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Tag 0CL0

Chapter 86: Morphisms of Algebraic Stacks > Section 86.6: Higher diagonals

Lemma 86.6.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

  1. The following are equivalent
    1. $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is separated,
    2. $\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is separated, and
    3. $\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion.
  2. The following are equivalent
    1. $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is quasi-separated,
    2. $\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is quasi-separated, and
    3. $\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a quasi-compact.
  3. The following are equivalent
    1. $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is locally separated,
    2. $\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is locally separated, and
    3. $\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an immersion.

Proof. Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then $G = U \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic space over $U$ (Lemma 86.5.1). In fact, $G$ is a group algebraic space over $U$ by the group law on relative inertia constructed in Remark 86.5.2. Moreover, $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is surjective and smooth as a base change of $U \to \mathcal{X}$. Finally, the base change of $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is the identity $U \to G$ of $G/U$. Thus the equivalence of (a) and (c) follows from Groupoids in Spaces, Lemma 67.6.1. Since $\Delta_{f, 2}$ is the diagonal of $\Delta_f$ we have (b) $\Leftrightarrow$ (c) by definition. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 1350–1391 (see updates for more information).

    \begin{lemma}
    \label{lemma-diagonal-diagonal}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
    \begin{enumerate}
    \item
    The following are equivalent
    \begin{enumerate}
    \item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$
    is separated,
    \item $\Delta_{f, 1} = \Delta_f :
    \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$
    is separated, and
    \item $\Delta_{f, 2} = e :
    \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is a closed immersion.
    \end{enumerate}
    \item
    The following are equivalent
    \begin{enumerate}
    \item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$
    is quasi-separated,
    \item $\Delta_{f, 1} = \Delta_f :
    \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$
    is quasi-separated, and
    \item $\Delta_{f, 2} = e :
    \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is a quasi-compact.
    \end{enumerate}
    \item
    The following are equivalent
    \begin{enumerate}
    \item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$
    is locally separated,
    \item $\Delta_{f, 1} = \Delta_f :
    \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$
    is locally separated, and
    \item $\Delta_{f, 2} = e :
    \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is an immersion.
    \end{enumerate}
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Choose an algebraic space $U$ and a surjective smooth morphism
    $U \to \mathcal{X}$. Then
    $G = U \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is an algebraic space over $U$ (Lemma \ref{lemma-inertia}).
    In fact, $G$ is a group algebraic space over $U$
    by the group law on relative
    inertia constructed in Remark \ref{remark-inertia-is-group-in-spaces}.
    Moreover, $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is surjective and smooth as a base change of $U \to \mathcal{X}$.
    Finally, the base change of
    $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    by $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is the identity $U \to G$ of $G/U$.
    Thus the equivalence of (a) and (c) follows from
    Groupoids in Spaces, Lemma
    \ref{spaces-groupoids-lemma-group-scheme-separated}.
    Since $\Delta_{f, 2}$ is the diagonal of $\Delta_f$ we have
    (b) $\Leftrightarrow$ (c) by definition.
    \end{proof}

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