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\input{preamble}
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\begin{document}
\title{More on Morphisms of Stacks}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we continue our study of properties of morphisms of algebraic
stacks. A reference in the case of quasi-separated algebraic stacks with
representable diagonal is \cite{LM-B}.
\section{Conventions and abuse of language}
\label{section-conventions}
\noindent
We continue to use the conventions and the abuse of language
introduced in
Properties of Stacks, Section \ref{stacks-properties-section-conventions}.
\section{Thickenings}
\label{section-thickenings}
\noindent
The following terminology may not be completely standard, but it is convenient.
If $\mathcal{Y}$ is a closed substack of an algebraic stack $\mathcal{X}$,
then the morphism $\mathcal{Y} \to \mathcal{X}$ is representable.
\begin{definition}
\label{definition-thickening}
Thickenings.
\begin{enumerate}
\item We say an algebraic stack $\mathcal{X}'$ is a {\it thickening}
of an algebraic stack $\mathcal{X}$ if $\mathcal{X}$ is a closed substack
of $\mathcal{X}'$ and the associated topological spaces are equal.
\item Given two thickenings $\mathcal{X} \subset \mathcal{X}'$ and
$\mathcal{Y} \subset \mathcal{Y}'$ a {\it morphism of thickenings}
is a morphism $f' : \mathcal{X}' \to \mathcal{Y}'$ of algebraic stacks
such that $f'|_\mathcal{X}$ factors through the closed
substack $\mathcal{Y}$. In this situation we set
$f = f'|_\mathcal{X} : \mathcal{X} \to \mathcal{Y}$ and we say that
$(f, f') : (\mathcal{X} \subset \mathcal{X}') \to
(\mathcal{Y} \subset \mathcal{Y}')$ is a morphism of thickenings.
\item Let $\mathcal{Z}$ be an algebraic stack. We similarly define
{\it thickenings over $\mathcal{Z}$} and
{\it morphisms of thickenings over $\mathcal{Z}$}.
This means that the algebraic stacks
$\mathcal{X}'$ and $\mathcal{Y}'$
are algebraic stack endowed with a structure
morphism to $\mathcal{Z}$ and that $f'$ fits into a suitable
$2$-commutative diagram of algebraic stacks.
\end{enumerate}
\end{definition}
\noindent
Let $\mathcal{X} \subset \mathcal{X}'$ be a thickening of algebraic stacks.
Let $U'$ be a scheme and let $U' \to \mathcal{X}'$ be a surjective smooth
morphism. Setting $U = \mathcal{X} \times_{\mathcal{X}'} U'$ we obtain
a morphism of thickenings
$$
(U \subset U') \longrightarrow (\mathcal{X} \subset \mathcal{X}')
$$
and $U \to \mathcal{X}$ is a surjective smooth morphism. We can often
deduce properties of the thickening $\mathcal{X} \subset \mathcal{X}'$
from the corresponding properties of the thickening $U \subset U'$.
Sometimes, by abuse of language, we say that a morphism
$\mathcal{X} \to \mathcal{X}'$ is a thickening if it is a closed
immersion inducing a bijection $|\mathcal{X}| \to |\mathcal{X}'|$.
\begin{lemma}
\label{lemma-thickening}
Let $i : \mathcal{X} \to \mathcal{X}'$ be a morphism of algebraic stacks.
The following are equivalent
\begin{enumerate}
\item $i$ is a thickening of algebraic stacks (abuse of language as above), and
\item $i$ is representable by algebraic spaces and
is a thickening in the sense of Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms}.
\end{enumerate}
In this case $i$ is a closed immersion and a universal homeomorphism.
\end{lemma}
\begin{proof}
By More on Morphisms of Spaces, Lemmas
\ref{spaces-more-morphisms-lemma-descending-property-thickening} and
\ref{spaces-more-morphisms-lemma-base-change-thickening}
the property $P$ that a morphism of algebraic spaces is a
(first order) thickening is fpqc local on the base and stable under base
change. Thus the discussion in Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms} indeed applies.
Having said this the equivalence of (1) and (2) follows from
the fact that $P = P_1 + P_2$ where $P_1$ is the property of being
a closed immersion and $P_2$ is the property of being surjective.
(Strictly speaking, the reader should also consult
More on Morphisms of Spaces, Definition
\ref{spaces-more-morphisms-definition-thickening},
Properties of Stacks, Definition \ref{stacks-properties-definition-immersion}
and the discussion following, Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-surjective-representable},
Properties of Stacks, Section \ref{stacks-properties-section-surjective}
to see that all the concepts all match up.)
The final assertion is clear from the foregoing.
\end{proof}
\noindent
We will use the lemma without further mention. Using the same references
More on Morphisms of Spaces, Lemmas
\ref{spaces-more-morphisms-lemma-descending-property-thickening} and
\ref{spaces-more-morphisms-lemma-base-change-thickening}
as used in the lemma, allows us to define a first order thickening as follows.
\begin{definition}
\label{definition-first-order-thickening}
We say an algebraic stack $\mathcal{X}'$ is a {\it first order thickening}
of an algebraic stack $\mathcal{X}$ if $\mathcal{X}$ is a closed substack
of $\mathcal{X}'$ and $\mathcal{X} \to \mathcal{X}'$ is a first order
thickening in the sense of Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms}.
\end{definition}
\noindent
If $(U \subset U') \to (\mathcal{X} \subset \mathcal{X}')$ is a smooth
cover by a scheme as above, then this simply means that $U \subset U'$
is a first order thickening. Next we formulate the obligatory lemmas.
\begin{lemma}
\label{lemma-base-change-thickening}
Let $\mathcal{Y} \subset \mathcal{Y}'$ be a thickening of algebraic stacks.
Let $\mathcal{X}' \to \mathcal{Y}'$ be a morphism of algebraic stacks
and set $\mathcal{X} = \mathcal{Y} \times_{\mathcal{Y}'} \mathcal{X}'$.
Then
$(\mathcal{X} \subset \mathcal{X}') \to (\mathcal{Y} \subset \mathcal{Y}')$
is a morphism of thickenings. If $\mathcal{Y} \subset \mathcal{Y}'$ is a first
order thickening, then $\mathcal{X} \subset \mathcal{X}'$ is a first
order thickening.
\end{lemma}
\begin{proof}
See discussion above, Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms}, and
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-base-change-thickening}.
\end{proof}
\begin{lemma}
\label{lemma-composition-thickening}
If $\mathcal{X} \subset \mathcal{X}'$ and $\mathcal{X}' \subset \mathcal{X}''$
are thickenings of algebraic stacks, then so is
$\mathcal{X} \subset \mathcal{X}''$.
\end{lemma}
\begin{proof}
See discussion above, Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms}, and
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-composition-thickening}
\end{proof}
\begin{example}
\label{example-reduction-thickening}
Let $\mathcal{X}'$ be an algebraic stack. Then $\mathcal{X}'$ is a thickening
of the reduction $\mathcal{X}'_{red}$, see
Properties of Stacks, Definition
\ref{stacks-properties-definition-reduced-induced-stack}.
Moreover, if $\mathcal{X} \subset \mathcal{X}'$ is a thickening
of algebraic stacks, then
$\mathcal{X}'_{red} = \mathcal{X}_{red} \subset \mathcal{X}$.
In other words, $\mathcal{X} = \mathcal{X}'_{red}$ if and only
if $\mathcal{X}$ is a reduced algebraic stack.
\end{example}
\begin{lemma}
\label{lemma-reduced-diagonal}
Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to
(\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings
of algebraic stacks. Then
$\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}'$
is a thickening and the canonical diagram
$$
\xymatrix{
\mathcal{X} \ar[r]_-\Delta \ar[d] &
\mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[d] \\
\mathcal{X}' \ar[r]^-{\Delta'} &
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}'
}
$$
is cartesian.
\end{lemma}
\begin{proof}
Since $\mathcal{X} \to \mathcal{Y}'$ factors through the closed
substack $\mathcal{Y}$ we see that
$\mathcal{X} \times_\mathcal{Y} \mathcal{X} =
\mathcal{X} \times_{\mathcal{Y}'} \mathcal{X}$.
Hence
$\mathcal{X} \times_\mathcal{Y} \mathcal{X} \to
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}'$
is isomorphic to the composition
$$
\mathcal{X} \times_{\mathcal{Y}'} \mathcal{X} \to
\mathcal{X} \times_{\mathcal{Y}'} \mathcal{X}' \to
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}'
$$
both of which are thickenings as base changes of thickenings
(Lemma \ref{lemma-base-change-thickening}).
Hence so is the composition
(Lemma \ref{lemma-composition-thickening}).
Since $\mathcal{X} \to \mathcal{X}'$ is a monomorphism,
the final statement of the lemma follows from
Properties of Stacks, Lemma
\ref{stacks-properties-lemma-monomorphism-diagonal}
applied to $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}'$.
\end{proof}
\begin{lemma}
\label{lemma-thickening-diagonals}
Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to
(\mathcal{Y} \subset \mathcal{Y}')$ be a morphism of thickenings
of algebraic stacks.
Let $\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ and
$\Delta' : \mathcal{X}' \to \mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}'$
be the corresponding diagonal morphisms.
Then each property from the following list is satisfied by $\Delta$ if
and only if it is satisfied by $\Delta'$:
(a) representable by schemes, (b) affine, (c) surjective, (d) quasi-compact,
(e) universally closed, (f) integral, (g) quasi-separated, (h) separated,
(i) universally injective, (j) universally open, (k) locally quasi-finite,
(l) finite, (m) unramified, (n) monomorphism, (o) immersion,
(p) closed immersion, and (q) proper.
\end{lemma}
\begin{proof}
Observe that
$$
(\Delta, \Delta') :
(\mathcal{X} \subset \mathcal{X}')
\longrightarrow
(\mathcal{X} \times_\mathcal{Y} \mathcal{X} \subset
\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}')
$$
is a morphism of thickenings (Lemma \ref{lemma-reduced-diagonal}).
Moreover $\Delta$ and $\Delta'$ are
representable by algebraic spaces by
Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-properties-diagonal}.
Hence, via the discussion in
Properties of Stacks, Section
\ref{stacks-properties-section-properties-morphisms}
the lemma follows for cases (a), (b), (c), (d),
(e), (f), (g), (h), (i), and (j) by using
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-thicken-property-morphisms}.
\medskip\noindent
Lemma \ref{lemma-reduced-diagonal} tells us that
$\mathcal{X} = (\mathcal{X} \times_\mathcal{Y} \mathcal{X})
\times_{(\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}')} \mathcal{X}'$.
Moreover, $\Delta$ and $\Delta'$ are locally of finite type by
the aforementioned
Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-properties-diagonal}.
Hence the result for cases (k), (l), (m), (n), (o), (p), and (q) by using
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-properties-that-extend-over-thickenings}.
\end{proof}
\noindent
As a consequence we obtain the following pleasing result.
\begin{lemma}
\label{lemma-thickening-properties}
\begin{reference}
\cite[Theorem 2.2.5]{Conrad-moduli}
\end{reference}
Let $\mathcal{X} \subset \mathcal{X}'$ be a thickening of algebraic
stacks. Then
\begin{enumerate}
\item $\mathcal{X}$ is an algebraic space if and only if $\mathcal{X}'$
is an algebraic space,
\item $\mathcal{X}$ is a scheme if and only if $\mathcal{X}'$ is a scheme,
\item $\mathcal{X}$ is DM if and only if $\mathcal{X}'$ is DM,
\item $\mathcal{X}$ is quasi-DM if and only if $\mathcal{X}'$ is quasi-DM,
\item $\mathcal{X}$ is separated if and only if $\mathcal{X}'$ is separated,
\item $\mathcal{X}$ is quasi-separated if and only if $\mathcal{X}'$ is
quasi-separated, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
In each case we reduce to a question about the diagonal and then
we use Lemma \ref{lemma-thickening-diagonals} applied to the
morphism of thickenings
$$
(\mathcal{X} \subset \mathcal{X}') \to
\left(\Spec(\mathbf{Z}) \subset \Spec(\mathbf{Z})\right)
$$
We do this after viewing
$\mathcal{X} \subset \mathcal{X}'$ as a thickening of algebraic stacks
over $\Spec(\mathbf{Z})$ via
Algebraic Stacks, Definition \ref{algebraic-definition-viewed-as}.
\medskip\noindent
Case (1). An algebraic stack is an algebraic space if and only if its
diagonal is a monomorphism, see
Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-hierarchy}
(this also follows immediately from Algebraic Stacks,
Proposition \ref{algebraic-proposition-algebraic-stack-no-automorphisms}).
\medskip\noindent
Case (2). By (1) we may assume that $\mathcal{X}$ and $\mathcal{X}'$
are algebraic spaces and then we can use
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-thickening-scheme}.
\medskip\noindent
Case (3) -- (6). Each of these cases corresponds to a condition
on the diagonal, see Morphisms of Stacks, Definitions
\ref{stacks-morphisms-definition-separated} and
\ref{stacks-morphisms-definition-absolute-separated}.
\end{proof}
\section{Morphisms of thickenings}
\label{section-morphisms-thickenings}
\noindent
If $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to
(\mathcal{Y} \subset \mathcal{Y}')$ is a morphism
of thickenings of algebraic stacks, then often properties of the morphism
$f$ are inherited by $f'$. There are several variants.
\begin{lemma}
\label{lemma-thicken-property-morphisms}
Let $(f, f') : (\mathcal{X} \subset \mathcal{X}') \to
(\mathcal{Y} \subset \mathcal{Y}')$
be a morphism of thickenings of algebraic stacks. Then
\begin{enumerate}
\item $f$ is an affine morphism if and only if $f'$ is an affine morphism,
\item $f$ is a surjective morphism if and only if $f'$ is a surjective morphism,
\item $f$ is quasi-compact if and only if $f'$ quasi-compact,
\item $f$ is universally closed if and only if $f'$ is universally closed,
\item $f$ is integral if and only if $f'$ is integral,
\item $f$ is universally injective if and only if $f'$ is universally injective,
\item $f$ is universally open if and only if $f'$ is universally open,
\item $f$ is quasi-DM if and only if $f'$ is quasi-DM,
\item $f$ is DM if and only if $f'$ is DM,
\item $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,
\item $f$ is representable if and only if $f'$ is representable,
\item $f$ is representable by algebraic spaces if and only if $f'$ is
representable by algebraic spaces,
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-thickening} the morphisms $\mathcal{X} \to \mathcal{X}'$
and $\mathcal{Y} \to \mathcal{Y}'$ are universal homeomorphisms. Thus any
condition on $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is equivalent with
the corresponding condition on $|f'| : |\mathcal{X}'| \to |\mathcal{Y}'|$
and the same is true after arbitrary base change by a morphism
$\mathcal{Z}' \to \mathcal{Y}'$. This proves that
(2), (3), (4), (6), (7) hold.
\medskip\noindent
In cases (8), (9), (10), (12) we can translate the conditions on
$f$ and $f'$ into conditions on the diagonals $\Delta$ and $\Delta'$
as in Lemma \ref{lemma-thickening-diagonals}. See
Morphisms of Stacks, Definition \ref{stacks-morphisms-definition-separated} and
Lemma \ref{stacks-morphisms-lemma-hierarchy}.
Hence these cases follow from Lemma \ref{lemma-thickening-diagonals}.
\medskip\noindent
Proof of (11). If $f'$ is representable, then so is $f$, because
for a scheme $T$ and a morphism $T \to \mathcal{Y}$ we have
$\mathcal{X} \times_\mathcal{Y} T =
\mathcal{X} \times_{\mathcal{X}'} (\mathcal{X}' \times_{\mathcal{Y}'} T)$
and $\mathcal{X} \to \mathcal{X}'$ is a closed immersion (hence representable).
Conversely, assume $f$ is representable, and let $T' \to \mathcal{Y}'$
be a morphism where $T'$ is a scheme. Then
$$
\mathcal{X} \times_{\mathcal{Y}}
(\mathcal{Y} \times_{\mathcal{Y}'} T') =
\mathcal{X} \times_{\mathcal{X}'}
(\mathcal{X}' \times_{\mathcal{Y}'} T') \to
\mathcal{X}' \times_{\mathcal{Y}'} T'
$$
is a thickening (by Lemma \ref{lemma-base-change-thickening})
and the source is a scheme. Hence the target is a scheme by
Lemma \ref{lemma-thickening-properties}.
\medskip\noindent
In cases (1) and (5) if either $f$ or $f'$ has the stated property,
then both $f$ and $f'$ are representable by (11). In this case
choose an algebraic space $V'$ and a surjective smooth morphism
$V' \to \mathcal{Y}'$. Set $V = \mathcal{Y} \times_{\mathcal{Y}'} V'$,
$U' = \mathcal{X}' \times_{\mathcal{Y}'} V'$, and
$U = \mathcal{X} \times_{\mathcal{Y}'} V'$. Then the desired
results follow from the corresponding results for
the morphism $(U \subset U') \to (V \subset V')$ of thickenings
of algebraic spaces via the principle of Properties of Stacks, Lemma
\ref{stacks-properties-lemma-check-property-covering}.
See More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-thicken-property-morphisms}
for the corresponding results in the case of algebraic spaces.
\end{proof}
\section{Infinitesimal deformations of algebraic stacks}
\label{section-deform}
\noindent
This section is the analogue of
More on Morphisms of Spaces, Section
\ref{spaces-more-morphisms-section-deform}.
\begin{lemma}
\label{lemma-flatness-morphism-thickenings-fp-over-ft}
Consider a commutative diagram
$$
\xymatrix{
(\mathcal{X} \subset \mathcal{X}') \ar[rr]_{(f, f')} \ar[rd] & &
(\mathcal{Y} \subset \mathcal{Y}') \ar[ld] \\
& (\mathcal{B} \subset \mathcal{B}')
}
$$
of thickenings of algebraic stacks. Assume
\begin{enumerate}
\item $\mathcal{Y}' \to \mathcal{B}'$ is locally of finite type,
\item $\mathcal{X}' \to \mathcal{B}'$ is
flat and locally of finite presentation,
\item $f$ is flat, and
\item $\mathcal{X} = \mathcal{B} \times_{\mathcal{B}'} \mathcal{X}'$ and
$\mathcal{Y} = \mathcal{B} \times_{\mathcal{B}'} \mathcal{Y}'$.
\end{enumerate}
Then $f'$ is flat and for all $y' \in |\mathcal{Y}'|$ in the image of $|f'|$
the morphism $\mathcal{Y}' \to \mathcal{B}'$ is flat at $y'$.
\end{lemma}
\begin{proof}
Choose an algebraic space $U'$ and a surjective smooth morphism
$U' \to \mathcal{B}'$.
Choose an algebraic space $V'$ and a surjective smooth morphism
$V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$.
Choose an algebraic space $W'$ and a surjective smooth morphism
$W' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$. Let $U, V, W$
be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$.
Then flatness of $f'$ is equivalent to flatness of $W' \to V'$ and
we are given that $W \to V$ is flat. Hence we may apply the lemma
in the case of algebraic spaces to the diagram
$$
\xymatrix{
(W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\
& (U \subset U')
}
$$
of thickenings of algebraic spaces. See
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-flatness-morphism-thickenings-fp-over-ft}.
The statement about flatness of $\mathcal{Y}'/\mathcal{B}'$ at points in the
image of $|f'|$ follows in the same manner.
\end{proof}
\begin{lemma}
\label{lemma-deform-property-fp-over-ft}
Consider a commutative diagram
$$
\xymatrix{
(\mathcal{X} \subset \mathcal{X}') \ar[rr]_{(f, f')} \ar[rd] & &
(\mathcal{Y} \subset \mathcal{Y}') \ar[ld] \\
& (\mathcal{B} \subset \mathcal{B}')
}
$$
of thickenings of algebraic stacks.
Assume $\mathcal{Y}' \to \mathcal{B}'$ locally of finite type,
$\mathcal{X}' \to \mathcal{B}'$ flat and locally of finite presentation,
$\mathcal{X} = \mathcal{B} \times_{\mathcal{B}'} \mathcal{X}'$, and
$\mathcal{Y} = \mathcal{B} \times_{\mathcal{B}'} \mathcal{Y}'$. Then
\begin{enumerate}
\item $f$ is flat if and only if $f'$ is flat,
\label{item-flat-fp-over-ft}
\item $f$ is an isomorphism if and only if $f'$ is an isomorphism,
\label{item-isomorphism-fp-over-ft}
\item $f$ is an open immersion if and only if $f'$ is an open immersion,
\label{item-open-immersion-fp-over-ft}
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\label{item-monomorphism-fp-over-ft}
\item $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,
\label{item-quasi-finite-fp-over-ft}
\item $f$ is syntomic if and only if $f'$ is syntomic,
\label{item-syntomic-fp-over-ft}
\item $f$ is smooth if and only if $f'$ is smooth,
\label{item-smooth-fp-over-ft}
\item $f$ is unramified if and only if $f'$ is unramified,
\label{item-unramified-fp-over-ft}
\item $f$ is \'etale if and only if $f'$ is \'etale,
\label{item-etale-fp-over-ft}
\item $f$ is finite if and only if $f'$ is finite, and
\label{item-finite-fp-over-ft}
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
In case (\ref{item-flat-fp-over-ft}) this follows from
Lemma \ref{lemma-flatness-morphism-thickenings-fp-over-ft}.
\medskip\noindent
In cases
(\ref{item-syntomic-fp-over-ft}), (\ref{item-smooth-fp-over-ft})
this can be proved by the method used in the proof of
Lemma \ref{lemma-flatness-morphism-thickenings-fp-over-ft}.
Namely, choose an algebraic space $U'$ and a surjective smooth morphism
$U' \to \mathcal{B}'$.
Choose an algebraic space $V'$ and a surjective smooth morphism
$V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$.
Choose an algebraic space $W'$ and a surjective smooth morphism
$W' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$. Let $U, V, W$
be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$.
Then the property for $f$, resp.\ $f'$
is equivalent to the property for of $W' \to V'$, resp.\ $W \to V$.
Hence we may apply the lemma in the case of algebraic spaces to the
diagram
$$
\xymatrix{
(W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\
& (U \subset U')
}
$$
of thickenings of algebraic spaces. See
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-deform-property-fp-over-ft}.
\medskip\noindent
In cases (\ref{item-unramified-fp-over-ft}) and (\ref{item-etale-fp-over-ft})
we first see that the assumption for $f$ or $f'$ implies that both
$f$ and $f'$ are DM morphisms of algebraic stacks, see
Lemma \ref{lemma-thicken-property-morphisms}. Then we can choose
an algebraic space $U'$ and a surjective smooth morphism
$U' \to \mathcal{B}'$.
Choose an algebraic space $V'$ and a surjective smooth morphism
$V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$.
Choose an algebraic space $W'$ and a surjective \'etale(!) morphism
$W' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$. Let $U, V, W$
be the base change of $U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$.
Then $W \to V \times_\mathcal{Y} \mathcal{X}$ is surjective
\'etale as well. Hence the property for $f$, resp.\ $f'$
is equivalent to the property for of $W' \to V'$, resp.\ $W \to V$.
Hence we may apply the lemma in the case of algebraic spaces to the
diagram
$$
\xymatrix{
(W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\
& (U \subset U')
}
$$
of thickenings of algebraic spaces. See
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-deform-property-fp-over-ft}.
\medskip\noindent
In cases (\ref{item-isomorphism-fp-over-ft}),
(\ref{item-open-immersion-fp-over-ft}),
(\ref{item-monomorphism-fp-over-ft}),
(\ref{item-finite-fp-over-ft})
we first conclude by Lemma \ref{lemma-thicken-property-morphisms}
that $f$ and $f'$ are representable by algebraic spaces. Thus we may choose
an algebraic space $U'$ and a surjective smooth morphism
$U' \to \mathcal{B}'$,
an algebraic space $V'$ and a surjective smooth morphism
$V' \to U' \times_{\mathcal{B}'} \mathcal{Y}'$, and then
$W' = V' \times_{\mathcal{Y}'} \mathcal{X}'$ will be an algebraic space.
Let $U, V, W$ be the base change of
$U', V', W'$ by $\mathcal{B} \to \mathcal{B}'$.
Then $W = V \times_\mathcal{Y} \mathcal{X}$ as well.
Then we have to see that $W' \to V'$ is an
isomorphism, resp.\ an open immersion, resp.\ a monomorphism,
resp.\ finite, if and only if $W \to V$ has the same property.
See Properties of Stacks, Lemma
\ref{stacks-properties-lemma-check-property-covering}.
Thus we conclude by applying the results
for algebraic spaces as above.
\medskip\noindent
In the case (\ref{item-quasi-finite-fp-over-ft}) we first
observe that $f$ and $f'$ are locally of finite type by
Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-finite-type-permanence}.
On the other hand, the morphism $f$ is quasi-DM if and only if
$f'$ is by
Lemma \ref{lemma-thicken-property-morphisms}.
The last thing to check to see if $f$ or $f'$ is locally quasi-finite
(Morphisms of Stacks, Definition
\ref{stacks-morphisms-definition-quasi-finite})
is a condition on underlying topological spaces
which holds for $f$ if and only if it holds for $f'$ by
the discussion in the first paragraph of the proof.
\end{proof}
\section{Lifting affines}
\label{section-lifting-affines}
\noindent
Consider a solid diagram
$$
\xymatrix{
W \ar[d] \ar@{..>}[r] & W' \ar@{..>}[d] \\
\mathcal{X} \ar[r] & \mathcal{X}'
}
$$
where $\mathcal{X} \subset \mathcal{X}'$ is a thickening of algebraic stacks,
$W$ is an affine scheme and $W \to \mathcal{X}$ is smooth. The question
we address in this section is whether we can find $W'$ and the dotted
arrows so that the square is cartesian and $W' \to \mathcal{X}'$ is smooth.
We do not know the answer in general, but if $\mathcal{X} \subset \mathcal{X}'$
is a first order thickening we will prove the answer is yes.
\medskip\noindent
To study this problem we introduce the following category.
\begin{remark}[Category of lifts]
\label{remark-gerbe-of-lifts}
Consider a diagram
$$
\xymatrix{
W \ar[d]_x \\
\mathcal{X} \ar[r] & \mathcal{X}'
}
$$
where $\mathcal{X} \subset \mathcal{X}'$ is a thickening of algebraic stacks,
$W$ is an algebraic space, and $W \to \mathcal{X}$ is smooth.
We will construct a category $\mathcal{C}$ and a functor
$$
p : \mathcal{C} \longrightarrow W_{spaces, \etale}
$$
(see Properties of Spaces, Definition
\ref{spaces-properties-definition-spaces-etale-site} for notation)
as follows. An object of $\mathcal{C}$ will be a system
$(U, U', a, i, y', \alpha)$
which forms a commutative diagram
\begin{equation}
\label{equation-object}
\vcenter{
\xymatrix{
U \ar[d]_a \ar[r]_i & U' \ar[dd]^{x'} \\
W \ar[d]_x & \\
\mathcal{X} \ar[r] & \mathcal{X}'
}
}
\end{equation}
with commutativity witnessed by the $2$-morphism
$\alpha : x \circ a \to x' \circ i$ such that
$U$ and $U'$ are algebraic spaces,
$a : U \to W$ is \'etale, $x' : U' \to \mathcal{X}'$ is smooth,
and such that $U = \mathcal{X} \times_{\mathcal{X}'} U'$.
In particular $U \subset U'$ is a thickening.
A morphism
$$
(U, U', a, i, x', \alpha) \to (V, V', b, j, y', \beta)
$$
is given by $(f, f', \gamma)$ where $f : U \to V$ is a morphism
over $W$, $f' : U' \to V'$ is a morphism whose restriction
to $U$ gives $f$, and $\gamma : x' \circ f' \to y'$ is a $2$-morphism
witnessing the commutativity in right triangle of the diagram below
\begin{equation}
\label{equation-morphism}
\vcenter{
\xymatrix{
& V \ar[ld]_f \ar[ldd]^b \ar[rr]_j & & V' \ar[ld]_{f'} \ar[lddd]^{y'} \\
U \ar[d]_a \ar[rr]_i & & U' \ar[dd]_{x'} \\
W \ar[d]_x & \\
\mathcal{X} \ar[rr] & & \mathcal{X}'
}
}
\end{equation}
Finally, we require that $\gamma$ is compatible with $\alpha$ and $\beta$:
in the calculus of $2$-categories of Categories, Sections
\ref{categories-section-formal-cat-cat} and
\ref{categories-section-2-categories} this reads
$$
\beta = (\gamma \star \text{id}_j) \circ (\alpha \star \text{id}_f)
$$
(more succinctly: $\beta = j^*\gamma \circ f^*\alpha$).
Another formulation is that objects are commutative diagrams
(\ref{equation-object}) with some additional properties and
morphisms are commutative diagrams
(\ref{equation-morphism}) in the category $\textit{Spaces}/\mathcal{X}'$
introduced in Properties of Stacks, Remark
\ref{stacks-properties-remark-representable-over}.
This makes it clear that $\mathcal{C}$ is a category
and that the rule $p : \mathcal{C} \to W_{spaces, \etale}$
sending $(U, U', a, i, x', \alpha)$ to $a : U \to W$
is a functor.
\end{remark}
\begin{lemma}
\label{lemma-morphisms-lifts-etale}
For any morphism (\ref{equation-morphism}) the map $f' : V' \to U'$ is \'etale.
\end{lemma}
\begin{proof}
Namely $f : V \to U$ is \'etale as a morphism in $W_{spaces, \etale}$
and we can apply
Lemma \ref{lemma-deform-property-fp-over-ft} because $U' \to \mathcal{X}'$
and $V' \to \mathcal{X}'$ are smooth and
$U = \mathcal{X} \times_{\mathcal{X}'} U'$ and
$V = \mathcal{X} \times_{\mathcal{X}'} V'$.
\end{proof}
\begin{lemma}
\label{lemma-gerbe-of-lifts-fibred}
The category $p : \mathcal{C} \to W_{spaces, \etale}$ constructed
in Remark \ref{remark-gerbe-of-lifts} is fibred in groupoids.
\end{lemma}
\begin{proof}
We claim the fibre categories of $p$ are groupoids.
If $(f, f', \gamma')$ as in (\ref{equation-morphism})
is a morphism such that $f : U \to V$ is an isomorphism, then
$f'$ is an isomorphism by Lemma \ref{lemma-deform-property-fp-over-ft}
and hence $(f, f', \gamma')$ is an isomorphism.
\medskip\noindent
Consider a morphism $f : V \to U$ in $W_{spaces, \etale}$
and an object $\xi = (U, U', a, i, x', \alpha)$
of $\mathcal{C}$ over $U$. We are going to construct the ``pullback''
$f^*\xi$ over $V$.
Namely, set $b = a \circ f$. Let $f' : V' \to U'$ be the \'etale morphism
whose restriction to $V$ is $f$ (More on Morphisms of Spaces,
Lemma \ref{spaces-more-morphisms-lemma-topological-invariance}).
Denote $j : V \to V'$ the corresponding thickening.
Let $y' = x' \circ f'$ and $\gamma = \text{id} : x' \circ f' \to y'$.
Set
$$
\beta = \alpha \star \text{id}_f :
x \circ b = x \circ a \circ f \to
x' \circ i \circ f = x' \circ f' \circ j = y' \circ j
$$
It is clear that $(f, f', \gamma) : (V, V', b, j, y', \beta) \to
(U, U', a, i, x', \alpha)$ is a morphism as in
(\ref{equation-morphism}). The morphisms $(f, f', \gamma)$
so constructed are strongly cartesian
(Categories, Definition \ref{categories-definition-cartesian-over-C}).
We omit the detailed proof, but essentially the reason is that
given a morphism
$(g, g', \epsilon) : (Y, Y', c, k, z', \delta) \to (U, U', a, i, x', \alpha)$
in $\mathcal{C}$ such that $g$ factors as $g = f \circ h$ for some
$h : Y \to V$, then we get a unique factorization $g' = f' \circ h'$
from More on Morphisms of Spaces,
Lemma \ref{spaces-more-morphisms-lemma-topological-invariance}
and after that one can produce the necessary $\zeta$ such that
$(h, h', \zeta) : (Y, Y', c, k, z', \delta) \to
(V, V', b, j, y', \beta)$ is a morphism of $\mathcal{C}$
with $(g, g', \epsilon) = (f, f', \gamma) \circ (h, h', \zeta)$.
\medskip\noindent
Therefore $p : \mathcal{C} \to W_\etale$ is a fibred
category (Categories, Definition \ref{categories-definition-fibred-category}).
Combined with the fact that the fibre categories are groupoids
seen above we conclude that $p : \mathcal{C} \to W_\etale$
is fibred in groupoids by Categories, Lemma
\ref{categories-lemma-fibred-groupoids}.
\end{proof}
\begin{lemma}
\label{lemma-gerbe-of-lifts-stack}
The category $p : \mathcal{C} \to W_{spaces, \etale}$ constructed
in Remark \ref{remark-gerbe-of-lifts} is a stack in groupoids.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-gerbe-of-lifts-fibred} we see the first condition
of Stacks, Definition \ref{stacks-definition-stack-in-groupoids} holds.
As is customary we check descent of objects and we leave it to the reader
to check descent of morphisms. Thus suppose we have $a : U \to W$
in $W_{spaces, \etale}$, a covering $\{U_k \to U\}_{k \in K}$ in
$W_{spaces, \etale}$, objects $\xi_k = (U_k, U'_k, a_k, i_k, x'_k, \alpha_k)$
of $\mathcal{C}$ over $U_k$, and morphisms
$$
\varphi_{kk'} = (f_{kk'}, f'_{kk'}, \gamma_{kk'}) :
\xi_k|_{U_k \times_U U_{k'}} \to
\xi_{k'}|_{U_k \times_U U_{k'}}
$$
between restrictions satisfying the cocycle condition. In order to prove
effectivity we may first refine the covering. Hence we may assume each
$U_k$ is a scheme (even an affine scheme if you like). Let us write
$$
\xi_k|_{U_k \times_U U_{k'}} =
(U_k \times_U U_{k'}, U'_{kk'}, a_{kk'}, x'_{kk'}, \alpha_{kk'})
$$
Then we get an \'etale (by Lemma \ref{lemma-morphisms-lifts-etale}) morphism
$s_{kk'} : U'_{kk'} \to U'_k$
as the second component of the morphism
$\xi_k|_{U_k \times_U U_{k'}} \to \xi_k$ of $\mathcal{C}$.
Similarly we obtain an \'etale morphism $t_{kk'} : U'_{kk'} \to U'_{k'}$
by looking at the second component of the composition
$$
\xi_k|_{U_k \times_U U_{k'}} \xrightarrow{\varphi_{kk'}}
\xi_{k'}|_{U_k \times_U U_{k'}} \to \xi_{k'}
$$
We claim that
$$
j :
\coprod\nolimits_{(k, k') \in K \times K} U'_{kk'}
\xrightarrow{(\coprod s_{kk'}, \coprod t_{kk'})}
(\coprod\nolimits_{k \in K} U'_k) \times (\coprod\nolimits_{k \in K} U'_k)
$$
is an \'etale equivalence relation. First, we have already seen
that the components $s, t$ of the displayed morphism are \'etale.
The base change of the morphism $j$ by
$(\coprod U_k) \times (\coprod U_k) \to (\coprod U'_k) \times (\coprod U'_k)$
is a monomorphism because it is the map
$$
\coprod\nolimits_{(k, k') \in K \times K} U_k \times_U U_{k'}
\longrightarrow
(\coprod\nolimits_{k \in K} U_k) \times (\coprod\nolimits_{k \in K} U_k)
$$
Hence $j$ is a monomorphism by More on Morphisms, Lemma
\ref{more-morphisms-lemma-properties-that-extend-over-thickenings}.
Finally, symmetry of the relation $j$ comes from the fact that
$\varphi_{kk'}^{-1}$ is the ``flip'' of $\varphi_{k'k}$ (see
Stacks, Remarks \ref{stacks-remarks-definition-descent-datum})
and transitivity comes from the cocycle condition (details omitted).
Thus the quotient of $\coprod U'_k$ by $j$ is an algebraic space $U'$
(Spaces, Theorem \ref{spaces-theorem-presentation}).
Above we have already shown that there is a thickening
$i : U \to U'$ as we saw that the restriction of $j$ on
$\coprod U_k$ gives $(\coprod U_k) \times_U (\coprod U_k)$.
Finally, if we temporarily view the $1$-morphisms
$x'_k : U'_k \to \mathcal{X}'$ as objects of the stack
$\mathcal{X}'$ over $U'_k$ then we see that these come endowed with a
descent datum with respect to the \'etale covering
$\{U'_k \to U'\}$ given by the third component $\gamma_{kk'}$
of the morphisms $\varphi_{kk'}$ in $\mathcal{C}$.
Since $\mathcal{X}'$ is a stack
this descent datum is effective and translating back we obtain
a $1$-morphism $x' : U' \to \mathcal{X}'$ such that the compositions
$U'_k \to U' \to \mathcal{X}'$ come equipped with isomorphisms to $x'_k$
compatible with $\gamma_{kk'}$. This means that the morphisms
$\alpha_k : x \circ a_k \to x'_k \circ i_k$ glue to a morphism
$\alpha : x \circ a \to x' \circ i$. Then $\xi = (U, U', a, i, x', \alpha)$
is the desired object over $U$.
\end{proof}
\begin{lemma}
\label{lemma-etale-local-lifts}
Let $\mathcal{X} \subset \mathcal{X}'$ be a thickening of algebraic stacks.
Let $W$ be an algebraic space and let $W \to \mathcal{X}$ be a smooth morphism.
There exists an \'etale covering $\{W_i \to W\}_{i \in I}$ and for each $i$
a cartesian diagram
$$
\xymatrix{
W_i \ar[r] \ar[d] & W_i' \ar[d] \\
\mathcal{X} \ar[r] & \mathcal{X}'
}
$$
with $W_i' \to \mathcal{X}'$ smooth.
\end{lemma}
\begin{proof}
Choose a scheme $U'$ and a surjective smooth morphism $U' \to \mathcal{X}'$.
As usual we set $U = \mathcal{X} \times_{\mathcal{X}'} U'$. Then
$U \to \mathcal{X}$ is a surjective smooth morphism. Therefore the base change
$$
V = W \times_{\mathcal{X}} U \longrightarrow W
$$
is a surjective smooth morphism of algebraic spaces.
By Topologies on Spaces, Lemma
\ref{spaces-topologies-lemma-etale-dominates-smooth}
we can find an \'etale covering $\{W_i \to W\}$ such
that $W_i \to W$ factors through $V \to W$.
After covering $W_i$ by affines (Properties of Spaces, Lemma
\ref{spaces-properties-lemma-cover-by-union-affines})
we may assume each $W_i$ is affine. We may and do replace $W$ by $W_i$
which reduces us to the situation discussed in the next paragraph.
\medskip\noindent
Assume $W$ is affine and the given morphism $W \to \mathcal{X}$ factors
through $U$. Picture
$$
W \xrightarrow{i} U \to \mathcal{X}
$$
Since $W$ and $U$ are smooth over $\mathcal{X}$ we see that
$i$ is locally of finite type (Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-finite-type-permanence}).
After replacing $U$ by $\mathbf{A}^n_U$ we may assume
that $i$ is an immersion, see
Morphisms, Lemma \ref{morphisms-lemma-quasi-affine-finite-type-over-S}.
By Morphisms of Stacks,
Lemma \ref{stacks-morphisms-lemma-lci-permanence}
the morphism $i$ is a local complete intersection.
Hence $i$ is a Koszul-regular immersion (as defined in
Divisors, Definition \ref{divisors-definition-regular-immersion})
by More on Morphisms, Lemma \ref{more-morphisms-lemma-lci}.
\medskip\noindent
We may still replace $W$ by an affine open covering.
For every point $w \in W$ we can choose an affine open
$U'_w \subset U'$ such that if $U_w \subset U$ is the
corresponding affine open, then $w \in i^{-1}(U_w)$ and
$i^{-1}(U_w) \to U_w$ is a closed immersion cut out
by a Koszul-regular sequence
$f_1, \ldots, f_r \in \Gamma(U_w, \mathcal{O}_{U_w})$.
This follows from the definition of Koszul-regular immersions
and Divisors, Lemma \ref{divisors-lemma-regular-ideal-sheaf-scheme}.
Set $W_w = i^{-1}(U_w)$; this is an affine open neighbourhood
of $w \in W$.
Choose lifts $f'_1, \ldots, f'_r \in \Gamma(U'_w, \mathcal{O}_{U'_w})$
of $f_1, \ldots, f_r$. This is possible as $U_w \to U'_w$
is a closed immersion of affine schemes.
Let $W'_w \subset U'_w$ be the closed subscheme cut out by
$f'_1, \ldots, f'_r$.
We claim that $W'_w \to \mathcal{X}'$ is smooth.
The claim finishes the proof as
$W_w = \mathcal{X} \times_{\mathcal{X}'} W'_w$
by construction.
\medskip\noindent
To check the claim it suffices to check that the base change
$W'_w \times_{\mathcal{X}'} X' \to X'$ is smooth for every
affine scheme $X'$ smooth over $\mathcal{X}'$. Choose an
\'etale morphism
$$
Y' \to U'_w \times_{\mathcal{X}'} X'
$$
with $Y'$ affine. Because $U'_w \times_{\mathcal{X}'} X'$ is covered
by the images of such morphisms, it is enough to show that the closed
subscheme $Z'$ of $Y'$ cut out by $f'_1, \ldots, f'_r$ is smooth over $X'$.
Picture
$$
\xymatrix{
Z' \ar[r] \ar[d] & Y' \ar[d] \\
W'_w \times_{\mathcal{X}'} X' \ar[d] \ar[r] &
U'_w \times_{\mathcal{X}'} X' \ar[d] \ar[r] & X' \\
W'_w = V(f'_1, \ldots, f'_r) \ar[r] & U'_w
}
$$
Set $X = \mathcal{X} \times_{\mathcal{X}'} X'$,
$Y = X \times_{X'} Y' = \mathcal{X} \times_{\mathcal{X}'} Y'$, and
$Z = Y \times_{Y'} Z' = X \times_{X'} Z' =
\mathcal{X} \times_{\mathcal{X}'} Z'$.
Then $(Z \subset Z') \to (Y \subset Y') \subset (X \subset X')$
are (cartesian) morphisms of thickenings of affine schemes and
we are given that $Z \to X$ and $Y' \to X'$ are smooth.
Finally, the sequence of functions $f'_1, \ldots, f'_r$
map to a Koszul-regular sequence in $\Gamma(Y', \mathcal{O}_{Y'})$ by
More on Algebra, Lemma \ref{more-algebra-lemma-koszul-regular-flat-base-change}
because $Y' \to U'_w$ is smooth and hence flat.
By More on Algebra, Lemma \ref{more-algebra-lemma-cut-by-koszul}
(and the fact that Koszul-regular sequences are quasi-regular sequences
by More on Algebra, Lemmas \ref{more-algebra-lemma-regular-koszul-regular},
\ref{more-algebra-lemma-koszul-regular-H1-regular}, and
\ref{more-algebra-lemma-H1-regular-quasi-regular})
we conclude that $Z' \to X'$ is smooth as desired.
\end{proof}
\begin{lemma}
\label{lemma-etale-local-lifts-isomorphic}
Let $\mathcal{X} \subset \mathcal{X}'$ be a thickening of algebraic stacks.
Consider a commutative diagram
$$
\xymatrix{
W'' \ar[d]_{x''} & W \ar[l] \ar[r] \ar[d]_x & W' \ar[d]^{x'} \\
\mathcal{X}' & \mathcal{X} \ar[l] \ar[r] & \mathcal{X}'
}
$$
with cartesian squares where $W', W, W''$ are algebraic spaces and
the vertical arrows are smooth. Then there exist
\begin{enumerate}
\item an \'etale covering $\{f'_k : W'_k \to W'\}_{k \in K}$,
\item \'etale morphisms $f''_k : W'_k \to W''$, and
\item $2$-morphisms $\gamma_k : x'' \circ f''_k \to x' \circ f'_k$
\end{enumerate}
such that (a) $(f'_k)^{-1}(W) = (f''_k)^{-1}(W)$, (b)
$f'_k|_{(f'_k)^{-1}(W)} = f''_k|_{(f''_k)^{-1}(W)}$, and
(c) pulling back $\gamma_k$ to the closed subscheme of (a)
agrees with the $2$-morphism given by the commutativity of
the initial diagram over $W$.
\end{lemma}
\begin{proof}
Denote $i : W \to W'$ and $i'' : W \to W''$ the given thickenings.
The commutativity of the diagram in the statement of the lemma
means there is a $2$-morphism $\delta : x' \circ i' \to x'' \circ i''$
This is the $2$-morphism referred to in part (c) of the statement.
Consider the algebraic space
$$
I' = W' \times_{x', \mathcal{X}', x''} W''
$$
with projections $p' : I' \to W'$ and $q' : I' \to W''$.
Observe that there is a ``universal'' $2$-morphism
$\gamma : x' \circ p' \to x'' \circ q'$ (we will use this later).
The choice of $\delta$ defines a morphism
$$
\xymatrix{
W \ar[rr]_\delta & & I' \ar[ld]^{p'} \ar[rd]_{q'} \\
& W' & & W''
}
$$
such that the compositions $W \to I' \to W'$ and $W \to I' \to W''$
are $i : W \to W'$ and $i' : W \to W''$.
Since $x''$ is smooth, the morphism $p' : I' \to W'$ is smooth
as a base change of $x''$.
\medskip\noindent
Suppose we can find an \'etale covering $\{f'_k : W'_k \to W'\}$
and morphisms $\delta_k : W'_k \to I'$ such that the restriction
of $\delta_k$ to $W_k = (f'_k)^{-1}$ is equal to $\delta \circ f_k$
where $f_k = f'_k|_{W_k}$. Picture
$$
\xymatrix{
W_k \ar[r]^{f_k} \ar[d] & W \ar[r]^\delta & I' \ar[d]^{p'} \\
W'_k \ar[rr]^{f'_k} \ar[rru]^{\delta_k} & & W'
}
$$
In other words, we want to be able to extend the given section
$\delta : W \to I'$ of $p'$ to a section over $W'$ after possibly
replacing $W'$ by an \'etale covering.
\medskip\noindent
If this is true, then we can set $f''_k = q' \circ \delta_k$
and $\gamma_k = \gamma \star \text{id}_{\delta_k}$ (more succinctly
$\gamma_k = \delta_k^*\gamma$). Namely, the only thing left to show
at this is that the morphism $f''_k$ is \'etale.
By construction the morphism $x' \circ p'$ is $2$-isomorphic
to $x'' \circ q'$. Hence $x'' \circ f''_k$ is $2$-isomorphic
to $x' \circ f'_k$. We conclude that the composition
$$
W'_k \xrightarrow{f''_k} W'' \xrightarrow{x''} \mathcal{X}'
$$
is smooth because $x' \circ f'_k$ is so.
As $f_k$ is \'etale we conclude $f''_k$ is \'etale
by Lemma \ref{lemma-deform-property-fp-over-ft}.
\medskip\noindent
If the thickening is a first order thickening, then we can
choose any \'etale covering $\{W'_k \to W'\}$ with $W_k'$ affine.
Namely, since $p'$ is smooth we see that $p'$ is formally smooth by the
infinitesimal lifting criterion (More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-smooth-formally-smooth}).
As $W_k$ is affine and as $W_k \to W'_k$ is a first order thickening
(as a base change of $\mathcal{X} \to \mathcal{X}'$, see
Lemma \ref{lemma-base-change-thickening}) we get $\delta_k$ as
desired.
\medskip\noindent
In the general case the existence of the covering and the morphisms
$\delta_k$ follows from More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-smooth-strong-lift}.
\end{proof}
\begin{lemma}
\label{lemma-gerbe-of-lifts}
The category $p : \mathcal{C} \to W_{spaces, \etale}$ constructed
in Remark \ref{remark-gerbe-of-lifts} is a gerbe.
\end{lemma}
\begin{proof}
In Lemma \ref{lemma-gerbe-of-lifts-stack}
we have seen that it is a stack in groupoids.
Thus it remains to check conditions (2) and (3) of
Stacks, Definition \ref{stacks-definition-gerbe}.
Condition (2) follows from
Lemma \ref{lemma-etale-local-lifts}.
Condition (3) follows from
Lemma \ref{lemma-etale-local-lifts-isomorphic}.
\end{proof}
\begin{lemma}
\label{lemma-gerbe-of-lifts-first-order}
In Remark \ref{remark-gerbe-of-lifts} assume $\mathcal{X} \subset \mathcal{X}'$
is a first order thickening. Then
\begin{enumerate}
\item the automorphism sheaves of objects of the gerbe
$p : \mathcal{C} \to W_{spaces, \etale}$ constructed
in Remark \ref{remark-gerbe-of-lifts} are abelian, and
\item the sheaf of groups $\mathcal{G}$ constructed in
Stacks, Lemma \ref{stacks-lemma-gerbe-abelian-auts}
is a quasi-coherent $\mathcal{O}_W$-module.
\end{enumerate}
\end{lemma}
\begin{proof}
We will prove both statements at the same time. Namely, given
an object $\xi = (U, U', a, i, x', \alpha)$ we will endow
$\mathit{Aut}(\xi)$ with the structure of a
quasi-coherent $\mathcal{O}_U$-module on $U_{spaces, \etale}$ and
we will show that this structure is compatible with pullbacks.
This will be sufficient by glueing of sheaves
(Sites, Section \ref{sites-section-glueing-sheaves})
and the construction of $\mathcal{G}$ in the proof of
Stacks, Lemma \ref{stacks-lemma-gerbe-abelian-auts}
as the glueing of the automorphism sheaves $\mathit{Aut}(\xi)$
and the fact that it suffices to check a module is
quasi-coherent after going to an \'etale covering
(Properties of Spaces, Lemma
\ref{spaces-properties-lemma-characterize-quasi-coherent}).
\medskip\noindent
We will describe the sheaf $\mathit{Aut}(\xi)$ using the
same method as used in the proof of
Lemma \ref{lemma-etale-local-lifts-isomorphic}.
Consider the algebraic space
$$
I' = U' \times_{x', \mathcal{X}', x'} U'
$$
with projections $p' : I' \to U'$ and $q' : I' \to U'$.
Over $I'$ there is a universal $2$-morphism
$\gamma : x' \circ p' \to x' \circ q'$.
The identity $x' \to x'$ defines a diagonal morphism
$$
\xymatrix{
U' \ar[rr]_{\Delta'} & & I' \ar[ld]^{p'} \ar[rd]_{q'} \\
& U' & & U'
}
$$
such that the compositions $U' \to I' \to U'$ and $U' \to I' \to U'$
are the identity morphisms. We will denote the base change of
$U', I', p', q', \Delta'$ to $\mathcal{X}$ by $U, I, p, q, \Delta$.
Since $W' \to \mathcal{X}'$ is smooth, we see that $p' : I' \to U'$
is smooth as a base change.
\medskip\noindent
A section of $\mathit{Aut}(\xi)$ over $U$ is a morphism $\delta' : U' \to I'$
such that $\delta'|_U = \Delta$ and such that
$p' \circ \delta' = \text{id}_{U'}$. To be explicit,
$(\text{id}_U, q' \circ \delta', (\delta')^*\gamma) : \xi \to \xi$
is a formula for the corresponding automorphism.
More generally, if
$f : V \to U$ is an \'etale morphism, then there is a thickening
$j : V \to V'$ and an \'etale morphism
$f' : V' \to U'$ whose restriction to $V$ is $f$ and $f^*\xi$
corresponds to $(V, V', a \circ f, j, x' \circ f', f^*\alpha)$, see proof of
Lemma \ref{lemma-gerbe-of-lifts-fibred}.
a section of $\mathit{Aut}(\xi)$ over $V$ is a morphism
$\delta' : V' \to I'$
such that $\delta'|_V = \Delta \circ f$
and $p' \circ \delta' = f'$\footnote{A formula for the corresponding
automorphism is $(\text{id}_V, h', (\delta')^*\gamma)$.
Here $h' : V' \to V'$ is the unique (iso)morphism such that
$h'|_V = \text{id}_V$ and such that
$$
\xymatrix{
V' \ar[r]_{h'} \ar[rd]_{q' \circ \delta'} & V' \ar[d]^{f'} \\
& U'
}
$$
commutes. Uniqueness and existence of $h'$ by topological invariance of
the \'etale site, see More on Morphisms of Spaces,
Theorem \ref{spaces-more-morphisms-theorem-topological-invariance}.
The reader may feel we should instead look at morphisms
$\delta'' : V' \to V' \times_{\mathcal{X}'} V'$ with
$\delta'' \circ j = \Delta_{V'/\mathcal{X}'}$ and
$\text{pr}_1 \circ \delta'' = \text{id}_{V'}$.
This would be fine too: as $V' \times_{\mathcal{X}'} V' \to I'$ is \'etale,
the same topological invariance tells us that sending
$\delta''$ to $\delta' = (V' \times_{\mathcal{X}'} V' \to I') \circ \delta''$
is a bijection between the two
sets of morphisms.}.
\medskip\noindent
We conclude that $\mathit{Aut}(\xi)$ as a sheaf of sets agrees with
the sheaf defined in
More on Morphisms of Spaces, Remark
\ref{spaces-more-morphisms-remark-another-special-case}
for the thickenings $(U \subset U')$ and $(I \subset I')$ over
$(U \subset U')$ via $\text{id}_{U'}$ and $p'$.
The diagonal $\Delta'$ is a section of this sheaf and by
acting on this section using More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-action-sheaf}
we get an isomorphism
\begin{equation}
\label{equation-isomorphism}
\SheafHom_{\mathcal{O}_U}(\Delta^*\Omega_{I/U}, \mathcal{C}_{U/U'})
\longrightarrow
\mathit{Aut}(\xi)
\end{equation}
on $U_{spaces, \etale}$. There three things left to check
\begin{enumerate}
\item the construction of (\ref{equation-isomorphism})
commutes with \'etale localization,
\item $\SheafHom_{\mathcal{O}_U}(\Delta^*\Omega_{I/U}, \mathcal{C}_{U/U'})$
is a quasi-coherent module on $U$,
\item the composition in $\mathit{Aut}(\xi)$ corresponds
to addition of sections in this quasi-coherent module.
\end{enumerate}
We will check these in order.
\medskip\noindent
To see (1) we have to show that if $f : V \to U$ is \'etale,
then (\ref{equation-isomorphism}) constructed using $\xi$ over $U$,
restricts to the map (\ref{equation-isomorphism})
$$
\SheafHom_{\mathcal{O}_V}(
\Delta_V^*\Omega_{V \times_\mathcal{X} V/V}, \mathcal{C}_{V/V'}) \to
\mathit{Aut}(\xi|_V)
$$
constructed using $\xi|_V$ over $V$ on $V_{spaces, \etale}$.
This follows from the discussion in the footnote above
and More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-action-by-derivations-etale-localization}.
\medskip\noindent
Proof of (2). Since $p'$ is smooth, the morphism $I \to U$ is smooth,
and hence the relative module of differentials $\Omega_{I/U}$
is finite locally free (More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-smooth-omega-finite-locally-free}).
On the other hand, $\mathcal{C}_{U/U'}$ is quasi-coherent
(More on Morphisms of Spaces, Definition
\ref{spaces-more-morphisms-definition-conormal-sheaf}).
By Properties of Spaces, Lemma
\ref{spaces-properties-lemma-properties-quasi-coherent}
we conclude.
\medskip\noindent
Proof of (3). There exists a morphism $c' : I' \times_{p', U', q'} I' \to I'$
such that $(U', I', p', q', c')$ is a groupoid in algebraic spaces
with identity $\Delta'$. See
Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack} for example.
Composition in $\mathit{Aut}(\xi)$ is induced by the morphism
$c'$ as follows. Suppose we have two morphisms
$$
\delta'_1, \delta'_2 : U' \longrightarrow I'
$$
corresponding to sections of $\mathit{Aut}(\xi)$ over $U$ as above,
in other words, we have $\delta'_i|U = \Delta_U$ and
$p' \circ \delta'_i = \text{id}_{U'}$. Then the composition
in $\mathit{Aut}(\xi)$ is
$$
\delta'_1 \circ \delta'_2 = c'(\delta'_1 \circ q' \circ \delta'_2, \delta'_2)
$$
We omit the detailed verification\footnote{The reader can see immediately
that it is necessary to precompose
$\delta'_1$ by $q' \circ \delta'_2$ to get a well defined $U'$-valued
point of the fibre product $I' \times_{p', U', q'} I'$.}.
Thus we are in the situation described in
More on Groupoids in Spaces, Section
\ref{spaces-more-groupoids-section-groupoid-sections}
and the desired result follows from
More on Groupoids in Spaces, Lemma
\ref{spaces-more-groupoids-lemma-composition-is-addition}.
\end{proof}
\begin{proposition}[Emerton]
\label{proposition-affine-smooth-lift-to-first-order}
\begin{reference}
Email of Matthew Emerton dated April 27, 2016.
\end{reference}
Let $\mathcal{X} \subset \mathcal{X}'$ be a first order thickening
of algebraic stacks. Let $W$ be an affine scheme and let
$W \to \mathcal{X}$ be a smooth morphism. Then there exists
a cartesian diagram
$$
\xymatrix{
W \ar[d] \ar[r] & W' \ar[d] \\
\mathcal{X} \ar[r] & \mathcal{X}'
}
$$
with $W' \to \mathcal{X}'$ smooth.
\end{proposition}
\begin{proof}
Consider the category $p : \mathcal{C} \to W_{spaces, \etale}$
introduced in Remark \ref{remark-gerbe-of-lifts}.
The proposition states that there exists an object of $\mathcal{C}$
lying over $W$. By Lemma \ref{lemma-gerbe-of-lifts}
$\mathcal{C}$ is a gerbe over $W_{spaces, \etale}$.
This means we can \'etale locally find a solution and
these local solutions are \'etale locally isomorphic;
this part does not require the assumption that the thickening is first order.
By Lemma \ref{lemma-gerbe-of-lifts-first-order}
the automorphism sheaves of objects of our gerbe are abelian and
fit together to form a quasi-coherent module $\mathcal{G}$
on $W_{spaces, \etale}$. We will verify conditions (1) and (2)
of Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-existence}
to conclude the existence of an object of $\mathcal{C}$ lying over $W$.
Condition (1) is true: the \'etale coverings $\{W_i \to W\}$
with each $W_i$ affine are cofinal in the collection of all coverings.
For such a covering $W_i$ and $W_i \times_W W_j$ are affine
and $H^1(W_i, \mathcal{G})$ and $H^1(W_i \times_W W_j, \mathcal{G})$
are zero: the cohomology of a quasi-coherent module over an affine
algebraic space is zero for example by Cohomology of Spaces, Proposition
\ref{spaces-cohomology-proposition-vanishing}.
Finally, condition (2) is that $H^2(W, \mathcal{G}) = 0$
for our quasi-coherent sheaf $\mathcal{G}$ which again follows
from Cohomology of Spaces, Proposition
\ref{spaces-cohomology-proposition-vanishing}.
This finishes the proof.
\end{proof}
\section{Blowing up and flatness}
\label{section-blowup-flat}
\noindent
This section quickly discusses what you can deduce from
More on Morphisms of Spaces, Sections
\ref{spaces-more-morphisms-section-blowup-flat} and
\ref{spaces-more-morphisms-section-applications-flattening-by-blowing-up}
for algebraic stacks over algebraic spaces.
\begin{lemma}
\label{lemma-flatten-stack}
Let $f : \mathcal{X} \to Y$ be a morphism from an algebraic stack
to an algebraic space. Let $V \subset Y$ be an open subspace. Assume
\begin{enumerate}
\item $Y$ is quasi-compact and quasi-separated,
\item $f$ is of finite type and quasi-separated,
\item $V$ is quasi-compact, and
\item $\mathcal{X}_V$ is flat and locally of finite presentation over $V$.
\end{enumerate}
Then there exists a $V$-admissible blowup $Y' \to Y$
and a closed substack $\mathcal{X}' \subset \mathcal{X}_{Y'}$
with $\mathcal{X}'_V = \mathcal{X}_V$ such that
$\mathcal{X}' \to Y'$ is flat and of finite presentation.
\end{lemma}
\begin{proof}
Observe that $\mathcal{X}$ is quasi-compact.
Choose an affine scheme $U$ and a surjective smooth
morphism $U \to \mathcal{X}$.
Let $R = U \times_\mathcal{X} U$ so that we obtain
a groupoid $(U, R, s, t, c)$ in algebraic spaces over $Y$ with
$\mathcal{X} = [U/R]$
(Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation}).
We may apply
More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-flat-after-blowing-up}
to $U \to Y$ and the open $V \subset Y$.
Thus we obtain a $V$-admissible blowup $Y' \to Y$
such that the strict transform $U' \subset U_{Y'}$
is flat and of finite presentation over $Y'$.
Let $R' \subset R_{Y'}$ be the strict transform of $R$.
Since $s$ and $t$ are smooth (and in particular flat)
it follows from
Divisors on Spaces, Lemma
\ref{spaces-divisors-lemma-strict-transform-flat}
that we have cartesian diagrams
$$
\vcenter{
\xymatrix{
R' \ar[r] \ar[d] & R_{Y'} \ar[d]^{s_{Y'}} \\
U' \ar[r] & U_{Y'}
}
}
\quad\text{and}\quad
\vcenter{
\xymatrix{
R' \ar[r] \ar[d] & R_{Y'} \ar[d]^{t_{Y'}} \\
U' \ar[r] & U_{Y'}
}
}
$$
In other words, $U'$ is an $R_{Y'}$-invariant closed subspace
of $U_{Y'}$. Thus $U'$ defines a closed substack
$\mathcal{X}' \subset \mathcal{X}_{Y'}$ by
Properties of Stacks, Lemma
\ref{stacks-properties-lemma-substacks-presentation}.
The morphism $\mathcal{X}' \to Y'$ is flat and
locally of finite presentation because this is
true for $U' \to Y'$. On the other hand,
we already know $\mathcal{X}' \to Y'$ is quasi-compact and quasi-separated
(by our assumptions on $f$ and because this is true for closed immersions).
This finishes the proof.
\end{proof}
\section{Chow's lemma for algebraic stacks}
\label{section-chow}
\noindent
In this section we discuss Chow's lemma for algebraic stacks.
\begin{lemma}
\label{lemma-finite-cover-factor}
Let $Y$ be a quasi-compact and quasi-separated algebraic space.
Let $V \subset Y$ be a quasi-compact open. Let $f : \mathcal{X} \to V$
be surjective, flat, and locally of finite presentation.
Then there exists a finite surjective morphism $g : Y' \to Y$ such that
$V' = g^{-1}(V) \to Y$ factors Zariski locally through $f$.
\end{lemma}
\begin{proof}
We first prove this when $Y$ is a scheme.
We may choose a scheme $U$ and a surjective smooth morphism
$U \to \mathcal{X}$. Then $\{U \to V\}$ is an fppf covering of schemes.
By More on Morphisms, Lemma \ref{more-morphisms-lemma-dominate-fppf-finite}
there exists a finite surjective morphism
$V' \to V$ such that $V' \to V$ factors Zariski locally
through $U$. By
More on Morphisms, Lemma
\ref{more-morphisms-lemma-extend-finite-surjective-morphisms}
we can find a finite surjective morphism $Y' \to Y$
whose restriction to $V$ is $V' \to V$ as desired.
\medskip\noindent
If $Y$ is an algebraic space, then we see the lemma is true by
first doing a finite base change by a finite surjective morphism
$Y' \to Y$ where $Y'$ is a scheme. See
Limits of Spaces, Proposition
\ref{spaces-limits-proposition-there-is-a-scheme-finite-over}.
\end{proof}
\begin{lemma}
\label{lemma-make-section}
Let $f : \mathcal{X} \to Y$ be a morphism from an algebraic stack
to an algebraic space. Let $V \subset Y$ be an open subspace.
Assume
\begin{enumerate}
\item $f$ is separated and of finite type,
\item $Y$ is quasi-compact and quasi-separated,
\item $V$ is quasi-compact, and
\item $\mathcal{X}_V$ is a gerbe over $V$.
\end{enumerate}
Then there exists a commutative diagram
$$
\xymatrix{
\overline{Z} \ar[rd]_{\overline{g}} &
Z \ar[l]^j \ar[d]_g \ar[r]_h & \mathcal{X} \ar[ld]^f \\
& Y
}
$$
with $j$ an open immersion, $\overline{g}$ and $h$ proper,
and such that $|V|$ is contained in the image of $|g|$.
\end{lemma}
\begin{proof}
Suppose we have a commutative diagram
$$
\xymatrix{
\mathcal{X}' \ar[d]_{f'} \ar[r] & \mathcal{X} \ar[d]^f \\
Y' \ar[r] & Y
}
$$
and a quasi-compact open $V' \subset Y'$, such that
$Y' \to Y$ is a proper morphism of algebraic spaces,
$\mathcal{X}' \to \mathcal{X}$ is a proper morphism of algebraic stacks,
$V' \subset Y'$ maps surjectively onto $V$, and
$\mathcal{X}'_{V'}$ is a gerbe over $V'$.
Then it suffices to prove the lemma for the pair
$(f' : \mathcal{X}' \to Y', V')$. Some details omitted.
\medskip\noindent
Overall strategy of the proof. We will reduce
to the case where the image of $f$ is open and $f$
has a section over this open by repeatedly applying the
above remark. Each step is straightforward, but there are
quite a few of them which makes the proof a bit involved.
\medskip\noindent
Using Limits of Spaces, Proposition
\ref{spaces-limits-proposition-there-is-a-scheme-finite-over}
we reduce to the case where $Y$ is a scheme.
(Let $Y' \to Y$ be a finite surjective morphism where $Y'$ is
a scheme. Set $\mathcal{X}' = \mathcal{X}_{Y'}$ and apply
the initial remark of the proof.)
\medskip\noindent
Using Lemma \ref{lemma-flatten-stack}
(and Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-fppf}
to see that a gerbe is flat and locally of finite presentation)
we reduce to the case where $f$ is flat and of finite presentation.
\medskip\noindent
Since $f$ is flat and locally of finite
presentation, we see that the image of $|f|$ is an open $W \subset Y$.
Since $\mathcal{X}$ is quasi-compact (as $f$ is of finite type
and $Y$ is quasi-compact) we see that $W$ is quasi-compact.
By Lemma \ref{lemma-finite-cover-factor}
we can find a finite surjective morphism $g : Y' \to Y$
such that $g^{-1}(W) \to Y$ factors Zariski locally
through $\mathcal{X} \to Y$.
After replacing $Y$ by $Y'$ and $\mathcal{X}$ by
$\mathcal{X} \times_Y Y'$ we reduce to the situation
described in the next paragraph.
\medskip\noindent
Assume there exists $n \geq 0$, quasi-compact opens
$W_i \subset Y$, $i = 1, \ldots, n$, and
morphisms $x_i : W_i \to \mathcal{X}$ such that
(a) $f \circ x_i = \text{id}_{W_i}$,
(b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$, and
(c) $W$ is the image of $|f|$.
We will use induction on $n$. The base case is $n = 0$: this
implies $V = \emptyset$ and in this case we can take
$\overline{Z} = \emptyset$.
If $n > 0$, then for $i = 1, \ldots, n$
consider the reduced closed subschemes
$Y_i$ with underlying topological space
$Y \setminus W_i$. Consider the finite morphism
$$
Y' = Y \amalg \coprod\nolimits_{i = 1, \ldots, n} Y_i \longrightarrow Y
$$
and the quasi-compact open
$$
V' = (W_1 \cap \ldots \cap W_n \cap V) \amalg
\coprod_{i = 1, \ldots, n} (V \cap Y_i).
$$
By the initial remark of the proof, if we can prove the lemma for the pairs
$$
(\mathcal{X} \to Y, W_1 \cap \ldots \cap W_n \cap V)
\quad\text{and}\quad
(\mathcal{X} \times_Y Y_i \to Y_i, V \cap Y_i),\quad
i = 1, \ldots, n
$$
then the result is true. Here we use the settheoretic equality
$V = (W_1 \cap \ldots \cap W_n \cap V) \cup
\bigcup\nolimits_{i = 1, \ldots n} (V \cap Y_i)$.
The induction hypothesis applies to the second type of
pairs above. Hence we reduce to the situation described in
the next paragraph.
\medskip\noindent
Assume there exists $n \geq 0$, quasi-compact opens
$W_i \subset Y$, $i = 1, \ldots, n$, and
morphisms $x_i : W_i \to \mathcal{X}$ such that
(a) $f \circ x_i = \text{id}_{W_i}$,
(b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$,
(c) $W$ is the image of $|f|$, and
(d) $V \subset W_1 \cap \ldots \cap W_n$.
The morphisms
$$
T_{ij} = \mathit{Isom}_\mathcal{X}(x_i|_{W_i \cap W_j \cap V},
x_j|_{W_i \cap W_j \cap V}) \longrightarrow W_i \cap W_j \cap V
$$
are surjective, flat, and locally of finite presentation
(Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-isom-fppf}).
We apply Lemma \ref{lemma-finite-cover-factor}
to each quasi-compact open $W_i \cap W_j \cap V$ and the morphisms
$T_{ij} \to W_i \cap W_j \cap V$ to get finite surjective morphisms
$Y'_{ij} \to Y$. After replacing $Y$ by the fibre product of all
of the $Y'_{ij}$ over $Y$ we reduce to the situation described
in the next paragraph.
\medskip\noindent
Assume there exists $n \geq 0$, quasi-compact opens
$W_i \subset Y$, $i = 1, \ldots, n$, and
morphisms $x_i : W_i \to \mathcal{X}$ such that
(a) $f \circ x_i = \text{id}_{W_i}$,
(b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$,
(c) $W$ is the image of $|f|$,
(d) $V \subset W_1 \cap \ldots \cap W_n$, and
(e) $x_i$ and $x_j$ are Zariski locally isomorphic over $W_i \cap W_j \cap V$.
Let $y \in V$ be arbitrary.
Suppose that we can find a quasi-compact open neighbourhood
$y \in V_y \subset V$ such that the lemma is true for
the pair $(\mathcal{X} \to Y, V_y)$, say with solution
$\overline{Z}_y, Z_y, \overline{g}_y, g_y, h_y$.
Since $V$ is quasi-compact, we can find a finite number
$y_1, \ldots, y_m$ such that $V = V_{y_1} \cup \ldots \cup V_{y_m}$.
Then it follows that setting
$$
\overline{Z} = \coprod \overline{Z}_{y_j},\quad
Z = \coprod Z_{y_j},\quad
\overline{g} = \coprod \overline{g}_{y_j},\quad
g = \coprod g_{y_j},\quad
h = \coprod h_{y_j}
$$
is a solution for the lemma. Given $y$ by condition (e)
we can choose a quasi-compact open neighbourhood $y \in V_y \subset V$
and isomorphisms $\varphi_i : x_1|_{V_y} \to x_i|_{V_y}$ for
$i = 2, \ldots, n$. Set $\varphi_{ij} = \varphi_j \circ \varphi_i^{-1}$.
This leads us to the situation described in the next paragraph.
\medskip\noindent
Assume there exists $n \geq 0$, quasi-compact opens
$W_i \subset Y$, $i = 1, \ldots, n$, and
morphisms $x_i : W_i \to \mathcal{X}$ such that
(a) $f \circ x_i = \text{id}_{W_i}$,
(b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$,
(c) $W$ is the image of $|f|$,
(d) $V \subset W_1 \cap \ldots \cap W_n$, and
(f) there are isomorphisms $\varphi_{ij} : x_i|_V \to x_j|_V$
satisfying $\varphi_{jk} \circ \varphi_{ij} = \varphi_{ik}$.
The morphisms
$$
I_{ij} = \mathit{Isom}_\mathcal{X}(x_i|_{W_i \cap W_j},
x_j|_{W_i \cap W_j}) \longrightarrow W_i \cap W_j
$$
are proper because $f$ is separated
(Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-separated-implies-isom}).
Observe that $\varphi_{ij}$ defines a section $V \to I_{ij}$
of $I_{ij} \to W_i \cap W_j$ over $V$.
By More on Morphisms of Spaces, Lemma
\ref{spaces-more-morphisms-lemma-get-section-after-blowup}
we can find $V$-admissible blowups
$p_{ij} : Y_{ij} \to Y$ such that $s_{ij}$
extends to $p_{ij}^{-1}(W_i \cap W_j)$.
After replacing $Y$ by the fibre product of all the $Y_{ij}$
over $Y$ we get to the situation described in the next paragraph.
\medskip\noindent
Assume there exists $n \geq 0$, quasi-compact opens
$W_i \subset Y$, $i = 1, \ldots, n$, and
morphisms $x_i : W_i \to \mathcal{X}$ such that
(a) $f \circ x_i = \text{id}_{W_i}$,
(b) $W = \bigcup_{i = 1, \ldots, n} W_i$ contains $V$,
(c) $W$ is the image of $|f|$,
(d) $V \subset W_1 \cap \ldots \cap W_n$, and
(g) there are isomorphisms
$\varphi_{ij} : x_i|_{W_i \cap W_j} \to x_j|_{W_i \cap W_j}$
satisfying
$$
\varphi_{jk}|_V \circ \varphi_{ij}|_V = \varphi_{ik}|_V.
$$
After replacing $Y$ by another $V$-admissible blowup if necessary
we may assume that $V$ is dense and scheme theoretically dense
in $Y$ and hence in any open subspace of $Y$ containing $V$.
After such a replacement we conclude that
$$
\varphi_{jk}|_{W_i \cap W_j \cap W_k} \circ
\varphi_{ij}|_{W_i \cap W_j \cap W_k} =
\varphi_{ik}|_{W_i \cap W_j \cap W_k}
$$
by appealing to Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-equality-of-morphisms}
and the fact that $I_{ik} \to W_i \cap W_j$ is proper
(hence separated).
Of course this means that $(x_i, \varphi_{ij})$
is a desent datum and we obtain a morphism
$x : W \to \mathcal{X}$ agreeing with $x_i$ over $W_i$
because $\mathcal{X}$ is a stack.
Since $x$ is a section of the separated morphism
$\mathcal{X} \to W$ we see that $x$ is proper
(Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-section-immersion}).
Thus the lemma now holds with $\overline{Z} = Y$,
$Z = W$, $\overline{g} = \text{id}_Y$, $g = \text{id}_W$,
$h = x$.
\end{proof}
\begin{theorem}[Chow's lemma]
\label{theorem-chow-finite-type}
\begin{reference}
This is a result due to Ofer Gabber, see
\cite[Theorem 1.1]{olsson_proper}
\end{reference}
Let $f : \mathcal{X} \to Y$ be a morphism from an algebraic stack
to an algebraic space. Assume
\begin{enumerate}
\item $Y$ is quasi-compact and quasi-separated,
\item $f$ is separated of finite type.
\end{enumerate}
Then there exists a commutative diagram
$$
\xymatrix{
\mathcal{X} \ar[rd] & X \ar[l] \ar[d] \ar[r] & \overline{X} \ar[ld] \\
& Y
}
$$
where $X \to \mathcal{X}$ is proper surjective,
$X \to \overline{X}$ is an open immersion, and
$\overline{X} \to Y$ is proper morphism of algebraic spaces.
\end{theorem}
\begin{proof}
The rough idea is to use that $\mathcal{X}$ has a dense open
which is a gerbe (Morphisms of Stacks, Proposition
\ref{stacks-morphisms-proposition-open-stratum})
and appeal to Lemma \ref{lemma-make-section}.
The reason this does not work is that the open may not be
quasi-compact and one runs into technical problems. Thus
we first do a (standard) reduction to the Noetherian case.
\medskip\noindent
First we choose a closed immersion
$\mathcal{X} \to \mathcal{X}'$ where $\mathcal{X}'$
is an algebraic stack separated and of finite type over $Y$.
See Limits of Stacks, Lemma
\ref{stacks-limits-lemma-separated-closed-in-finite-presentation}.
Clearly it suffices to prove the theorem for
$\mathcal{X}'$, hence we may assume $\mathcal{X} \to Y$
is separated and of finite presentation.
\medskip\noindent
Assume $\mathcal{X} \to Y$ is separated and of finite presentation.
By Limits of Spaces, Proposition \ref{spaces-limits-proposition-approximate}
we can write $Y = \lim Y_i$ as the directed limit of a system
of Noetherian algebraic spaces with affine transition morphisms.
By Limits of Stacks, Lemma \ref{stacks-limits-lemma-descend-a-stack-down}
there is an $i$ and a morphism $\mathcal{X}_i \to Y_i$ of finite presentation
from an algebraic stack to $Y_i$ such that
$\mathcal{X} = Y \times_{Y_i} \mathcal{X}_i$.
After increasing $i$ we may assume that $\mathcal{X}_i \to Y_i$
is separated, see Limits of Stacks, Lemma
\ref{stacks-limits-lemma-eventually-separated}.
Then it suffices to prove the theorem
for $\mathcal{X}_i \to Y_i$. This reduces us to the case discussed
in the next paragraph.
\medskip\noindent
Assume $Y$ is Noetherian. We may replace $\mathcal{X}$ by its
reduction (Properties of Stacks, Definition
\ref{stacks-properties-definition-reduced-induced-stack}).
This reduces us to the case discussed
in the next paragraph.
\medskip\noindent
Assume $Y$ is Noetherian and $\mathcal{X}$ is reduced.
Since $\mathcal{X} \to Y$ is separated and $Y$ quasi-separated,
we see that $\mathcal{X}$ is quasi-separated as an algebraic stack.
Hence the inertia $\mathcal{I}_\mathcal{X} \to \mathcal{X}$
is quasi-compact. Thus by Morphisms of Stacks, Proposition
\ref{stacks-morphisms-proposition-open-stratum}
there exists a dense open substack $\mathcal{V} \subset \mathcal{X}$
which is a gerbe. Let $\mathcal{V} \to V$ be the morphism
which expresses $\mathcal{V}$ as a gerbe over the algebraic space $V$.
See
Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-over-iso-classes}
for a construction of $\mathcal{V} \to V$.
This construction in particular shows that the morphism
$\mathcal{V} \to Y$ factors as $\mathcal{V} \to V \to Y$.
Picture
$$
\xymatrix{
\mathcal{V} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\
V \ar[r] & Y
}
$$
Since the morphism $\mathcal{V} \to V$ is surjective, flat, and
of finite presentation
(Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-gerbe-fppf})
and since $\mathcal{V} \to Y$ is locally of finite presentation,
it follows that $V \to Y$ is locally of finite presentation
(Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-flat-finite-presentation-permanence}).
Note that $\mathcal{V} \to V$ is a universal homeomorphism
(Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-gerbe-bijection-points}).
Since $\mathcal{V}$ is quasi-compact (see
Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-locally-closed-in-noetherian})
we see that $V$ is quasi-compact.
Finally, since $\mathcal{V} \to Y$ is separated the same is true
for $V \to Y$ by
Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-check-separated-on-ui-cover}
applied to $\mathcal{V} \to V \to Y$
(whose assumptions are satisfied as we've already seen).
\medskip\noindent
All of the above means that the assumptions of
Limits of Spaces, Lemma \ref{spaces-limits-lemma-embedding-into-affine-over-qs}
apply to the morphism $V \to Y$. Thus we can find a dense open
subspace $V' \subset V$ and an immersion $V' \to \mathbf{P}^n_Y$
over $Y$. Clearly we may replace $V$ by $V'$ and $\mathcal{V}$
by the inverse image of $V'$ in $\mathcal{V}$ (recall that
$|\mathcal{V}| = |V|$ as we've seen above).
Thus we may assume we have a diagram
$$
\xymatrix{
\mathcal{V} \ar[rr] \ar[d] & & \mathcal{X} \ar[d] \\
V \ar[r] & \mathbf{P}^n_Y \ar[r] & Y
}
$$
where the arrow $V \to \mathbf{P}^n_Y$ is an immersion.
Let $\mathcal{X}'$ be the scheme theoretic image of the morphism
$$
j : \mathcal{V} \longrightarrow \mathbf{P}^n_Y \times_Y \mathcal{X}
$$
and let $Y'$ be the scheme theoretic image of the morphism
$V \to \mathbf{P}^n_Y$. We obtain a commutative diagram
$$
\xymatrix{
\mathcal{V} \ar[r] \ar[d] &
\mathcal{X}' \ar[r] \ar[d] &
\mathbf{P}^n_Y \times_Y \mathcal{X} \ar[d] \ar[r] &
\mathcal{X} \ar[d] \\
V \ar[r] &
Y' \ar[r] &
\mathbf{P}^n_Y \ar[r] &
Y
}
$$
(See Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-factor-factor}).
We claim that $\mathcal{V} = V \times_{Y'} \mathcal{X}'$ and that
Lemma \ref{lemma-make-section} applies to the morphism $\mathcal{X}' \to Y'$
and the open subspace $V \subset Y'$. If the claim is true, then we obtain
$$
\xymatrix{
\overline{X} \ar[rd]_{\overline{g}} &
X \ar[l] \ar[d]_g \ar[r]_h & \mathcal{X}' \ar[ld]^f \\
& Y'
}
$$
with $X \to \overline{X}$ an open immersion, $\overline{g}$ and $h$ proper,
and such that $|V|$ is contained in the image of $|g|$.
Then the composition $X \to \mathcal{X}' \to \mathcal{X}$ is
proper (as a composition of proper morphisms) and its image
contains $|\mathcal{V}|$, hence this composition is surjective.
As well, $\overline{X} \to Y' \to Y$ is proper as a composition
of proper morphisms.
\medskip\noindent
The last step is to prove the claim.
Observe that $\mathcal{X}' \to Y'$ is separated and of finite type,
that $Y'$ is quasi-compact and quasi-separated, and that $V$ is quasi-compact
(we omit checking all the details completely).
Next, we observe that
$b : \mathcal{X}' \to \mathcal{X}$ is an isomorphism over
$\mathcal{V}$ by Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-scheme-theoretic-image-of-partial-section}.
In particular $\mathcal{V}$ is identified with an open substack
of $\mathcal{X}'$.
The morphism $j$ is quasi-compact
(source is quasi-compact and target is quasi-separated), so formation
of the scheme theoretic image of $j$ commutes with flat base change by
Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-existence-plus-flat-base-change}.
In particular we see that $V \times_{Y'} \mathcal{X}'$ is the
scheme theoretic image of
$\mathcal{V} \to V \times_{Y'} \mathcal{X}'$.
However, by Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-universally-closed-permanence}
the image of $|\mathcal{V}| \to |V \times_{Y'} \mathcal{X}'|$
is closed (use that $\mathcal{V} \to V$ is a universal homeomorphism
as we've seen above and hence is universally closed).
Also the image is dense (combine what we just said with
Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-topology-scheme-theoretic-image})
we conclude $|\mathcal{V}| = |V \times_{Y'} \mathcal{X}'|$.
Thus $\mathcal{V} \to V \times_{Y'} \mathcal{X}'$ is an
isomorphism and the proof of the claim is complete.
\end{proof}
\section{Noetherian valuative criterion}
\label{section-Noetherian-valuative-criterion}
\noindent
In this section we will discuss (refined) valuative criteria for
morphisms of algebraic stacks using only discrete valuation rings
in the Noetherian setting. There are many different variants
and we will add more here over time as needed.
\medskip\noindent
Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks
(or algebraic spaces or schemes). A {\it refined} valuative criterion
is one where we are given a morphism $\mathcal{U} \to \mathcal{X}$
(with some properties) and we only look at existence or uniqueness
of dotted arrows in solid diagrams of the form
$$
\xymatrix{
\Spec(K) \ar[d] \ar[r] & \mathcal{U} \ar[r] & \mathcal{X} \ar[d] \\
\Spec(A) \ar[rr] \ar@{..>}[rru] & & \mathcal{Y}
}
$$
We use this terminology below to describe the results we have obtained sofar.
\medskip\noindent
Non-Noetherian valuative criteria for morphisms of algebraic stacks
\begin{enumerate}
\item Morphisms of Stacks, Section
\ref{stacks-morphisms-section-valuative-second}
(for separatedness of the diagonal),
\item Morphisms of Stacks, Section
\ref{stacks-morphisms-section-valuative-diagonal}
(for separatedness),
\item Morphisms of Stacks, Section
\ref{stacks-morphisms-section-valutive-criterion}
(for universal closedness),
\item Morphisms of Stacks, Section
\ref{stacks-morphisms-section-valutive-criterion-properness}
(for properness).
\end{enumerate}
For algebraic spaces we have the following valuative criteria
\begin{enumerate}
\item Morphisms of Spaces, Section
\ref{spaces-morphisms-section-valuative-criterion-universally-closed}
(for universal closedness),
\item Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-refined-valuative-criterion-universally-closed}
(refined for universal closedness)
\item Morphisms of Spaces, Section
\ref{spaces-morphisms-section-valuative-separatedness}
(for separatedness),
\item Morphisms of Spaces, Section
\ref{spaces-morphisms-section-valuative-criterion-properness}
(for properness),
\item Decent Spaces, Section
\ref{decent-spaces-section-valuative-criterion-universally-closed}
(for universal closedness for decent spaces),
\item Decent Spaces, Lemma
\ref{decent-spaces-lemma-re-characterize-universally-closed}
(for universal closedness for decent morphisms between algebraic spaces),
\item Cohomology of Spaces, Section
\ref{spaces-cohomology-section-Noetherian-valuative-criterion}
contains Noetherian valuative criteria
\begin{enumerate}
\item Cohomology of Spaces, Lemma
\ref{spaces-cohomology-lemma-check-separated-dvr}
(for separatedness using discrete valuation rings),
\item Cohomology of Spaces, Lemma
\ref{spaces-cohomology-lemma-check-proper-dvr}
(for properness using discrete valuation rings),
\item Cohomology of Spaces, Remark
\ref{spaces-cohomology-remark-variant}
(discusses how to reduce to complete discrete valuation rings),
\end{enumerate}
\item Limits of Spaces, Section
\ref{spaces-limits-section-Noetherian-valuative-criterion}
discussing Noetherian valuative criteria
\begin{enumerate}
\item Limits of Spaces, Lemma
\ref{spaces-limits-lemma-refined-valuative-criterion-proper}
(refined for properness using discrete valuation rings),
\item Limits of Spaces, Lemma
\ref{spaces-limits-lemma-refined-valuative-criterion-separated}
(refined for separatedness using discrete valuation rings),
\item Limits of Spaces, Lemma
\ref{spaces-limits-lemma-refined-valuative-criterion-universally-closed}
(refined for universal closedness using discrete valuation rings).
\end{enumerate}
\end{enumerate}
For schemes we have the following valuative criteria
\begin{enumerate}
\item Schemes, Section
\ref{schemes-section-valuative-criterion-universal-closedness}
(for universal closedness)
\item Schemes, Section \ref{schemes-section-valuative-separatedness}
(for separatedness),
\item Morphisms, Section \ref{morphisms-section-valuative-criteria}
(for properness)
\item Morphisms, Lemma
\ref{morphisms-lemma-refined-valuative-criterion-universally-closed}
(refined for universal closedness),
\item Limits, Section \ref{limits-section-Noetherian-valuative-criterion}
discussing Noetherian valuative criteria
\begin{enumerate}
\item Limits, Lemma \ref{limits-lemma-Noetherian-dvr-valuative-separation}
(for separatedness using discrete valuation rings and generic points)
\item Limits, Lemma \ref{limits-lemma-Noetherian-dvr-valuative-proper}
(for properness using discrete valuation rings and generic points)
\item Limits, Lemma \ref{limits-lemma-check-universally-closed-Noetherian}
(for universal closedness using discrete valuation rings),
\item Limits, Lemma \ref{limits-lemma-refined-valuative-criterion-proper}
(refined for properness using discrete valuation rings),
\item Limits, Lemma \ref{limits-lemma-refined-valuative-criterion-separated}
(refined for separatedness using discrete valuation rings),
\item Limits, Lemma
\ref{limits-lemma-refined-valuative-criterion-universally-closed}
(refined for universal closedness using discrete valuation rings).
\end{enumerate}
\end{enumerate}
This ends our list of previous results.
\begin{lemma}
\label{lemma-refined-valuative-criterion-proper}
Let $f : \mathcal{X} \to \mathcal{Y}$ and $h : \mathcal{U} \to \mathcal{X}$
be morphisms of algebraic stacks. Assume that $\mathcal{Y}$ is
locally Noetherian, that $f$ and $h$ are of finite type,
that $f$ is separated, and that the image of
$|h| : |\mathcal{U}| \to |\mathcal{X}|$ is dense in $|\mathcal{X}|$.
If given any $2$-commutative diagram
$$
\xymatrix{
\Spec(K) \ar[r]_-u \ar[d]_j & \mathcal{U} \ar[r]_h & \mathcal{X} \ar[d]^f \\
\Spec(A) \ar[rr]^-y & & \mathcal{Y}
}
$$
where $A$ is a discrete valuation ring with field of fractions $K$
and $\gamma : y \circ j \to f \circ h \circ u$ there
exist an extension $K'/K$ of fields, a valuation ring $A' \subset K'$
dominating $A$ such that the category of dotted arrows for the
induced diagram
$$
\xymatrix{
\Spec(K') \ar[r]_-{x'} \ar[d]_{j'} & \mathcal{X} \ar[d]^f \\
\Spec(A') \ar[r]^-{y'} \ar@{..>}[ru] & \mathcal{Y}
}
$$
with induced $2$-arrow $\gamma' : y' \circ j' \to f \circ x'$
is nonempty (Morphisms of Stacks, Definition
\ref{stacks-morphisms-definition-fill-in-diagram}), then $f$ is proper.
\end{lemma}
\begin{proof}
It suffices to prove that $f$ is universally closed.
Let $V \to \mathcal{Y}$ be a smooth morphism where $V$ is an affine scheme.
By Properties of Stacks, Lemma
\ref{stacks-properties-lemma-points-cartesian}
the image $I$ of
$|\mathcal{U} \times_\mathcal{Y} V| \to |\mathcal{X} \times_\mathcal{Y} V|$
is the inverse image of the image of $|h|$. Since
$|\mathcal{X} \times_\mathcal{Y} V| \to |\mathcal{X}|$ is open
(Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-fppf-open})
we conclude that $I$ is dense in $|\mathcal{X} \times_\mathcal{Y} V|$.
Also since the category of dotted arrows behaves well with respect
to base change (Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-cat-dotted-arrows-base-change})
the assumption on existence of dotted arrows (after extension)
is inherited by the morphisms
$\mathcal{U} \times_\mathcal{Y} V \to \mathcal{X} \times_\mathcal{Y} V \to V$.
Therefore the assumptions of the lemma are
satisfied for the morphisms
$\mathcal{U} \times_\mathcal{Y} V \to \mathcal{X} \times_\mathcal{Y} V \to V$.
Hence we may assume $\mathcal{Y}$ is an affine scheme.
\medskip\noindent
Assume $\mathcal{Y} = Y$ is an affine scheme.
(From now on we no longer have to worry about the
$2$-arrows $\gamma$ and $\gamma'$, see Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-cat-dotted-arrows-independent}.)
Then $\mathcal{U}$ is quasi-compact. Choose an affine scheme $U$ and a
surjective smooth morphism $U \to \mathcal{U}$.
Then we may and do replace $\mathcal{U}$ by $U$.
Thus we may assume that $\mathcal{U}$ is an affine scheme.
\medskip\noindent
Assume $\mathcal{Y} = Y$ and $\mathcal{U} = U$ are affine schemes.
By Chow's lemma (Theorem \ref{theorem-chow-finite-type})
we can choose a surjective proper morphism $X \to \mathcal{X}$
where $X$ is an algebraic space. We will use below that $X \to Y$ is separated
as a composition of separated morphisms. Consider the
algebraic space $W = X \times_\mathcal{X} U$. The projection morphism
$W \to X$ is of finite type.
We may replace $X$ by the scheme theoretic image of
$W \to X$ and hence we may assume that the image of $|W|$ in $|X|$
is dense in $|X|$ (here we use that the image of $|h|$
is dense in $|\mathcal{X}|$, so after this replacement, the
morphism $X \to \mathcal{X}$ is still surjective).
We claim that for every solid commutative diagram
$$
\xymatrix{
\Spec(K) \ar[r] \ar[d] & W \ar[r] & X \ar[d] \\
\Spec(A) \ar[rr] \ar@{..>}[rru] & & Y
}
$$
where $A$ is a discrete valuation ring with field of fractions $K$, there
exists a dotted arrow making the diagram commute. First, it is enough to
prove there exists a dotted arrow after replacing $K$ by an extension
and $A$ by a valuation ring in this extension dominating $A$, see
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-usual-enough}.
By the assumption of the lemma we get an extension $K'/K$ and a valuation
ring $A' \subset K'$ dominating $A$ and
an arrow $\Spec(A') \to \mathcal{X}$ lifting the composition
$\Spec(A') \to \Spec(A) \to Y$ and compatible with the composition
$\Spec(K') \to \Spec(K) \to W \to X$. Because $X \to \mathcal{X}$
is proper, we can use the valuative criterion of properness
(Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-criterion-proper})
to find an extension $K''/K'$ and a valuation ring $A'' \subset K''$
dominating $A'$ and a morphism $\Spec(A'') \to X$ lifting the composition
$\Spec(A'') \to \Spec(A') \to \mathcal{X}$ and compatible with the composition
$\Spec(K'') \to \Spec(K') \to \Spec(K) \to X$.
Then $K''/K$ and $A'' \subset K''$ and the morphism $\Spec(A'') \to X$
is a solution to the problem posed above and the claim holds.
\medskip\noindent
The claim implies the morphism $X \to Y$ is proper by the
case of the lemma for algebraic spaces
(Limits of Spaces, Lemma
\ref{spaces-limits-lemma-refined-valuative-criterion-proper}).
By Morphisms of Stacks, Lemma
\ref{stacks-morphisms-lemma-image-proper-is-proper}
we conclude that $\mathcal{X} \to Y$ is proper as desired.
\end{proof}
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