# The Stacks Project

## Tag 0CLC

Lemma 91.38.2. In the situation of Definition 91.38.1 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 91.6.1 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 8637–8642 (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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