \end{lemma}
    
    \begin{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
    $(a, \theta) : \Spec(A) \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$.
    The second condition
    $\alpha = \alpha \circ (\theta \star \text{id}_j)$
    implies that the restriction of $(a, \theta)$ to $\Spec(K)$
    is the identity. Picture
    $$
    \xymatrix{
    \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] & &
    \Spec(K) \ar[d]^j \ar[ll]_{(a \circ j, \text{id})} \\
    \mathcal{X} & & \Spec(A) \ar[ll]_a \ar[llu]_{(a, \theta)}
    }
    $$
    In other words, if $G \to \Spec(A)$ 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 $e : \Spec(A) \to G$
    is a closed immersion, then this can happen only if
    $\theta$ is the identity.
    Looking at Lemma \ref{lemma-diagonal-diagonal}
    we obtain the result we want.
    \end{proof}

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