# The Stacks Project

## Tag 0CL0

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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