# The Stacks Project

## Tag 0CL0

Lemma 91.6.1. 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.
4. The following are equivalent
1. $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is unramified,
2. $f$ is DM.
5. The following are equivalent
1. $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is locally quasi-finite,
2. $f$ is quasi-DM.

Proof. Proof of (1), (2), and (3). 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 91.5.1). In fact, $G$ is a group algebraic space over $U$ by the group law on relative inertia constructed in Remark 91.5.2. 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 69.6.1. Since $\Delta_{f, 2}$ is the diagonal of $\Delta_f$ we have (b) $\Leftrightarrow$ (c) by definition.

Proof of (4) and (5). Recall that (4)(b) means $\Delta_f$ is unramified and (5)(b) means that $\Delta_f$ is locally quasi-finite. Choose a scheme $Z$ and a morphism $a : Z \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$. Then $a = (x_1, x_2, \alpha)$ where $x_i : Z \to \mathcal{X}$ and $\alpha : f \circ x_1 \to f \circ x_2$ is a $2$-morphism. Recall that $$\vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) \ar[d] \ar[r] & Z \ar[d] \\ \mathcal{X} \ar[r]^{\Delta_f} & \mathcal{X} \times_\mathcal{Y} \mathcal{X} } } \quad\text{and}\quad \vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \ar[d] \ar[r] & Z \ar[d]^{x_2} \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} } }$$ are cartesian squares. By Lemma 91.5.4 the algebraic space $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)$ is a pseudo torsor for $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ over $Z$. Thus the equivalences in (4) and (5) follow from Groupoids in Spaces, Lemma 69.9.5. $\square$

The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 1350–1405 (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}
\item
The following are equivalent
\begin{enumerate}
\item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$
is unramified,
\item $f$ is DM.
\end{enumerate}
\item
The following are equivalent
\begin{enumerate}
\item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$
is locally quasi-finite,
\item $f$ is quasi-DM.
\end{enumerate}
\end{enumerate}
\end{lemma}

\begin{proof}
Proof of (1), (2), and (3).
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.

\medskip\noindent
Proof of (4) and (5). Recall that (4)(b) means $\Delta_f$ is
unramified and (5)(b) means that $\Delta_f$ is locally quasi-finite.
Choose a scheme $Z$ and a morphism
$a : Z \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$.
Then $a = (x_1, x_2, \alpha)$ where $x_i : Z \to \mathcal{X}$
and $\alpha : f \circ x_1 \to f \circ x_2$ is a $2$-morphism.
Recall that
$$\vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) \ar[d] \ar[r] & Z \ar[d] \\ \mathcal{X} \ar[r]^{\Delta_f} & \mathcal{X} \times_\mathcal{Y} \mathcal{X} } } \quad\text{and}\quad \vcenter{ \xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \ar[d] \ar[r] & Z \ar[d]^{x_2} \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} } }$$
are cartesian squares. By Lemma \ref{lemma-isom-pseudo-torsor-aut-over-space}
the
algebraic space $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)$
is a pseudo torsor for $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$
over $Z$. Thus the equivalences in (4) and (5) follow from
Groupoids in Spaces, Lemma
\ref{spaces-groupoids-lemma-pseudo-torsor-implications}.
\end{proof}

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