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\input{preamble}
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\begin{document}
\title{Morphisms of Algebraic Spaces}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we introduce some types of morphisms of algebraic spaces.
A reference is \cite{Kn}.
\medskip\noindent
The goal is to extend the definition of each of the types of morphisms of
schemes defined in the chapters on schemes, and on morphisms of schemes
to the category of algebraic spaces. Each case is slightly different and
it seems best to treat them all separately.
\section{Conventions}
\label{section-conventions}
\noindent
The standing assumption is that all schemes are contained in
a big fppf site $\Sch_{fppf}$. And all rings $A$ considered
have the property that $\Spec(A)$ is (isomorphic) to an
object of this big site.
\medskip\noindent
Let $S$ be a scheme and let $X$ be an algebraic space over $S$.
In this chapter and the following we will write $X \times_S X$
for the product of $X$ with itself (in the category of algebraic
spaces over $S$), instead of $X \times X$.
\section{Properties of representable morphisms}
\label{section-representable}
\noindent
Let $S$ be a scheme.
Let $f : X \to Y$ be a representable morphism of algebraic spaces. In
Spaces, Section \ref{spaces-section-representable-properties}
we defined what it means for $f$ to
have property $\mathcal{P}$ in case $\mathcal{P}$ is a property
of morphisms of schemes which
\begin{enumerate}
\item is preserved under any base change,
see Schemes, Definition \ref{schemes-definition-preserved-by-base-change},
and
\item is fppf local on the base, see
Descent, Definition \ref{descent-definition-property-morphisms-local}.
\end{enumerate}
Namely, in this case we say $f$ has property $\mathcal{P}$ if and only
if for every scheme $U$ and any morphism $U \to Y$ the morphism of schemes
$X \times_Y U \to U$ has property $\mathcal{P}$.
\medskip\noindent
According to the lists in
Spaces, Section \ref{spaces-section-lists}
this applies to the following properties:
(1)(a) closed immersions,
(1)(b) open immersions,
(1)(c) quasi-compact immersions,
(2) quasi-compact,
(3) universally-closed,
(4) (quasi-)separated,
(5) monomorphism,
(6) surjective,
(7) universally injective,
(8) affine,
(9) quasi-affine,
(10) (locally) of finite type,
(11) (locally) quasi-finite,
(12) (locally) of finite presentation,
(13) locally of finite type of relative dimension $d$,
(14) universally open,
(15) flat,
(16) syntomic,
(17) smooth,
(18) unramified (resp.\ G-unramified),
(19) \'etale,
(20) proper,
(21) finite or integral,
(22) finite locally free,
(23) universally submersive,
(24) universal homeomorphism, and
(25) immersion.
\medskip\noindent
In this chapter we will redefine these notions for not necessarily
representable morphisms of algebraic spaces. Whenever we do this we will make
sure that the new definition agrees with the old one, in order to avoid
ambiguity.
\medskip\noindent
Note that the definition above applies whenever $X$ is a scheme,
since a morphism from a scheme to an algebraic space is representable.
And in particular it applies when both $X$ and $Y$ are schemes.
In
Spaces, Lemma
\ref{spaces-lemma-morphism-schemes-gives-representable-transformation-property}
we have seen that in this case the definitions
match, and no ambiguity arise.
\medskip\noindent
Furthermore, in
Spaces, Lemma
\ref{spaces-lemma-base-change-representable-transformations-property}
we have seen that the property of
representable morphisms of algebraic spaces so defined is stable under
arbitrary base change by a morphism of algebraic spaces.
And finally, in
Spaces, Lemmas
\ref{spaces-lemma-composition-representable-transformations-property} and
\ref{spaces-lemma-product-representable-transformations-property}
we have seen that if $\mathcal{P}$ is stable under compositions,
which holds for the properties
(1)(a), (1)(b), (1)(c), (2) -- (25), except (13) above, then
taking products of representable morphisms preserves property $\mathcal{P}$
and compositions of representable morphisms preserves property $\mathcal{P}$.
\medskip\noindent
We will use these facts below, and whenever we do we will simply refer
to this section as a reference.
\section{Separation axioms}
\label{section-separation-axioms}
\noindent
It makes sense to list some a priori properties of the diagonal of
a morphism of algebraic spaces.
\begin{lemma}
\label{lemma-properties-diagonal}
Let $S$ be a scheme contained in $\Sch_{fppf}$.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $\Delta_{X/Y} : X \to X \times_Y X$ be the diagonal morphism.
Then
\begin{enumerate}
\item $\Delta_{X/Y}$ is representable,
\item $\Delta_{X/Y}$ is locally of finite type,
\item $\Delta_{X/Y}$ is a monomorphism,
\item $\Delta_{X/Y}$ is separated, and
\item $\Delta_{X/Y}$ is locally quasi-finite.
\end{enumerate}
\end{lemma}
\begin{proof}
We are going to use the fact that $\Delta_{X/S}$ is
representable (by definition of an algebraic space) and that
it satisfies properties (2) -- (5), see
Spaces, Lemma \ref{spaces-lemma-properties-diagonal}.
Note that we have a factorization
$$
X
\longrightarrow
X \times_Y X
\longrightarrow
X \times_S X
$$
of the diagonal $\Delta_{X/S} : X \to X \times_S X$. Since
$X \times_Y X \to X \times_S X$ is a monomorphism, and since
$\Delta_{X/S}$ is representable, it follows formally that
$\Delta_{X/Y}$ is representable. In particular, the rest of
the statements now make sense, see
Section \ref{section-representable}.
\medskip\noindent
Choose a surjective \'etale morphism $U \to X$, with $U$ a scheme.
Consider the diagram
$$
\xymatrix{
R = U \times_X U \ar[r] \ar[d] &
U \times_Y U \ar[d] \ar[r] &
U \times_S U \ar[d] \\
X \ar[r] & X \times_Y X \ar[r] & X \times_S X
}
$$
Both squares are cartesian, hence so is the outer rectangle.
The top row consists of schemes, and the vertical arrows
are surjective \'etale morphisms. By
Spaces, Lemma \ref{spaces-lemma-representable-morphisms-spaces-property}
the properties (2) -- (5) for $\Delta_{X/Y}$ are equivalent to those of
$R \to U \times_Y U$. In the proof of
Spaces, Lemma \ref{spaces-lemma-properties-diagonal}
we have seen that $R \to U \times_S U$ has properties (2) -- (5).
The morphism $U \times_Y U \to U \times_S U$ is a monomorphism
of schemes. These facts imply that $R \to U \times_Y U$ have
properties (2) -- (5).
\medskip\noindent
Namely: For (3), note that $R \to U \times_Y U$
is a monomorphism as the composition
$R \to U \times_S U$ is a monomorphism. For (2), note that
$R \to U \times_Y U$ is locally of finite type, as the
composition $R \to U \times_S U$ is locally of finite type
(Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}).
A monomorphism which is locally of finite type is locally quasi-finite
because it has finite fibres
(Morphisms, Lemma \ref{morphisms-lemma-finite-fibre}), hence (5).
A monomorphism is separated
(Schemes, Lemma \ref{schemes-lemma-monomorphism-separated}), hence (4).
\end{proof}
\begin{definition}
\label{definition-separated}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $\Delta_{X/Y} : X \to X \times_Y X$ be the diagonal morphism.
\begin{enumerate}
\item We say $f$ is {\it separated} if $\Delta_{X/Y}$ is a closed immersion.
\item We say $f$ is {\it locally separated}\footnote{In the literature
this term often refers to quasi-separated and locally separated morphisms.}
if $\Delta_{X/Y}$ is an immersion.
\item We say $f$ is {\it quasi-separated} if $\Delta_{X/Y}$ is quasi-compact.
\end{enumerate}
\end{definition}
\noindent
This definition makes sense since $\Delta_{X/Y}$ is representable,
and hence we know what it means for it to have one of the properties
described in the definition. We will see below
(Lemma \ref{lemma-match-separated}) that this definition matches the ones
we already have for morphisms of schemes and representable morphisms.
\begin{lemma}
\label{lemma-trivial-implications}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. If $f$ is separated, then $f$ is locally separated and
$f$ is quasi-separated.
\end{lemma}
\begin{proof}
This is true, via the general principle
Spaces,
Lemma \ref{spaces-lemma-representable-transformations-property-implication},
because a closed immersion of schemes is an immersion and is quasi-compact.
\end{proof}
\begin{lemma}
\label{lemma-base-change-separated}
All of the separation axioms listed in Definition \ref{definition-separated}
are stable under base change.
\end{lemma}
\begin{proof}
Let $f : X \to Y$ and $Y' \to Y$ be morphisms of algebraic spaces.
Let $f' : X' \to Y'$ be the base change of $f$ by $Y' \to Y$. Then
$\Delta_{X'/Y'}$ is the base change of $\Delta_{X/Y}$ by
the morphism $X' \times_{Y'} X' \to X \times_Y X$. By the results of
Section \ref{section-representable}
each of the properties of the diagonal used in
Definition \ref{definition-separated}
is stable under base change. Hence the lemma is true.
\end{proof}
\begin{lemma}
\label{lemma-fibre-product-after-map}
\begin{slogan}
The top arrow of a ``magic diagram'' of algebraic spaces has nice
immersion-like properties, and under separatedness hypotheses
these get stronger.
\end{slogan}
Let $S$ be a scheme. Let $f : X \to Z$, $g : Y \to Z$ and $Z \to T$
be morphisms of algebraic spaces over $S$. Consider the induced morphism
$i : X \times_Z Y \to X \times_T Y$. Then
\begin{enumerate}
\item $i$ is representable, locally of finite type, locally quasi-finite,
separated and a monomorphism,
\item if $Z \to T$ is locally separated, then $i$ is an immersion,
\item if $Z \to T$ is separated, then $i$ is a closed immersion, and
\item if $Z \to T$ is quasi-separated, then $i$ is quasi-compact.
\end{enumerate}
\end{lemma}
\begin{proof}
By general category theory the following diagram
$$
\xymatrix{
X \times_Z Y \ar[r]_i \ar[d] & X \times_T Y \ar[d] \\
Z \ar[r]^-{\Delta_{Z/T}} \ar[r] & Z \times_T Z
}
$$
is a fibre product diagram. Hence $i$ is the base change of the
diagonal morphism $\Delta_{Z/T}$. Thus the lemma follows
from Lemma \ref{lemma-properties-diagonal}, and the material in
Section \ref{section-representable}.
\end{proof}
\begin{lemma}
\label{lemma-semi-diagonal}
\begin{slogan}
Properties of the graph of a morphism of algebraic spaces
as a consequence of separation properties of the target.
\end{slogan}
Let $S$ be a scheme. Let $T$ be an algebraic space over $S$.
Let $g : X \to Y$ be a morphism of algebraic spaces over $T$.
Consider the graph $i : X \to X \times_T Y$ of $g$. Then
\begin{enumerate}
\item $i$ is representable, locally of finite type, locally quasi-finite,
separated and a monomorphism,
\item if $Y \to T$ is locally separated, then $i$ is an immersion,
\item if $Y \to T$ is separated, then $i$ is a closed immersion, and
\item if $Y \to T$ is quasi-separated, then $i$ is quasi-compact.
\end{enumerate}
\end{lemma}
\begin{proof}
This is a special case of Lemma \ref{lemma-fibre-product-after-map}
applied to the morphism $X = X \times_Y Y \to X \times_T Y$.
\end{proof}
\begin{lemma}
\label{lemma-section-immersion}
Let $S$ be a scheme.
Let $f : X \to T$ be a morphism of algebraic spaces over $S$.
Let $s : T \to X$ be a section of $f$ (in a formula
$f \circ s = \text{id}_T$). Then
\begin{enumerate}
\item $s$ is representable, locally of finite type, locally quasi-finite,
separated and a monomorphism,
\item if $f$ is locally separated, then $s$ is an immersion,
\item if $f$ is separated, then $s$ is a closed immersion, and
\item if $f$ is quasi-separated, then $s$ is quasi-compact.
\end{enumerate}
\end{lemma}
\begin{proof}
This is a special case of Lemma \ref{lemma-semi-diagonal} applied to
$g = s$ so the morphism $i = s : T \to T \times_T X$.
\end{proof}
\begin{lemma}
\label{lemma-composition-separated}
All of the separation axioms listed in Definition \ref{definition-separated}
are stable under composition of morphisms.
\end{lemma}
\begin{proof}
Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces
to which the axiom in question applies.
The diagonal $\Delta_{X/Z}$ is the composition
$$
X \longrightarrow X \times_Y X \longrightarrow X \times_Z X.
$$
Our separation axiom is defined by requiring the diagonal
to have some property $\mathcal{P}$. By
Lemma \ref{lemma-fibre-product-after-map} above we see that
the second arrow also has this property. Hence the lemma follows
since the composition of (representable) morphisms with property
$\mathcal{P}$ also is a morphism with property $\mathcal{P}$, see
Section \ref{section-representable}.
\end{proof}
\begin{lemma}
\label{lemma-separated-over-separated}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
\begin{enumerate}
\item If $Y$ is separated and $f$ is separated, then $X$ is separated.
\item If $Y$ is quasi-separated and $f$ is quasi-separated, then
$X$ is quasi-separated.
\item If $Y$ is locally separated and $f$ is locally separated, then
$X$ is locally separated.
\item If $Y$ is separated over $S$ and $f$ is separated, then
$X$ is separated over $S$.
\item If $Y$ is quasi-separated over $S$ and $f$ is quasi-separated, then
$X$ is quasi-separated over $S$.
\item If $Y$ is locally separated over $S$ and $f$ is locally separated, then
$X$ is locally separated over $S$.
\end{enumerate}
\end{lemma}
\begin{proof}
Parts (4), (5), and (6) follow immediately from
Lemma \ref{lemma-composition-separated}
and
Spaces, Definition \ref{spaces-definition-separated}.
Parts (1), (2), and (3) reduce to parts (4), (5), and (6) by thinking
of $X$ and $Y$ as algebraic spaces over $\Spec(\mathbf{Z})$, see
Properties of Spaces, Definition \ref{spaces-properties-definition-separated}.
\end{proof}
\begin{lemma}
\label{lemma-compose-after-separated}
Let $S$ be a scheme.
Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$.
\begin{enumerate}
\item If $g \circ f$ is separated then so is $f$.
\item If $g \circ f$ is locally separated then so is $f$.
\item If $g \circ f$ is quasi-separated then so is $f$.
\end{enumerate}
\end{lemma}
\begin{proof}
Consider the factorization
$$
X \to X \times_Y X \to X \times_Z X
$$
of the diagonal morphism of $g \circ f$. In any case the last morphism
is a monomorphism. Hence for any scheme $T$ and morphism
$T \to X \times_Y X$ we have the equality
$$
X \times_{(X \times_Y X)} T = X \times_{(X \times_Z X)} T.
$$
Hence the result is clear.
\end{proof}
\begin{lemma}
\label{lemma-separated-implies-morphism-separated}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
\begin{enumerate}
\item If $X$ is separated then $X$ is separated over $S$.
\item If $X$ is locally separated then $X$ is locally separated over $S$.
\item If $X$ is quasi-separated then $X$ is quasi-separated over $S$.
\end{enumerate}
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
\begin{enumerate}
\item[(4)] If $X$ is separated over $S$ then $f$ is separated.
\item[(5)] If $X$ is locally separated over $S$ then $f$ is locally separated.
\item[(6)] If $X$ is quasi-separated over $S$ then $f$ is quasi-separated.
\end{enumerate}
\end{lemma}
\begin{proof}
Parts (4), (5), and (6) follow immediately from
Lemma \ref{lemma-compose-after-separated}
and
Spaces, Definition \ref{spaces-definition-separated}.
Parts (1), (2), and (3) follow from parts (4), (5), and (6) by
thinking of $X$ and $Y$ as algebraic spaces over
$\Spec(\mathbf{Z})$, see
Properties of Spaces, Definition \ref{spaces-properties-definition-separated}.
\end{proof}
\begin{lemma}
\label{lemma-separated-local}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $\mathcal{P}$ be any of the separation
axioms of Definition \ref{definition-separated}.
The following are equivalent
\begin{enumerate}
\item $f$ is $\mathcal{P}$,
\item for every scheme $Z$ and morphism $Z \to Y$ the
base change $Z \times_Y X \to Z$ of $f$ is $\mathcal{P}$,
\item for every affine scheme $Z$ and every morphism $Z \to Y$ the
base change $Z \times_Y X \to Z$ of $f$ is $\mathcal{P}$,
\item for every affine scheme $Z$ and every morphism $Z \to Y$ the
algebraic space $Z \times_Y X$ is $\mathcal{P}$ (see
Properties of Spaces, Definition \ref{spaces-properties-definition-separated}),
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that the base change $V \times_Y X \to V$ has
$\mathcal{P}$, and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that each
of the morphisms $f^{-1}(Y_i) \to Y_i$ has $\mathcal{P}$.
\end{enumerate}
\end{lemma}
\begin{proof}
We will repeatedly use
Lemma \ref{lemma-base-change-separated}
without further mention. In particular, it is clear that
(1) implies (2) and (2) implies (3).
\medskip\noindent
Let us prove that (3) and (4) are equivalent. Note that if $Z$ is an affine
scheme, then the morphism $Z \to \Spec(\mathbf{Z})$ is a separated
morphism as a morphism of algebraic spaces over $\Spec(\mathbf{Z})$.
If $Z \times_Y X \to Z$ is $\mathcal{P}$, then
$Z \times_Y X \to \Spec(\mathbf{Z})$ is $\mathcal{P}$
as a composition (see
Lemma \ref{lemma-composition-separated}). Hence the algebraic
space $Z \times_Y X$ is $\mathcal{P}$. Conversely, if the algebraic
space $Z \times_Y X$ is $\mathcal{P}$, then
$Z \times_Y X \to \Spec(\mathbf{Z})$ is $\mathcal{P}$, and
hence by
Lemma \ref{lemma-compose-after-separated}
we see that $Z \times_Y X \to Z$ is $\mathcal{P}$.
\medskip\noindent
Let us prove that (3) implies (5). Assume (3). Let $V$ be a scheme
and let $V \to Y$ be \'etale surjective. We have to show that
$V \times_Y X \to V$ has property $\mathcal{P}$. In other words,
we have to show that the morphism
$$
V \times_Y X \longrightarrow
(V \times_Y X) \times_V (V \times_Y X) = V \times_Y X \times_Y X
$$
has the corresponding property (i.e., is a closed immersion, immersion,
or quasi-compact). Let $V = \bigcup V_j$ be an
affine open covering of $V$. By assumption we know that each of the morphisms
$$
V_j \times_Y X \longrightarrow V_j \times_Y X \times_Y X
$$
does have the corresponding property. Since being a closed immersion,
immersion, quasi-compact immersion, or quasi-compact is Zariski local
on the target, and since the $V_j$ cover $V$ we get the desired conclusion.
\medskip\noindent
Let us prove that (5) implies (1). Let $V \to Y$ be as in (5).
Then we have the fibre product diagram
$$
\xymatrix{
V \times_Y X \ar[r] \ar[d] &
X \ar[d] \\
V \times_Y X \times_Y X \ar[r] &
X \times_Y X
}
$$
By assumption the left vertical arrow is a closed immersion,
immersion, quasi-compact immersion, or quasi-compact. It follows from
Spaces, Lemma \ref{spaces-lemma-descent-representable-transformations-property}
that also the right vertical arrow is a closed immersion,
immersion, quasi-compact immersion, or quasi-compact.
\medskip\noindent
It is clear that (1) implies (6) by taking the covering $Y = Y$.
Assume $Y = \bigcup Y_i$ is as in (6). Choose schemes $V_i$ and
surjective \'etale morphisms $V_i \to Y_i$. Note that the morphisms
$V_i \times_Y X \to V_i$ have $\mathcal{P}$ as they are base changes
of the morphisms $f^{-1}(Y_i) \to Y_i$. Set $V = \coprod V_i$.
Then $V \to Y$ is a morphism as in (5) (details omitted). Hence
(6) implies (5) and we are done.
\end{proof}
\begin{lemma}
\label{lemma-match-separated}
Let $S$ be a scheme.
Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$.
\begin{enumerate}
\item The morphism $f$ is locally separated.
\item The morphism $f$ is (quasi-)separated in the sense of
Definition \ref{definition-separated}
above if and only if $f$ is (quasi-)separated in the sense of
Section \ref{section-representable}.
\end{enumerate}
In particular, if $f : X \to Y$ is a morphism of schemes over $S$, then
$f$ is (quasi-)separated in the sense of
Definition \ref{definition-separated}
if and only if $f$ is (quasi-)separated as a morphism of schemes.
\end{lemma}
\begin{proof}
This is the equivalence of (1) and (2) of
Lemma \ref{lemma-separated-local}
combined with the fact that any morphism of schemes is locally separated, see
Schemes, Lemma \ref{schemes-lemma-diagonal-immersion}.
\end{proof}
\section{Surjective morphisms}
\label{section-surjective}
\noindent
We have already defined in Section \ref{section-representable}
what it means for a representable morphism of algebraic spaces
to be surjective.
\begin{lemma}
\label{lemma-surjective-representable}
Let $S$ be a scheme. Let $f : X \to Y$ be a representable
morphism of algebraic spaces over $S$. Then
$f$ is surjective (in the sense of Section \ref{section-representable})
if and only if $|f| : |X| \to |Y|$ is surjective.
\end{lemma}
\begin{proof}
Namely, if $f : X \to Y$ is representable, then it is surjective if and only if
for every scheme $T$ and every morphism $T \to Y$ the base change
$f_T : T \times_Y X \to T$ of $f$ is a surjective morphism of schemes,
in other words, if and only if $|f_T|$ is surjective. By
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}
the map $|T \times_Y X| \to |T| \times_{|Y|} |X|$ is always surjective.
Hence $|f_T| : |T \times_Y X| \to |T|$ is surjective if $|f| : |X| \to |Y|$
is surjective. Conversely, if $|f_T|$ is surjective
for every $T \to Y$ as above, then by taking $T$ to be the spectrum of a
field we conclude that $|X| \to |Y|$ is surjective.
\end{proof}
\noindent
This clears the way for the following definition.
\begin{definition}
\label{definition-surjective}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic
spaces over $S$. We say $f$ is {\it surjective}
if the map $|f| : |X| \to |Y|$ of associated topological spaces
is surjective.
\end{definition}
\begin{lemma}
\label{lemma-surjective-local}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
The following are equivalent:
\begin{enumerate}
\item $f$ is surjective,
\item for every scheme $Z$ and any morphism $Z \to Y$ the morphism
$Z \times_Y X \to Z$ is surjective,
\item for every affine scheme $Z$ and any morphism
$Z \to Y$ the morphism $Z \times_Y X \to Z$ is surjective,
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that $V \times_Y X \to V$ is a surjective morphism,
\item there exists a scheme $U$ and a surjective \'etale morphism
$\varphi : U \to X$ such that the composition $f \circ \varphi$
is surjective,
\item there exists a commutative diagram
$$
\xymatrix{
U \ar[d] \ar[r] & V \ar[d] \\
X \ar[r] & Y
}
$$
where $U$, $V$ are schemes and the vertical arrows are surjective \'etale
such that the top horizontal arrow is surjective, and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that
each of the morphisms $f^{-1}(Y_i) \to Y_i$ is surjective.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-composition-surjective}
The composition of surjective morphisms is surjective.
\end{lemma}
\begin{proof}
This is immediate from the definition.
\end{proof}
\begin{lemma}
\label{lemma-base-change-surjective}
The base change of a surjective morphism is surjective.
\end{lemma}
\begin{proof}
Follows immediately from
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}.
\end{proof}
\section{Open morphisms}
\label{section-open}
\noindent
For a representable morphism of algebraic spaces we have already defined (in
Section \ref{section-representable})
what it means to be universally open. Hence before we give the natural
definition we check that it agrees with this in the representable case.
\begin{lemma}
\label{lemma-characterize-representable-universally-open}
Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of
algebraic spaces over $S$. The following are equivalent
\begin{enumerate}
\item $f$ is universally open
(in the sense of Section \ref{section-representable}), and
\item for every morphism of algebraic spaces $Z \to Y$ the morphism of
topological spaces $|Z \times_Y X| \to |Z|$ is open.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and
a surjective \'etale morphism $V \to Y$. By assumption the morphism
of schemes $V \times_Y X \to V$ is universally open. By
Properties of Spaces, Section \ref{spaces-properties-section-points}
in the commutative diagram
$$
\xymatrix{
|V \times_Y X| \ar[r] \ar[d] & |Z \times_Y X| \ar[d] \\
|V| \ar[r] & |Z|
}
$$
the horizontal arrows are open and surjective, and moreover
$$
|V \times_Y X| \longrightarrow |V| \times_{|Z|} |Z \times_Y X|
$$
is surjective. Hence as the left
vertical arrow is open it follows that the right vertical arrow is
open. This proves (2). The implication (2) $\Rightarrow$ (1) is
immediate from the definitions.
\end{proof}
\noindent
Thus we may use the following natural definition.
\begin{definition}
\label{definition-open}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$.
\begin{enumerate}
\item We say $f$ is {\it open} if the map of topological spaces
$|f| : |X| \to |Y|$ is open.
\item We say $f$ is {\it universally open} if for every morphism
of algebraic spaces $Z \to Y$ the morphism of topological spaces
$$
|Z \times_Y X| \to |Z|
$$
is open, i.e., the base change $Z \times_Y X \to Z$ is open.
\end{enumerate}
\end{definition}
\noindent
Note that an \'etale morphism of algebraic spaces is universally open,
see
Properties of Spaces, Definition \ref{spaces-properties-definition-etale} and
Lemmas \ref{spaces-properties-lemma-etale-open} and
\ref{spaces-properties-lemma-base-change-etale}.
\begin{lemma}
\label{lemma-base-change-universally-open}
The base change of a universally open morphism of algebraic spaces
by any morphism of algebraic spaces is universally open.
\end{lemma}
\begin{proof}
This is immediate from the definition.
\end{proof}
\begin{lemma}
\label{lemma-composition-open}
The composition of a pair of (universally) open morphisms of algebraic spaces
is (universally) open.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-universally-open-local}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. The following are equivalent
\begin{enumerate}
\item $f$ is universally open,
\item for every scheme $Z$ and every morphism $Z \to Y$
the projection $|Z \times_Y X| \to |Z|$ is open,
\item for every affine scheme $Z$ and every morphism $Z \to Y$
the projection $|Z \times_Y X| \to |Z|$ is open, and
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that $V \times_Y X \to V$ is a universally open morphism
of algebraic spaces, and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that
each of the morphisms $f^{-1}(Y_i) \to Y_i$ is universally open.
\end{enumerate}
\end{lemma}
\begin{proof}
We omit the proof that (1) implies (2), and that (2) implies (3).
\medskip\noindent
Assume (3). Choose a surjective \'etale morphism $V \to Y$.
We are going to show that $V \times_Y X \to V$ is a universally
open morphism of algebraic spaces. Let $Z \to V$ be a morphism
from an algebraic space to $V$. Let $W \to Z$ be a surjective \'etale
morphism where $W = \coprod W_i$ is a disjoint union of affine schemes, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-cover-by-union-affines}.
Then we have the following commutative diagram
$$
\xymatrix{
\coprod_i |W_i \times_Y X| \ar@{=}[r] \ar[d] &
|W \times_Y X| \ar[r] \ar[d] &
|Z \times_Y X| \ar[d] \ar@{=}[r] &
|Z \times_V (V \times_Y X)| \ar[ld] \\
\coprod |W_i| \ar@{=}[r] &
|W| \ar[r] &
|Z|
}
$$
We have to show the south-east arrow is open. The middle horizontal
arrows are surjective and open
(Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-open}).
By assumption (3), and the fact that
$W_i$ is affine we see that the left vertical arrows are open. Hence
it follows that the right vertical arrow is open.
\medskip\noindent
Assume $V \to Y$ is as in (4). We will show that $f$ is universally open.
Let $Z \to Y$ be a morphism of algebraic spaces. Consider the
diagram
$$
\xymatrix{
|(V \times_Y Z) \times_V (V \times_Y X)| \ar@{=}[r] \ar[rd] &
|V \times_Y X| \ar[r] \ar[d] &
|Z \times_Y X| \ar[d] \\
&
|V \times_Y Z| \ar[r] &
|Z|
}
$$
The south-west arrow is open by assumption. The horizontal arrows are
surjective and open because the corresponding morphisms of
algebraic spaces are \'etale (see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-open}).
It follows that the right vertical arrow is open.
\medskip\noindent
Of course (1) implies (5) by taking the covering $Y = Y$.
Assume $Y = \bigcup Y_i$ is as in (5). Then for any $Z \to Y$
we get a corresponding Zariski covering $Z = \bigcup Z_i$ such that
the base change of $f$ to $Z_i$ is open. By a simple topological
argument this implies that $Z \times_Y X \to Z$ is open. Hence (1) holds.
\end{proof}
\begin{lemma}
\label{lemma-space-over-field-universally-open}
Let $S$ be a scheme. Let $p : X \to \Spec(k)$ be a morphism of
algebraic spaces over $S$ where $k$ is a field. Then
$p : X \to \Spec(k)$ is universally open.
\end{lemma}
\begin{proof}
Choose a scheme $U$ and a surjective \'etale morphism $U \to X$.
The composition $U \to \Spec(k)$ is universally open (as a morphism
of schemes) by
Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}.
Let $Z \to \Spec(k)$ be a morphism of schemes. Then
$U \times_{\Spec(k)} Z \to X \times_{\Spec(k)} Z$ is surjective,
see
Lemma \ref{lemma-base-change-surjective}.
Hence the first of the maps
$$
|U \times_{\Spec(k)} Z| \to |X \times_{\Spec(k)} Z| \to |Z|
$$
is surjective. Since the composition is open by the above we conclude that
the second map is open as well. Whence $p$ is universally open by
Lemma \ref{lemma-universally-open-local}.
\end{proof}
\section{Submersive morphisms}
\label{section-submersive}
\noindent
For a representable morphism of algebraic spaces we have already defined (in
Section \ref{section-representable})
what it means to be universally submersive. Hence before we give the natural
definition we check that it agrees with this in the representable case.
\begin{lemma}
\label{lemma-characterize-representable-universally-submersive}
Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of
algebraic spaces over $S$. The following are equivalent
\begin{enumerate}
\item $f$ is universally submersive
(in the sense of Section \ref{section-representable}), and
\item for every morphism of algebraic spaces $Z \to Y$ the morphism of
topological spaces $|Z \times_Y X| \to |Z|$ is submersive.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and
a surjective \'etale morphism $V \to Y$. By assumption the morphism
of schemes $V \times_Y X \to V$ is universally submersive. By
Properties of Spaces, Section \ref{spaces-properties-section-points}
in the commutative diagram
$$
\xymatrix{
|V \times_Y X| \ar[r] \ar[d] & |Z \times_Y X| \ar[d] \\
|V| \ar[r] & |Z|
}
$$
the horizontal arrows are open and surjective, and moreover
$$
|V \times_Y X| \longrightarrow |V| \times_{|Z|} |Z \times_Y X|
$$
is surjective. Hence as the left vertical arrow is submersive
it follows that the right vertical arrow is submersive.
This proves (2). The implication (2) $\Rightarrow$ (1) is
immediate from the definitions.
\end{proof}
\noindent
Thus we may use the following natural definition.
\begin{definition}
\label{definition-submersive}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
\begin{enumerate}
\item We say $f$ is {\it submersive}\footnote{This is very different
from the notion of a submersion of differential manifolds.}
if the continuous map $|X| \to |Y|$ is submersive, see
Topology, Definition \ref{topology-definition-submersive}.
\item We say $f$ is {\it universally submersive} if for every
morphism of algebraic spaces $Y' \to Y$ the base change
$Y' \times_Y X \to Y'$ is submersive.
\end{enumerate}
\end{definition}
\noindent
We note that a submersive morphism is in particular surjective.
\begin{lemma}
\label{lemma-base-change-universally-submersive}
The base change of a universally submersive morphism of algebraic spaces
by any morphism of algebraic spaces is universally submersive.
\end{lemma}
\begin{proof}
This is immediate from the definition.
\end{proof}
\begin{lemma}
\label{lemma-composition-universally-submersive}
The composition of a pair of (universally) submersive morphisms of
algebraic spaces is (universally) submersive.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Quasi-compact morphisms}
\label{section-quasi-compact}
\noindent
By Section \ref{section-representable} we know what it means for
a representable morphism of algebraic spaces to be quasi-compact.
In order to formulate the definition for a general morphism
of algebraic spaces we make the following observation.
\begin{lemma}
\label{lemma-characterize-representable-quasi-compact}
Let $S$ be a scheme.
Let $f : X \to Y$ be a representable morphism of algebraic spaces over $S$.
The following are equivalent:
\begin{enumerate}
\item $f$ is quasi-compact
(in the sense of Section \ref{section-representable}), and
\item for every quasi-compact algebraic space $Z$ and any morphism
$Z \to Y$ the algebraic space $Z \times_Y X$ is quasi-compact.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1), and let $Z \to Y$ be a morphism of algebraic spaces with
$Z$ quasi-compact. By
Properties of Spaces,
Definition \ref{spaces-properties-definition-quasi-compact}
there exists a quasi-compact scheme $U$ and a surjective \'etale
morphism $U \to Z$. Since $f$ is representable and quasi-compact
we see by definition that $U \times_Y X$ is a scheme, and that
$U \times_Y X \to U$ is quasi-compact. Hence $U \times_Y X$ is
a quasi-compact scheme. The morphism $U \times_Y X \to Z \times_Y X$
is \'etale and surjective (as the base change of the representable
\'etale and surjective morphism $U \to Z$, see
Section \ref{section-representable}).
Hence by definition $Z \times_Y X$ is quasi-compact.
\medskip\noindent
Assume (2). Let $Z \to Y$ be a morphism, where $Z$ is a scheme.
We have to show that $p : Z \times_Y X \to Z$ is quasi-compact.
Let $U \subset Z$ be affine open. Then $p^{-1}(U) = U \times_Y Z$
and the scheme $U \times_Y Z$ is quasi-compact by assumption (2).
Hence $p$ is quasi-compact, see
Schemes, Section \ref{schemes-section-quasi-compact}.
\end{proof}
\noindent
This motivates the following definition.
\begin{definition}
\label{definition-quasi-compact}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
We say $f$ is {\it quasi-compact} if for every quasi-compact
algebraic space $Z$ and morphism $Z \to Y$ the fibre product
$Z \times_Y X$ is quasi-compact.
\end{definition}
\noindent
By Lemma \ref{lemma-characterize-representable-quasi-compact}
above this agrees with the already existing notion
for representable morphisms of algebraic spaces.
\begin{lemma}
\label{lemma-base-change-quasi-compact}
The base change of a quasi-compact morphism of algebraic spaces
by any morphism of algebraic spaces is quasi-compact.
\end{lemma}
\begin{proof}
Omitted. Hint: Transitivity of fibre products.
\end{proof}
\begin{lemma}
\label{lemma-composition-quasi-compact}
The composition of a pair of quasi-compact morphisms of algebraic spaces
is quasi-compact.
\end{lemma}
\begin{proof}
Omitted. Hint: Transitivity of fibre products.
\end{proof}
\begin{lemma}
\label{lemma-surjection-from-quasi-compact}
\begin{slogan}
The image of a quasi-compact algebraic space under a surjective morphism
is quasi-compact.
\end{slogan}
Let $S$ be a scheme.
\begin{enumerate}
\item If $X \to Y$ is a surjective morphism of algebraic spaces over $S$,
and $X$ is quasi-compact then $Y$ is quasi-compact.
\item If
$$
\xymatrix{
X \ar[rr]_f \ar[rd]_p & &
Y \ar[dl]^q \\
& Z
}
$$
is a commutative diagram of morphisms of algebraic spaces over $S$
and $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $X$ is quasi-compact and $X \to Y$ is surjective. By
Definition \ref{definition-surjective}
the map $|X| \to |Y|$ is surjective, hence we see $Y$ is quasi-compact by
Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-space}
and the topological fact that the image of a quasi-compact space under a
continuous map is quasi-compact, see
Topology, Lemma \ref{topology-lemma-image-quasi-compact}.
Let $f, p, q$ be as in (2).
Let $T \to Z$ be a morphism whose source is a quasi-compact algebraic space.
By assumption $T \times_Z X$ is quasi-compact. By
Lemma \ref{lemma-base-change-surjective}
the morphism $T \times_Z X \to T \times_Z Y$ is surjective.
Hence by part (1) we see $T \times_Z Y$ is quasi-compact too.
Thus $q$ is quasi-compact.
\end{proof}
\begin{lemma}
\label{lemma-descent-quasi-compact}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $g : Y' \to Y$ be a universally open and surjective morphism of
algebraic spaces such that the base change $f' : X' \to Y'$ is quasi-compact.
Then $f$ is quasi-compact.
\end{lemma}
\begin{proof}
Let $Z \to Y$ be a morphism of algebraic spaces with $Z$ quasi-compact.
As $g$ is universally open and surjective, we see that
$Y' \times_Y Z \to Z$ is open and surjective. As every point of
$|Y' \times_Y Z|$ has a fundamental system of quasi-compact open
neighbourhoods (see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-space-locally-quasi-compact})
we can find a quasi-compact open $W \subset |Y' \times_Y Z|$
which surjects onto $Z$. Denote
$f'' : W \times_Y X \to W$ the base change of $f'$ by $W \to Y'$.
By assumption $W \times_Y X$ is quasi-compact. As $W \to Z$ is surjective
we see that $W \times_Y X \to Z \times_Y X$ is surjective.
Hence $Z \times_Y X$ is quasi-compact by
Lemma \ref{lemma-surjection-from-quasi-compact}.
Thus $f$ is quasi-compact.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-local}
\begin{slogan}
Quasi-compact morphisms of algebraic spaces are preserved under pullback
and local on the target.
\end{slogan}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
The following are equivalent:
\begin{enumerate}
\item $f$ is quasi-compact,
\item for every scheme $Z$ and any morphism $Z \to Y$ the morphism of
algebraic spaces $Z \times_Y X \to Z$ is quasi-compact,
\item for every affine scheme $Z$ and any morphism
$Z \to Y$ the algebraic space $Z \times_Y X$ is quasi-compact,
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that $V \times_Y X \to V$ is a quasi-compact morphism
of algebraic spaces, and
\item there exists a surjective \'etale morphism
$Y' \to Y$ of algebraic spaces such that $Y' \times_Y X \to Y'$
is a quasi-compact morphism of algebraic spaces, and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that
each of the morphisms $f^{-1}(Y_i) \to Y_i$ is quasi-compact.
\end{enumerate}
\end{lemma}
\begin{proof}
We will use Lemma \ref{lemma-base-change-quasi-compact}
without further mention.
It is clear that (1) implies (2) and that (2) implies (3).
Assume (3). Let $Z$ be a quasi-compact algebraic space over $S$,
and let $Z \to Y$ be a morphism. By
Properties of Spaces, Lemma
\ref{spaces-properties-lemma-quasi-compact-affine-cover}
there exists an affine scheme $U$ and a surjective \'etale morphism
$U \to Z$. Then $U \times_Y X \to Z \times_Y X$ is a surjective
morphism of algebraic spaces, see
Lemma \ref{lemma-base-change-surjective}.
By assumption $|U \times_Y X|$ is quasi-compact. It surjects
onto $|Z \times_Y X|$, hence we conclude that $|Z \times_Y X|$
is quasi-compact, see
Topology, Lemma \ref{topology-lemma-image-quasi-compact}.
This proves that (3) implies (1).
\medskip\noindent
The implications (1) $\Rightarrow$ (4), (4) $\Rightarrow$ (5) are clear.
The implication (5) $\Rightarrow$ (1) follows from
Lemma \ref{lemma-descent-quasi-compact}
and the fact that an \'etale morphism of algebraic spaces is universally open
(see discussion following
Definition \ref{definition-open}).
\medskip\noindent
Of course (1) implies (6) by taking the covering $Y = Y$.
Assume $Y = \bigcup Y_i$ is as in (6). Let $Z$ be affine and let
$Z \to Y$ be a morphism. Then there exists a finite standard affine
covering $Z = Z_1 \cup \ldots \cup Z_n$ such that each $Z_j \to Y$
factors through $Y_{i_j}$ for some $i_j$. Hence the algebraic space
$$
Z_j \times_Y X = Z_j \times_{Y_{i_j}} f^{-1}(Y_{i_j})
$$
is quasi-compact. Since
$Z \times_Y X = \bigcup_{j = 1, \ldots, n} Z_j \times_Y X$
is a Zariski covering we see that
$|Z \times_Y X| = \bigcup_{j = 1, \ldots, n} |Z_j \times_Y X|$
(see Properties of Spaces, Lemma \ref{spaces-properties-lemma-open-subspaces})
is a finite union of quasi-compact spaces, hence quasi-compact.
Thus we see that (6) implies (3).
\end{proof}
\noindent
The following (and the next) lemma guarantees in particular that a morphism
$X \to \Spec(A)$ is quasi-compact as soon as
$X$ is a quasi-compact algebraic space
\begin{lemma}
\label{lemma-quasi-compact-permanence}
Let $S$ be a scheme.
Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$.
If $g \circ f$ is quasi-compact and $g$ is quasi-separated
then $f$ is quasi-compact.
\end{lemma}
\begin{proof}
This is true because $f$ equals the composition
$(1, f) : X \to X \times_Z Y \to Y$. The first map
is quasi-compact by Lemma \ref{lemma-section-immersion}
because it is a section of the quasi-separated morphism $X \times_Z Y \to X$
(a base change of $g$, see Lemma \ref{lemma-base-change-separated}).
The second map is quasi-compact as it
is the base change of $f$, see
Lemma \ref{lemma-base-change-quasi-compact}.
And compositions of quasi-compact
morphisms are quasi-compact, see Lemma \ref{lemma-composition-quasi-compact}.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-quasi-separated-permanence}
Let $f : X \to Y$ be a morphism of algebraic spaces
over a scheme $S$.
\begin{enumerate}
\item If $X$ is quasi-compact and $Y$ is quasi-separated, then $f$ is
quasi-compact.
\item If $X$ is quasi-compact and quasi-separated and $Y$ is quasi-separated,
then $f$ is quasi-compact and quasi-separated.
\item A fibre product of quasi-compact and quasi-separated algebraic spaces
is quasi-compact and quasi-separated.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) follows from
Lemma \ref{lemma-quasi-compact-permanence}
with $Z = S = \Spec(\mathbf{Z})$.
Part (2) follows from (1) and
Lemma \ref{lemma-compose-after-separated}.
For (3) let $X \to Y$ and $Z \to Y$ be morphisms
of quasi-compact and quasi-separated algebraic spaces. Then
$X \times_Y Z \to Z$ is quasi-compact and quasi-separated as a
base change of $X \to Y$ using (2) and
Lemmas \ref{lemma-base-change-quasi-compact} and
\ref{lemma-base-change-separated}.
Hence $X \times_Y Z$ is quasi-compact and quasi-separated as
an algebraic space quasi-compact and quasi-separated over
$Z$, see
Lemmas \ref{lemma-separated-over-separated} and
\ref{lemma-composition-quasi-compact}.
\end{proof}
\section{Universally closed morphisms}
\label{section-universally-closed}
\noindent
For a representable morphism of algebraic spaces we have already defined (in
Section \ref{section-representable})
what it means to be universally closed. Hence before we give the natural
definition we check that it agrees with this in the representable case.
\begin{lemma}
\label{lemma-characterize-representable-universally-closed}
Let $S$ be a scheme. Let $f : X \to Y$ be a representable morphism of
algebraic spaces over $S$. The following are equivalent
\begin{enumerate}
\item $f$ is universally closed
(in the sense of Section \ref{section-representable}), and
\item for every morphism of algebraic spaces $Z \to Y$ the morphism of
topological spaces $|Z \times_Y X| \to |Z|$ is closed.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1), and let $Z \to Y$ be as in (2). Choose a scheme $V$ and
a surjective \'etale morphism $V \to Y$. By assumption the morphism
of schemes $V \times_Y X \to V$ is universally closed. By
Properties of Spaces, Section \ref{spaces-properties-section-points}
in the commutative diagram
$$
\xymatrix{
|V \times_Y X| \ar[r] \ar[d] & |Z \times_Y X| \ar[d] \\
|V| \ar[r] & |Z|
}
$$
the horizontal arrows are open and surjective, and moreover
$$
|V \times_Y X| \longrightarrow |V| \times_{|Z|} |Z \times_Y X|
$$
is surjective. Hence as the left
vertical arrow is closed it follows that the right vertical arrow is
closed. This proves (2). The implication (2) $\Rightarrow$ (1) is
immediate from the definitions.
\end{proof}
\noindent
Thus we may use the following natural definition.
\begin{definition}
\label{definition-closed}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$.
\begin{enumerate}
\item We say $f$ is {\it closed} if the map of topological
spaces $|X| \to |Y|$ is closed.
\item We say $f$ is {\it universally closed} if for every morphism
of algebraic spaces $Z \to Y$ the morphism of topological spaces
$$
|Z \times_Y X| \to |Z|
$$
is closed, i.e., the base change $Z \times_Y X \to Z$ is closed.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-base-change-universally-closed}
The base change of a universally closed morphism of algebraic spaces
by any morphism of algebraic spaces is universally closed.
\end{lemma}
\begin{proof}
This is immediate from the definition.
\end{proof}
\begin{lemma}
\label{lemma-composition-universally-closed}
The composition of a pair of (universally) closed morphisms of algebraic spaces
is (universally) closed.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-universally-closed-local}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. The following are equivalent
\begin{enumerate}
\item $f$ is universally closed,
\item for every scheme $Z$ and every morphism $Z \to Y$
the projection $|Z \times_Y X| \to |Z|$ is closed,
\item for every affine scheme $Z$ and every morphism $Z \to Y$
the projection $|Z \times_Y X| \to |Z|$ is closed,
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that $V \times_Y X \to V$ is a universally closed morphism
of algebraic spaces, and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that
each of the morphisms $f^{-1}(Y_i) \to Y_i$ is universally closed.
\end{enumerate}
\end{lemma}
\begin{proof}
We omit the proof that (1) implies (2), and that (2) implies (3).
\medskip\noindent
Assume (3). Choose a surjective \'etale morphism $V \to Y$.
We are going to show that $V \times_Y X \to V$ is a universally
closed morphism of algebraic spaces. Let $Z \to V$ be a morphism
from an algebraic space to $V$. Let $W \to Z$ be a surjective \'etale
morphism where $W = \coprod W_i$ is a disjoint union of affine schemes, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-cover-by-union-affines}.
Then we have the following commutative diagram
$$
\xymatrix{
\coprod_i |W_i \times_Y X| \ar@{=}[r] \ar[d] &
|W \times_Y X| \ar[r] \ar[d] &
|Z \times_Y X| \ar[d] \ar@{=}[r] &
|Z \times_V (V \times_Y X)| \ar[ld] \\
\coprod |W_i| \ar@{=}[r] &
|W| \ar[r] &
|Z|
}
$$
We have to show the south-east arrow is closed. The middle horizontal
arrows are surjective and open
(Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-open}).
By assumption (3), and the fact that
$W_i$ is affine we see that the left vertical arrows are closed. Hence
it follows that the right vertical arrow is closed.
\medskip\noindent
Assume (4). We will show that $f$ is universally closed.
Let $Z \to Y$ be a morphism of algebraic spaces. Consider the
diagram
$$
\xymatrix{
|(V \times_Y Z) \times_V (V \times_Y X)| \ar@{=}[r] \ar[rd] &
|V \times_Y X| \ar[r] \ar[d] &
|Z \times_Y X| \ar[d] \\
&
|V \times_Y Z| \ar[r] &
|Z|
}
$$
The south-west arrow is closed by assumption. The horizontal arrows are
surjective and open because the corresponding morphisms of
algebraic spaces are \'etale (see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-open}).
It follows that the right vertical arrow is closed.
\medskip\noindent
Of course (1) implies (5) by taking the covering $Y = Y$.
Assume $Y = \bigcup Y_i$ is as in (5). Then for any $Z \to Y$
we get a corresponding Zariski covering $Z = \bigcup Z_i$ such that
the base change of $f$ to $Z_i$ is closed. By a simple topological
argument this implies that $Z \times_Y X \to Z$ is closed. Hence (1) holds.
\end{proof}
\begin{example}
\label{example-strange-universally-closed}
Strange example of a universally closed morphism.
Let $\mathbf{Q} \subset k$ be a field of characteristic zero.
Let $X = \mathbf{A}^1_k/\mathbf{Z}$ as in
Spaces, Example \ref{spaces-example-affine-line-translation}.
We claim the structure morphism $p : X \to \Spec(k)$
is universally closed.
Namely, if $Z/k$ is a scheme, and $T \subset |X \times_k Z|$ is closed,
then $T$ corresponds to a $\mathbf{Z}$-invariant closed subset of
$T' \subset |\mathbf{A}^1 \times Z|$. It is easy to see that
this implies that $T'$ is the inverse image of a subset $T''$ of
$Z$. By
Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}
we have that $T'' \subset Z$ is closed.
Of course $T''$ is the image of $T$. Hence $p$ is universally
closed by Lemma \ref{lemma-universally-closed-local}.
\end{example}
\begin{lemma}
\label{lemma-universally-closed-quasi-compact}
Let $S$ be a scheme.
A universally closed morphism of algebraic spaces over $S$ is quasi-compact.
\end{lemma}
\begin{proof}
This proof is a repeat of the proof in the case of schemes, see
Morphisms, Lemma \ref{morphisms-lemma-universally-closed-quasi-compact}.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Assume that $f$ is not quasi-compact.
Our goal is to show that $f$ is not universally closed. By
Lemma \ref{lemma-quasi-compact-local}
there exists an affine scheme $Z$ and a morphism $Z \to Y$
such that $Z \times_Y X \to Z$ is not quasi-compact. To achieve our goal
it suffices to show that $Z \times_Y X \to Z$ is not universally closed,
hence we may assume that $Y = \Spec(B)$ for some ring $B$.
\medskip\noindent
Write $X = \bigcup_{i \in I} X_i$ where the $X_i$ are quasi-compact
open subspaces of $X$. For example, choose a surjective \'etale morphism
$U \to X$ where $U$ is a scheme, choose an affine open covering
$U = \bigcup U_i$ and let $X_i \subset X$ be the image of $U_i$.
We will use later that the morphisms $X_i \to Y$ are quasi-compact, see
Lemma \ref{lemma-quasi-compact-permanence}.
Let $T = \Spec(B[a_i ; i \in I])$. Let $T_i = D(a_i) \subset T$.
Let $Z \subset T \times_Y X$ be the reduced closed subspace whose underlying
closed set of points is
$|T \times_Y Z| \setminus \bigcup_{i \in I} |T_i \times_Y X_i|$, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-reduced-closed-subspace}.
(Note that $T_i \times_Y X_i$ is an open subspace of $T \times_Y X$ as
$T_i \to T$ and $X_i \to X$ are open immersions, see
Spaces, Lemmas \ref{spaces-lemma-base-change-immersions} and
\ref{spaces-lemma-composition-immersions}.) Here is a diagram
$$
\xymatrix{
Z \ar[r] \ar[rd] &
T \times_Y X \ar[d]^{f_T} \ar[r]_q &
X \ar[d]^f \\
& T \ar[r]^p & Y
}
$$
It suffices to prove that the image $f_T(|Z|)$ is not closed in $|T|$.
\medskip\noindent
We claim there exists a point $y \in Y$ such that there is no
affine open neighborhood $V$ of $y$ in $Y$ such that $X_V$ is quasi-compact.
If not then we can cover $Y$ with finitely many such $V$ and for
each $V$ the morphism $Y_V \to V$ is quasi-compact by
Lemma \ref{lemma-quasi-compact-permanence}
and then
Lemma \ref{lemma-quasi-compact-local}
implies $f$ quasi-compact, a contradiction. Fix a $y \in Y$ as in the claim.
\medskip\noindent
Let $t \in T$ be the point lying over $y$ with $\kappa(t) = \kappa(y)$
such that $a_i = 1$ in $\kappa(t)$ for all $i$. Suppose $z \in |Z|$ with
$f_T(z) = t$. Then $q(t) \in X_i$ for some $i$. Hence $f_T(z) \not \in T_i$
by construction of $Z$, which contradicts the fact that $t \in T_i$ by
construction. Hence we see that $t \in |T| \setminus f_T(|Z|)$.
\medskip\noindent
Assume $f_T(|Z|)$ is closed in $|T|$. Then there exists an element
$g \in B[a_i; i \in I]$ with $f_T(|Z|) \subset V(g)$ but $t \not \in V(g)$.
Hence the image of $g$ in $\kappa(t)$ is nonzero. In particular some
coefficient of $g$ has nonzero image in $\kappa(y)$. Hence this coefficient is
invertible on some affine open neighborhood $V$ of $y$. Let $J$ be the finite
set of $j \in I$ such that the variable $a_j$ appears in $g$.
Since $X_V$ is not quasi-compact and each $X_{i, V}$ is quasi-compact,
we may choose a point $x \in |X_V| \setminus \bigcup_{j \in J} |X_{j, V}|$.
In other words, $x \in |X| \setminus \bigcup_{j \in J} |X_j|$ and $x$ lies
above some $v \in V$. Since $g$ has a coefficient that is invertible on $V$,
we can find a point $t' \in T$ lying above $v$ such that $t' \not \in V(g)$ and
$t' \in V(a_i)$ for all $i \notin J$. This is true because
$V(a_i; i \in I \setminus J) = \Spec(B[a_j; j\in J])$
and the set of points of this scheme lying over $v$ is bijective
with $\Spec(\kappa(v)[a_j; j \in J])$ and $g$ restricts to
a nonzero element of this polynomial ring by construction.
In other words $t' \not \in T_i$ for each $i \not \in J$. By
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}
we can find a point $z$ of $X \times_Y T$ mapping to $x \in X$ and to
$t' \in T$. Since $x \not \in |X_j|$ for $j \in J$ and $t' \not \in T_i$
for $i \in I \setminus J$ we see that $z \in |Z|$. On the other hand
$f_T(z) = t' \not \in V(g)$ which contradicts $f_T(Z) \subset V(g)$.
Thus the assumption ``$f_T(|Z|)$ closed'' is wrong and we conclude indeed
that $f_T$ is not closed as desired.
\end{proof}
\noindent
The target of a separated algebraic space under a surjective
universally closed morphism is separated.
\begin{lemma}
\label{lemma-image-universally-closed-separated}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $f : X \to Y$ be a surjective universally closed
morphism of algebraic spaces over $B$.
\begin{enumerate}
\item If $X$ is quasi-separated, then $Y$ is quasi-separated.
\item If $X$ is separated, then $Y$ is separated.
\item If $X$ is quasi-separated over $B$, then $Y$ is quasi-separated over $B$.
\item If $X$ is separated over $B$, then $Y$ is separated over $B$.
\end{enumerate}
\end{lemma}
\begin{proof}
Parts (1) and (2) are a consequence of (3) and (4) for
$S = B = \Spec(\mathbf{Z})$ (see
Properties of Spaces,
Definition \ref{spaces-properties-definition-separated}).
Consider the commutative diagram
$$
\xymatrix{
X \ar[d] \ar[rr]_{\Delta_{X/B}} & & X \times_B X \ar[d] \\
Y \ar[rr]^{\Delta_{Y/B}} & & Y \times_B Y
}
$$
The left vertical arrow is surjective (i.e., universally surjective).
The right vertical arrow is universally closed as a composition
of the universally closed morphisms
$X \times_B X \to X \times_B Y \to Y \times_B Y$. Hence it is also
quasi-compact, see
Lemma \ref{lemma-universally-closed-quasi-compact}.
\medskip\noindent
Assume $X$ is quasi-separated over $B$, i.e., $\Delta_{X/B}$ is
quasi-compact. Then if $Z$ is quasi-compact and $Z \to Y \times_B Y$ is
a morphism, then $Z \times_{Y \times_B Y} X \to Z \times_{Y \times_B Y} Y$
is surjective and $Z \times_{Y \times_B Y} X$ is quasi-compact by our remarks
above. We conclude that $\Delta_{Y/B}$ is quasi-compact, i.e., $Y$
is quasi-separated over $B$.
\medskip\noindent
Assume $X$ is separated over $B$, i.e., $\Delta_{X/B}$ is a closed
immersion. Then if $Z$ is affine, and $Z \to Y \times_B Y$ is
a morphism, then $Z \times_{Y \times_B Y} X \to Z \times_{Y \times_B Y} Y$
is surjective and $Z \times_{Y \times_B Y} X \to Z$ is universally closed
by our remarks above. We conclude that $\Delta_{Y/B}$ is universally closed.
It follows that $\Delta_{Y/B}$ is representable, locally of finite type, a
monomorphism (see
Lemma \ref{lemma-properties-diagonal})
and universally closed, hence a closed immersion, see
\'Etale Morphisms,
Lemma \ref{etale-lemma-characterize-closed-immersion}
(and also the abstract principle
Spaces, Lemma
\ref{spaces-lemma-representable-transformations-property-implication}).
Thus $Y$ is separated over $B$.
\end{proof}
\section{Monomorphisms}
\label{section-monomorphisms}
\noindent
A representable morphism $X \to Y$ of algebraic spaces is a monomorphism
according to Section \ref{section-representable} if for every scheme
$Z$ and morphism $Z \to Y$ the morphism $Z \times_Y X \to Z$
is representable by a monomorphism of schemes.
This means exactly that $Z \times_Y X \to Z$
is an injective map of sheaves on $(\Sch/S)_{fppf}$. Since this
is supposed to hold for all $Z$ and all maps $Z \to Y$ this is in turn
equivalent to the map $X \to Y$ being an injective map of sheaves on
$(\Sch/S)_{fppf}$. Thus we may define a monomorphism of a (possibly
nonrepresentable\footnote{We do not know whether any monomorphism
of algebraic spaces is representable. For a discussion see
More on Morphisms of Spaces, Section
\ref{spaces-more-morphisms-section-monomorphisms}.})
morphism of algebraic spaces as follows.
\begin{definition}
\label{definition-monomorphism}
Let $S$ be a scheme.
A morphism of algebraic spaces over $S$ is called a {\it monomorphism}
if it is an injective map of sheaves, i.e., a monomorphism in the category
of sheaves on $(\Sch/S)_{fppf}$.
\end{definition}
\noindent
The following lemma shows that this also means that it is a monomorphism
in the category of algebraic spaces over $S$.
\begin{lemma}
\label{lemma-monomorphism}
Let $S$ be a scheme.
Let $j : X \to Y$ be a morphism of algebraic spaces over $S$.
The following are equivalent:
\begin{enumerate}
\item $j$ is a monomorphism (as in Definition \ref{definition-monomorphism}),
\item $j$ is a monomorphism in the category of algebraic spaces over $S$, and
\item the diagonal morphism $\Delta_{X/Y} : X \to X \times_Y X$ is
an isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Note that $X \times_Y X$ is both the fibre product in the category of
sheaves on $(\Sch/S)_{fppf}$ and the fibre product in the category
of algebraic spaces over $S$, see
Spaces, Lemma \ref{spaces-lemma-fibre-product-spaces}.
The equivalence of (1) and (3) is a general characterization
of injective maps of sheaves on any site.
The equivalence of (2) and (3) is a characterization of monomorphisms
in any category with fibre products.
\end{proof}
\begin{lemma}
\label{lemma-monomorphism-separated}
A monomorphism of algebraic spaces is separated.
\end{lemma}
\begin{proof}
This is true because an isomorphism is a closed immersion,
and Lemma \ref{lemma-monomorphism} above.
\end{proof}
\begin{lemma}
\label{lemma-composition-monomorphism}
A composition of monomorphisms is a monomorphism.
\end{lemma}
\begin{proof}
True because a composition of injective sheaf maps is injective.
\end{proof}
\begin{lemma}
\label{lemma-base-change-monomorphism}
The base change of a monomorphism is a monomorphism.
\end{lemma}
\begin{proof}
This is a general fact about fibre products in a category of sheaves.
\end{proof}
\begin{lemma}
\label{lemma-monomorphism-local}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
The following are equivalent
\begin{enumerate}
\item $f$ is a monomorphism,
\item for every scheme $Z$ and morphism $Z \to Y$ the
base change $Z \times_Y X \to Z$ of $f$ is a monomorphism,
\item for every affine scheme $Z$ and every morphism $Z \to Y$ the
base change $Z \times_Y X \to Z$ of $f$ is a monomorphism,
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that the base change $V \times_Y X \to V$ is a
monomorphism, and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that each
of the morphisms $f^{-1}(Y_i) \to Y_i$ is a monomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
We will use without further mention that a base change of a monomorphism
is a monomorphism, see
Lemma \ref{lemma-base-change-monomorphism}.
In particular it is clear that
(1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4)
(by taking $V$ to be a disjoint union of affine schemes \'etale over $Y$, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-cover-by-union-affines}).
Let $V$ be a scheme, and let $V \to Y$ be a surjective \'etale morphism.
If $V \times_Y X \to V$ is a monomorphism, then it
follows that $X \to Y$ is a monomorphism. Namely, given any
cartesian diagram of sheaves
$$
\vcenter{
\xymatrix{
\mathcal{F} \ar[r]_a \ar[d]_b & \mathcal{G} \ar[d]^c \\
\mathcal{H} \ar[r]^d & \mathcal{I}
}
}
\quad
\quad
\mathcal{F} = \mathcal{H} \times_\mathcal{I} \mathcal{G}
$$
if $c$ is a surjection of sheaves, and $a$ is injective, then also
$d$ is injective. Thus (4) implies (1). Proof of the equivalence of
(5) and (1) is omitted.
\end{proof}
\begin{lemma}
\label{lemma-immersions-monomorphisms}
An immersion of algebraic spaces is a monomorphism.
In particular, any immersion is separated.
\end{lemma}
\begin{proof}
Let $f : X \to Y$ be an immersion of algebraic spaces.
For any morphism $Z \to Y$ with $Z$ representable the base
change $Z \times_Y X \to Z$ is an immersion of schemes, hence
a monomorphism, see
Schemes, Lemma \ref{schemes-lemma-immersions-monomorphisms}.
Hence $f$ is representable, and a monomorphism.
\end{proof}
\noindent
We will improve on the following lemma in
Decent Spaces, Lemma
\ref{decent-spaces-lemma-monomorphism-toward-disjoint-union-dim-0-rings}.
\begin{lemma}
\label{lemma-monomorphism-toward-field}
Let $S$ be a scheme. Let $k$ be a field and let $Z \to \Spec(k)$
be a monomorphism of algebraic spaces over $S$. Then either
$Z = \emptyset$ or $Z = \Spec(k)$.
\end{lemma}
\begin{proof}
By
Lemmas \ref{lemma-monomorphism-separated} and
\ref{lemma-separated-over-separated}
we see that $Z$ is a separated algebraic space. Hence there exists an
open dense subspace $Z' \subset Z$ which is a scheme, see
Properties of Spaces, Proposition
\ref{spaces-properties-proposition-locally-quasi-separated-open-dense-scheme}.
By
Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}
we see that either $Z' = \emptyset$ or $Z' \cong \Spec(k)$.
In the first case we conclude that $Z = \emptyset$ and in the
second case we conclude that $Z' = Z = \Spec(k)$
as $Z \to \Spec(k)$ is a monomorphism which is an
isomorphism over $Z'$.
\end{proof}
\begin{lemma}
\label{lemma-monomorphism-injective-points}
Let $S$ be a scheme. If $X \to Y$ is a monomorphism of algebraic spaces
over $S$, then $|X| \to |Y|$ is injective.
\end{lemma}
\begin{proof}
Immediate from the definitions.
\end{proof}
\section{Pushforward of quasi-coherent sheaves}
\label{section-pushforward}
\noindent
We first prove a simple lemma that relates pushforward of sheaves of modules
for a morphism of algebraic spaces to pushforward of sheaves of modules for
a morphism of schemes.
\begin{lemma}
\label{lemma-compute-pushforward}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $U \to X$ be a surjective \'etale morphism from a scheme to $X$.
Set $R = U \times_X U$ and denote $t, s : R \to U$ the projection
morphisms as usual. Denote $a : U \to Y$ and $b : R \to Y$ the induced
morphisms. For any object $\mathcal{F}$ of $\textit{Mod}(\mathcal{O}_X)$
there exists an exact sequence
$$
0 \to f_*\mathcal{F} \to a_*(\mathcal{F}|_U) \to b_*(\mathcal{F}|_R)
$$
where the second arrow is the difference $t^* - s^*$.
\end{lemma}
\begin{proof}
We denote $\mathcal{F}$ also its extension to a sheaf of modules on
$X_{spaces, \etale}$, see
Properties of Spaces,
Remark \ref{spaces-properties-remark-explain-equivalence}.
Let $V \to Y$ be an object of $Y_\etale$. Then $V \times_Y X$ is an
object of $X_{spaces, \etale}$, and by definition
$f_*\mathcal{F}(V) = \mathcal{F}(V \times_Y X)$. Since $U \to X$ is
surjective \'etale, we see that $\{V \times_Y U \to V \times_Y X\}$
is a covering. Also, we have
$(V \times_Y U) \times_X (V \times_Y U) = V \times_Y R$.
Hence, by the sheaf condition of $\mathcal{F}$ on
$X_{spaces, \etale}$ we have a short exact sequence
$$
0 \to \mathcal{F}(V \times_Y X)
\to \mathcal{F}(V \times_Y U) \to \mathcal{F}(V \times_Y R)
$$
where the second arrow is the difference of restricting via $t$ or $s$.
This exact sequence is functorial in $V$ and hence we obtain the lemma.
\end{proof}
\noindent
Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact and
quasi-separated morphism of representable algebraic spaces $X$
and $Y$ over $S$. By
Descent,
Proposition \ref{descent-proposition-equivalence-quasi-coherent-functorial}
the functor
$f_* : \QCoh(\mathcal{O}_X) \to \QCoh(\mathcal{O}_Y)$
agrees with the usual functor if we think of $X$ and $Y$ as schemes.
\medskip\noindent
More generally, suppose $f : X \to Y$ is a representable, quasi-compact, and
quasi-separated morphism of algebraic spaces over $S$. Let $V$ be a scheme
and let $V \to Y$ be an \'etale surjective morphism. Let $U = V \times_Y X$
and let $f' : U \to V$ be the base change of $f$. Then for any
quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$ we have
\begin{equation}
\label{equation-representable-pushforward}
f'_*(\mathcal{F}|_U) = (f_*\mathcal{F})|_V,
\end{equation}
see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-pushforward-etale-base-change-modules}.
And because $f' : U \to V$ is a quasi-compact and quasi-separated
morphism of schemes, by the remark of the preceding paragraph we may
compute $f'_*(\mathcal{F}|_U)$ by thinking of $\mathcal{F}|_U$ as a
quasi-coherent sheaf on the scheme $U$, and $f'$ as a morphism of schemes.
We will frequently use this without further mention.
\medskip\noindent
The next level of generality is to consider an arbitrary
quasi-compact and quasi-separated morphism of algebraic spaces.
\begin{lemma}
\label{lemma-pushforward}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
If $f$ is quasi-compact and quasi-separated, then $f_*$ transforms
quasi-coherent $\mathcal{O}_X$-modules into
quasi-coherent $\mathcal{O}_Y$-modules.
\end{lemma}
\begin{proof}
Let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. We have to show that
$f_*\mathcal{F}$ is a quasi-coherent sheaf on $Y$. For this it suffices
to show that for any affine scheme $V$ and \'etale morphism $V \to Y$
the restriction of $f_*\mathcal{F}$ to $V$ is quasi-coherent, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-characterize-quasi-coherent}.
Let $f' : V \times_Y X \to V$
be the base change of $f$ by $V \to Y$. Note that $f'$ is also
quasi-compact and quasi-separated, see
Lemmas \ref{lemma-base-change-quasi-compact} and
\ref{lemma-base-change-separated}.
By (\ref{equation-representable-pushforward})
we know that the restriction of $f_*\mathcal{F}$ to $V$ is $f'_*$ of the
restriction of $\mathcal{F}$ to $V \times_Y X$. Hence
we may replace $f$ by $f'$, and assume that $Y$ is an affine scheme.
\medskip\noindent
Assume $Y$ is an affine scheme. Since $f$ is quasi-compact we see that $X$
is quasi-compact. Thus we may choose an affine scheme $U$ and a surjective
\'etale morphism $U \to X$, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover}.
By Lemma \ref{lemma-compute-pushforward} we get an exact sequence
$$
0 \to f_*\mathcal{F} \to a_*(\mathcal{F}|_U) \to b_*(\mathcal{F}|_R).
$$
where $R = U \times_X U$.
As $X \to Y$ is quasi-separated we see that $R \to U \times_Y U$
is a quasi-compact monomorphism. This implies that $R$ is a quasi-compact
separated scheme (as $U$ and $Y$ are affine at this point).
Hence $a : U \to Y$ and $b : R \to Y$ are quasi-compact and
quasi-separated morphisms of schemes. Thus by
Descent,
Proposition \ref{descent-proposition-equivalence-quasi-coherent-functorial}
the sheaves $a_*(\mathcal{F}|_U)$ and $b_*(\mathcal{F}|_R)$
are quasi-coherent (see also the discussion preceding this lemma).
This implies that $f_*\mathcal{F}$ is a kernel of
quasi-coherent modules, and hence itself quasi-coherent, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-properties-quasi-coherent}.
\end{proof}
\noindent
Higher direct images are discussed in
Cohomology of Spaces, Section
\ref{spaces-cohomology-section-higher-direct-image}.
\section{Immersions}
\label{section-immersions}
\noindent
Open, closed and locally closed immersions of algebraic spaces were defined in
Spaces, Section \ref{spaces-section-Zariski}.
Namely, a morphism of algebraic spaces is a
{\it closed immersion} (resp. {\it open immersion}, resp.\ {\it immersion})
if it is representable and a closed immersion (resp.\ open immersion,
resp.\ immersion) in the sense of Section \ref{section-representable}.
\medskip\noindent
In particular these types of morphisms are stable under base change
and compositions of morphisms in the category of algebraic
spaces over $S$, see
Spaces, Lemmas \ref{spaces-lemma-composition-immersions} and
\ref{spaces-lemma-base-change-immersions}.
\begin{lemma}
\label{lemma-closed-immersion-local}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. The following are equivalent:
\begin{enumerate}
\item $f$ is a closed immersion (resp.\ open immersion, resp.\ immersion),
\item for every scheme $Z$ and any morphism $Z \to Y$ the morphism
$Z \times_Y X \to Z$ is a closed immersion (resp.\ open immersion,
resp.\ immersion),
\item for every affine scheme $Z$ and any morphism
$Z \to Y$ the morphism $Z \times_Y X \to Z$ is a closed immersion
(resp.\ open immersion, resp.\ immersion),
\item there exists a scheme $V$ and a surjective \'etale morphism
$V \to Y$ such that $V \times_Y X \to V$ is a closed immersion
(resp.\ open immersion, resp.\ immersion), and
\item there exists a Zariski covering $Y = \bigcup Y_i$ such that
each of the morphisms $f^{-1}(Y_i) \to Y_i$ is a closed immersion
(resp.\ open immersion, resp.\ immersion).
\end{enumerate}
\end{lemma}
\begin{proof}
Using that a base change of a
closed immersion (resp.\ open immersion, resp.\ immersion)
is another one it is clear that (1) implies (2) and (2) implies (3).
Also (3) implies (4) since we can take $V$ to be a disjoint union of
affines, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-cover-by-union-affines}.
\medskip\noindent
Assume $V \to Y$ is as in (4).
Let $\mathcal{P}$ be the property
closed immersion (resp.\ open immersion, resp.\ immersion)
of morphisms of schemes. Note that property $\mathcal{P}$
is preserved under any base change and fppf local on the
base (see Section \ref{section-representable}).
Moreover, morphisms of type $\mathcal{P}$ are separated and
locally quasi-finite (in each of the three cases, see
Schemes, Lemma \ref{schemes-lemma-immersions-monomorphisms}, and
Morphisms, Lemma \ref{morphisms-lemma-immersion-locally-quasi-finite}).
Hence by
More on Morphisms, Lemma
\ref{more-morphisms-lemma-separated-locally-quasi-finite-morphisms-fppf-descend}
the morphisms of type $\mathcal{P}$ satisfy descent for fppf covering. Thus
Spaces, Lemma \ref{spaces-lemma-morphism-sheaves-with-P-effective-descent-etale}
applies and we see that $X \to Y$ is representable and has property
$\mathcal{P}$, in other words (1) holds.
\medskip\noindent
The equivalence of (1) and (5) follows from the fact that
$\mathcal{P}$ is Zariski local on the target (since we saw
above that $\mathcal{P}$ is in fact fppf local on the target).
\end{proof}
\begin{lemma}
\label{lemma-immersion-permanence}
Let $S$ be a scheme.
Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$.
\begin{enumerate}
\item If $Z \to X$ is representable, locally of finite type, locally
quasi-finite, separated, and a monomorphism, then $Z \to Y$ is
representable, locally of finite type, locally quasi-finite,
separated, and a monomorphism.
\item If $Z \to X$ is an immersion and $Y \to X$ is locally separated,
then $Z \to Y$ is an immersion.
\item If $Z \to X$ is a closed immersion and $Y \to X$ is separated,
then $Z \to Y$ is a closed immersion.
\end{enumerate}
\end{lemma}
\begin{proof}
In each case the proof is to contemplate the commutative diagram
$$
\xymatrix{
Z \ar[r] \ar[rd] & Y \times_X Z \ar[r] \ar[d] & Z \ar[d] \\
& Y \ar[r] & X
}
$$
where the composition of the top horizontal arrows is the identity.
Let us prove (1). The first horizontal arrow is a section of
$Y \times_X Z \to Z$, whence representable, locally of finite type,
locally quasi-finite, separated, and a monomorphism by
Lemma \ref{lemma-section-immersion}.
The arrow $Y \times_X Z \to Y$ is a base change of $Z \to X$
hence is representable, locally of finite type,
locally quasi-finite, separated, and a monomorphism
(as each of these properties of morphisms of schemes is stable
under base change, see
Spaces, Remark \ref{spaces-remark-list-properties-stable-base-change}).
Hence the same is true for the composition (as each of these properties of
morphisms of schemes is stable under composition, see Spaces, Remark
\ref{spaces-remark-list-properties-stable-composition}).
This proves (1). The other results are proved in exactly the same manner.
\end{proof}
\begin{lemma}
\label{lemma-immersion-when-closed}
Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic
spaces over $S$. Then $|i| : |Z| \to |X|$ is a homeomorphism onto a
locally closed subset, and $i$ is a closed immersion if and only if
the image $|i|(|Z|) \subset |X|$ is a closed subset.
\end{lemma}
\begin{proof}
The first statement is Properties of Spaces, Lemma
\ref{spaces-properties-lemma-subspace-induced-topology}.
Let $U$ be a scheme and let $U \to X$ be a surjective \'etale morphism.
By assumption $T = U \times_X Z$ is a scheme and the morphism $j : T \to U$
is an immersion of schemes. By Lemma \ref{lemma-closed-immersion-local}
the morphism $i$ is a closed immersion if and only if $j$ is a closed
immersion. By Schemes, Lemma \ref{schemes-lemma-immersion-when-closed}
this is true if and only if $j(T)$ is closed in $U$.
However, the subset $j(T) \subset U$ is the inverse image of
$|i|(|Z|) \subset |X|$, see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}.
This finishes the proof.
\end{proof}
\begin{remark}
\label{remark-immersion}
Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic
spaces over $S$. Since $i$ is a monomorphism we may think of $|Z|$ as
a subset of $|X|$; in the rest of this remark we do so.
Let $\partial |Z|$ be the boundary of $|Z|$ in
the topological space $|X|$. In a formula
$$
\partial |Z| = \overline{|Z|} \setminus |Z|.
$$
Let $\partial Z$ be the reduced closed subspace of $X$ with
$|\partial Z| = \partial |Z|$
obtained by taking the reduced induced closed subspace structure, see
Properties of Spaces,
Definition \ref{spaces-properties-definition-reduced-induced-space}.
By construction we see that $|Z|$ is closed in
$|X| \setminus |\partial Z| = |X \setminus \partial Z|$.
Hence it is true that any immersion of algebraic spaces can be
factored as a closed immersion followed by an open immersion
(but not the other way in general, see
Morphisms, Example \ref{morphisms-example-thibaut}).
\end{remark}
\begin{remark}
\label{remark-space-structure-locally-closed-subset}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $T \subset |X|$ be a locally closed subset.
Let $\partial T$ be the boundary of $T$ in
the topological space $|X|$. In a formula
$$
\partial T = \overline{T} \setminus T.
$$
Let $U \subset X$ be the open subspace of $X$ with
$|U| = |X| \setminus \partial T$, see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-open-subspaces}.
Let $Z$ be the reduced closed subspace of $U$ with
$|Z| = T$ obtained by taking the reduced induced
closed subspace structure, see
Properties of Spaces,
Definition \ref{spaces-properties-definition-reduced-induced-space}.
By construction $Z \to U$ is a closed immersion of algebraic spaces
and $U \to X$ is an open immersion, hence
$Z \to X$ is an immersion of algebraic spaces over $S$ (see
Spaces, Lemma \ref{spaces-lemma-composition-immersions}).
Note that $Z$ is a reduced algebraic space and that
$|Z| = T$ as subsets of $|X|$. We sometimes say
$Z$ is the {\it reduced induced subspace structure} on $T$.
\end{remark}
\begin{lemma}
\label{lemma-factor-the-other-way}
Let $S$ be a scheme. Let $Z \to X$ be an immersion of algebraic spaces over
$S$. Assume $Z \to X$ is quasi-compact.
There exists a factorization $Z \to \overline{Z} \to X$ where
$Z \to \overline{Z}$ is an open immersion and $\overline{Z} \to X$
is a closed immersion.
\end{lemma}
\begin{proof}
Let $U$ be a scheme and let $U \to X$ be surjective \'etale.
As usual denote $R = U \times_X U$ with projections
$s, t : R \to U$. Set $T = Z \times_U X$. Let $\overline{T} \subset U$
be the scheme theoretic image of $T \to U$. Note that
$s^{-1}\overline{T} = t^{-1}\overline{T}$ as taking
scheme theoretic images of quasi-compact morphisms commute
with flat base change, see
Morphisms, Lemma \ref{morphisms-lemma-flat-base-change-scheme-theoretic-image}.
Hence we obtain a closed subspace $\overline{Z} \subset X$ whose
pullback to $U$ is $\overline{T}$, see
Properties of Spaces, Lemma
\ref{spaces-properties-lemma-subspaces-presentation}.
By Morphisms, Lemma \ref{morphisms-lemma-quasi-compact-immersion}
the morphism $T \to \overline{T}$
is an open immersion. It follows that $Z \to \overline{Z}$ is
an open immersion and we win.
\end{proof}
\section{Closed immersions}
\label{section-closed-immersions}
\noindent
In this section we elucidate some of the results obtained previously on
immersions of algebraic spaces. See
Spaces, Section \ref{spaces-section-Zariski}
and
Section \ref{section-immersions} in this chapter.
This section is the analogue of
Morphisms, Section \ref{morphisms-section-closed-immersions}
for algebraic spaces.
\begin{lemma}
\label{lemma-closed-immersion-ideals}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
For every closed immersion $i : Z \to X$ the sheaf
$i_*\mathcal{O}_Z$ is a quasi-coherent $\mathcal{O}_X$-module, the map
$i^\sharp : \mathcal{O}_X \to i_*\mathcal{O}_Z$ is surjective and its
kernel is a quasi-coherent sheaf of ideals. The rule
$Z \mapsto \Ker(\mathcal{O}_X \to i_*\mathcal{O}_Z)$
defines an inclusion reversing bijection
$$
\begin{matrix}
\text{closed subspaces}\\
Z \subset X
\end{matrix}
\longrightarrow
\begin{matrix}
\text{quasi-coherent sheaves}\\
\text{of ideals }\mathcal{I} \subset \mathcal{O}_X
\end{matrix}
$$
Moreover, given a closed subscheme $Z$ corresponding to the quasi-coherent
sheaf of ideals $\mathcal{I} \subset \mathcal{O}_X$ a morphism of algebraic
spaces $h : Y \to X$ factors through $Z$ if and only if the map
$h^*\mathcal{I} \to h^*\mathcal{O}_X = \mathcal{O}_Y$ is zero.
\end{lemma}
\begin{proof}
Let $U \to X$ be a surjective \'etale morphism whose source is a scheme.
Consider the diagram
$$
\xymatrix{
U \times_X Z \ar[r] \ar[d]_{i'} & Z \ar[d]^i \\
U \ar[r] & X
}
$$
By
Lemma \ref{lemma-closed-immersion-local}
we see that $i$ is a closed immersion
if and only if $i'$ is a closed immersion. By
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-pushforward-etale-base-change-modules}
we see that $i'_*\mathcal{O}_{U \times_X Z}$ is the restriction of
$i_*\mathcal{O}_Z$ to $U$. Hence the assertions on
$\mathcal{O}_X \to i_*\mathcal{O}_Z$ are equivalent to the
corresponding assertions on
$\mathcal{O}_U \to i'_*\mathcal{O}_{U \times_X Z}$.
And since $i'$ is a closed immersion of schemes, these results follow from
Morphisms, Lemma \ref{morphisms-lemma-closed-immersion}.
\medskip\noindent
Let us prove that given a quasi-coherent
sheaf of ideals $\mathcal{I} \subset \mathcal{O}_X$ the formula
$$
Z(T) = \{h : T \to X \mid h^*\mathcal{I} \to \mathcal{O}_T
\text{ is zero}\}
$$
defines a closed subspace of $X$. It is clearly a subfunctor of $X$.
To show that $Z \to X$ is representable by closed immersions, let
$\varphi : U \to X$ be a morphism from a scheme towards $X$. Then
$Z \times_X U$ is represented by the analogous subfunctor of $U$ corresponding
to the sheaf of ideals $\Im(\varphi^*\mathcal{I} \to \mathcal{O}_U)$. By
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-pullback-quasi-coherent}
the $\mathcal{O}_U$-module $\varphi^*\mathcal{I}$ is quasi-coherent on
on $U$, and hence $\Im(\varphi^*\mathcal{I} \to \mathcal{O}_U)$
is a quasi-coherent sheaf of ideals on $U$. By
Schemes, Lemma \ref{schemes-lemma-characterize-closed-subspace}
we conclude that $Z \times_X U$ is represented by the closed subscheme
of $U$ associated to $\Im(\varphi^*\mathcal{I} \to \mathcal{O}_U)$.
Thus $Z$ is a closed subspace of $X$.
\medskip\noindent
In the formula for $Z$ above the inputs $T$ are schemes since algebraic
spaces are sheaves on $(\Sch/S)_{fppf}$. We omit the verification
that the same formula remains true if $T$ is an algebraic space.
\end{proof}
\begin{definition}
\label{definition-inverse-image-closed-subspace}
Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces
over $S$. Let $Z \subset X$ be a closed subspace. The
{\it inverse image $f^{-1}(Z)$ of the closed subspace $Z$}
is the closed subspace $Z \times_X Y$ of $Y$.
\end{definition}
\noindent
This definition makes sense by Lemma \ref{lemma-closed-immersion-local}.
If $\mathcal{I} \subset \mathcal{O}_X$ is the quasi-coherent sheaf of
ideals corresponding to $Z$ via Lemma \ref{lemma-closed-immersion-ideals} then
$f^{-1}\mathcal{I}\mathcal{O}_Y = \Im(f^*\mathcal{I} \to \mathcal{O}_Y)$
is the sheaf of ideals corresponding to $f^{-1}(Z)$.
\begin{lemma}
\label{lemma-closed-immersion-quasi-compact}
A closed immersion of algebraic spaces is quasi-compact.
\end{lemma}
\begin{proof}
This follows from
Schemes, Lemma \ref{schemes-lemma-closed-immersion-quasi-compact}
by general principles, see
Spaces, Lemma
\ref{spaces-lemma-representable-transformations-property-implication}.
\end{proof}
\begin{lemma}
\label{lemma-closed-immersion-separated}
A closed immersion of algebraic spaces is separated.
\end{lemma}
\begin{proof}
This follows from
Schemes, Lemma \ref{schemes-lemma-immersions-monomorphisms}
by general principles, see
Spaces, Lemma
\ref{spaces-lemma-representable-transformations-property-implication}.
\end{proof}
\begin{lemma}
\label{lemma-closed-immersion-push-pull}
Let $S$ be a scheme. Let $i : Z \to X$ be a closed immersion of algebraic
spaces over $S$.
\begin{enumerate}
\item The functor
$$
i_{small, *} :
\Sh(Z_\etale)
\longrightarrow
\Sh(X_\etale)
$$
is fully faithful and its essential image is those sheaves of sets
$\mathcal{F}$ on $X_\etale$ whose restriction to $X \setminus Z$ is
isomorphic to $*$, and
\item the functor
$$
i_{small, *} :
\textit{Ab}(Z_\etale)
\longrightarrow
\textit{Ab}(X_\etale)
$$
is fully faithful and its essential image is those abelian sheaves on
$X_\etale$ whose support is contained in $|Z|$.
\end{enumerate}
In both cases $i_{small}^{-1}$ is a left inverse to the functor
$i_{small, *}$.
\end{lemma}
\begin{proof}
Let $U$ be a scheme and let $U \to X$ be surjective \'etale.
Set $V = Z \times_X U$. Then $V$ is a scheme and $i' : V \to U$ is
a closed immersion of schemes. By
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-pushforward-etale-base-change}
for any sheaf $\mathcal{G}$ on $Z$ we have
$$
(i_{small}^{-1}i_{small, *}\mathcal{G})|_V =
(i')_{small}^{-1}i'_{small, *}(\mathcal{G}|_V)
$$
By
\'Etale Cohomology, Proposition
\ref{etale-cohomology-proposition-closed-immersion-pushforward}
the map
$(i')_{small}^{-1}i'_{small, *}(\mathcal{G}|_V) \to \mathcal{G}|_V$
is an isomorphism. Since $V \to Z$ is surjective and \'etale this implies
that $i_{small}^{-1}i_{small, *}\mathcal{G} \to \mathcal{G}$ is an
isomorphism. This clearly implies that $i_{small, *}$ is fully faithful, see
Sites, Lemma \ref{sites-lemma-exactness-properties}.
To prove the statement on the essential image, consider a sheaf of sets
$\mathcal{F}$ on $X_\etale$ whose restriction to $X \setminus Z$ is
isomorphic to $*$. As in the proof of
\'Etale Cohomology, Proposition
\ref{etale-cohomology-proposition-closed-immersion-pushforward}
we consider the adjunction mapping
$$
\mathcal{F} \longrightarrow i_{small, *}i_{small}^{-1}\mathcal{F}.
$$
As in the first part we see that the restriction of this map to
$U$ is an isomorphism by the corresponding result for the case of
schemes. Since $U$ is an \'etale covering of $X$ we
conclude it is an isomorphism.
\end{proof}
\noindent
The following lemma holds more generally in the setting of a closed
immersion of topoi (insert future reference here).
\begin{lemma}
\label{lemma-closed-immersion-rings}
Let $S$ be a scheme. Let $i : Z \to X$ be a closed immersion of algebraic
spaces over $S$. Let $\mathcal{A}$ be a sheaf of rings on $X_\etale$.
Let $\mathcal{B}$ be a sheaf of rings on $Z_\etale$.
Let $\varphi : \mathcal{A} \to i_{small, *}\mathcal{B}$
(or what is the same thing
$\varphi : i_{small}^{-1}\mathcal{A} \to \mathcal{B}$)
be a homomorphism of sheaves of rings. Then for any sheaf of
$\mathcal{A}$-modules $\mathcal{F}$ the adjunction mapping
$\mathcal{F} \to i_{small, *}i_{small}^{-1}\mathcal{F}$
induces an isomorphism
$$
\mathcal{F} \otimes_\mathcal{A} i_{small, *}\mathcal{B}
\longrightarrow
i_{small, *}(i^{-1}_{small}\mathcal{F}
\otimes_{i_{small}^{-1}\mathcal{A}}
\mathcal{B}).
$$
\end{lemma}
\begin{proof}
During this proof we drop the subscript ${}_{small}$ from the notation.
There is a map
$i^{-1}\mathcal{F} \to
i^{-1}\mathcal{F} \otimes_{i^{-1}\mathcal{A}} \mathcal{B}$
to which we can apply $i_*$ and compose with the adjunction map:
$$
\mathcal{F} \longrightarrow
i_*(i^{-1}\mathcal{F}) \longrightarrow
i_*(i^{-1}\mathcal{F} \otimes_{i^{-1}\mathcal{A}} \mathcal{B}).
$$
The composition is $\mathcal{A}$-linear where $\mathcal{A}$ acts on the
target via $\varphi$. Note that this target
$i_*(i^{-1}\mathcal{F} \otimes_{i^{-1}\mathcal{A}} \mathcal{B})$
has a canonical $i_*\mathcal{B}$-module structure.
Hence by the universal property of tensor product we obtain
a map as in the lemma.
\medskip\noindent
Let $\mathcal{G}$ be a sheaf of $i_*\mathcal{B}$-modules on $X_\etale$.
Since the support of the sheaf of rings $i_*\mathcal{B}$ is contained in
$|Z|$ we see that the support of $\mathcal{G}$ is contained in $|Z|$.
Hence by
Lemma \ref{lemma-closed-immersion-push-pull}
we conclude that there exists a unique sheaf of $\mathcal{B}$-modules
$\mathcal{H}$ and an isomorphism $i_*\mathcal{H} = \mathcal{G}$
as $i_*\mathcal{B}$-modules.
To show that the map of the lemma is an isomorphism we show that the
right hand side of the arrow satisfies the universal property enjoyed
by the tensor product on the left (i.e., we will use
Yoneda's lemma, see
Categories, Lemma \ref{categories-lemma-yoneda}).
To see this we have to show that maps into $\mathcal{G}$ agree.
This can be seen using the following sequence of canonical isomorphisms
\begin{align*}
\Hom_{i_*\mathcal{B}}(
\mathcal{F} \otimes_\mathcal{A} i_*\mathcal{B},
\mathcal{G})
& =
\Hom_\mathcal{A}(
\mathcal{F},
\mathcal{G}) \\
& =
\Hom_\mathcal{A}(
\mathcal{F},
i_*(\mathcal{H})) \\
& =
\Hom_{i^{-1}\mathcal{A}}(
i^{-1}\mathcal{F},
\mathcal{H}) \\
& =
\Hom_\mathcal{B}(
i^{-1}\mathcal{F} \otimes_{i^{-1}\mathcal{A}} \mathcal{B},
\mathcal{H}) \\
& =
\Hom_{i_*\mathcal{B}}(
i_*(i^{-1}\mathcal{F} \otimes_{i^{-1}\mathcal{A}} \mathcal{B}),
i_*\mathcal{H}) \\
& =
\Hom_{i_*\mathcal{B}}(
i_*(i^{-1}\mathcal{F} \otimes_{i^{-1}\mathcal{A}} \mathcal{B}),
\mathcal{G})
\end{align*}
The fifth equality holds because of the equivalence of categories in
Lemma \ref{lemma-closed-immersion-push-pull}.
\end{proof}
\section{Closed immersions and quasi-coherent sheaves}
\label{section-closed-immersions-quasi-coherent}
\noindent
This section is the analogue of
Morphisms, Section \ref{morphisms-section-closed-immersions-quasi-coherent}.
\begin{lemma}
\label{lemma-i-star-equivalence}
Let $S$ be a scheme. Let $i : Z \to X$ be a closed immersion of algebraic
spaces over $S$. Let $\mathcal{I} \subset \mathcal{O}_X$ be the quasi-coherent
sheaf of ideals cutting out $Z$.
\begin{enumerate}
\item For any $\mathcal{O}_X$-module $\mathcal{F}$ the adjunction map
$\mathcal{F} \to i_*i^*\mathcal{F}$ induces an isomorphism
$\mathcal{F}/\mathcal{I}\mathcal{F} \cong i_*i^*\mathcal{F}$.
\item The functor $i^*$ is a left inverse to $i_*$, i.e., for any
$\mathcal{O}_Z$-module $\mathcal{G}$ the adjunction map
$i^*i_*\mathcal{G} \to \mathcal{G}$ is an isomorphism.
\item The functor
$$
i_* :
\QCoh(\mathcal{O}_Z)
\longrightarrow
\QCoh(\mathcal{O}_X)
$$
is exact, fully faithful, with essential image those quasi-coherent
$\mathcal{O}_X$-modules $\mathcal{F}$ such that $\mathcal{I}\mathcal{F} = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
During this proof we work exclusively with sheaves on
the small \'etale sites, and we use $i_*, i^{-1}, \ldots$
to denote pushforward and pullback of sheaves of abelian groups
instead of $i_{small, *}, i_{small}^{-1}$.
\medskip\noindent
Let $\mathcal{F}$ be an $\mathcal{O}_X$-module. By
Lemma \ref{lemma-closed-immersion-rings}
we see that
$i_*i^*\mathcal{F} = \mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{O}_Z$.
By
Lemma \ref{lemma-closed-immersion-ideals}
we see that we have a short exact sequence
$$
0 \to \mathcal{I} \to \mathcal{O}_X \to i_*\mathcal{O}_Z \to 0
$$
It follows from properties of the tensor product that
$\mathcal{F} \otimes_{\mathcal{O}_X} i_*\mathcal{O}_Z
= \mathcal{F}/\mathcal{I}\mathcal{F}$. This proves (1) (except
that we omit the verification that the map is induced by the
adjunction mapping).
\medskip\noindent
Let $\mathcal{G}$ be any $\mathcal{O}_Z$-module. By
Lemma \ref{lemma-closed-immersion-push-pull}
we see that $i^{-1}i_*\mathcal{G} = \mathcal{G}$.
Hence to prove (2) we have to show that the canonical map
$\mathcal{G} \otimes_{i^{-1}\mathcal{O}_X} \mathcal{O}_Z \to \mathcal{G}$
is an isomorphism. This follows from general properties of tensor products
if we can show that $i^{-1}\mathcal{O}_X \to \mathcal{O}_Z$ is surjective. By
Lemma \ref{lemma-closed-immersion-push-pull}
it suffices to prove that
$i_*i^{-1}\mathcal{O}_X \to i_*\mathcal{O}_Z$
is surjective. Since the surjective map $\mathcal{O}_X \to i_*\mathcal{O}_Z$
factors through this map we see that (2) holds.
\medskip\noindent
Finally we prove the most interesting part of the lemma, namely part (3).
A closed immersion is quasi-compact and separated, see
Lemmas \ref{lemma-closed-immersion-quasi-compact} and
\ref{lemma-closed-immersion-separated}. Hence
Lemma \ref{lemma-pushforward}
applies and the pushforward of a quasi-coherent
sheaf on $Z$ is indeed a quasi-coherent sheaf on $X$.
Thus we obtain our functor
$i^{QCoh}_* : \QCoh(\mathcal{O}_Z)
\to \QCoh(\mathcal{O}_X)$.
It is clear from part (2) that $i^{QCoh}_*$ is fully faithful since
it has a left inverse, namely $i^*$.
\medskip\noindent
Now we turn to the description of the essential image of the
functor $i_*$. It is clear that $\mathcal{I}(i_*\mathcal{G}) = 0$
for any $\mathcal{O}_Z$-module, since $\mathcal{I}$ is the kernel
of the map $\mathcal{O}_X \to i_*\mathcal{O}_Z$ which is the map
we use to put an $\mathcal{O}_X$-module structure on $i_*\mathcal{G}$.
Next, suppose that $\mathcal{F}$ is any quasi-coherent
$\mathcal{O}_X$-module such that $\mathcal{I}\mathcal{F} = 0$.
Then we see that $\mathcal{F}$ is an $i_*\mathcal{O}_Z$-module
because $i_*\mathcal{O}_Z = \mathcal{O}_X/\mathcal{I}$. Hence in
particular its support is contained in $|Z|$. We apply
Lemma \ref{lemma-closed-immersion-push-pull}
to see that $\mathcal{F} \cong i_*\mathcal{G}$ for some
$\mathcal{O}_Z$-module $\mathcal{G}$. The only small detail left over
is to see why $\mathcal{G}$ is quasi-coherent. This is true
because $\mathcal{G} \cong i^*\mathcal{F}$ by part (2) and
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-pullback-quasi-coherent}.
\end{proof}
\noindent
Let $i : Z \to X$ be a closed immersion of algebraic spaces.
Because of the lemma above we often,
by abuse of notation, denote $\mathcal{F}$ the sheaf $i_*\mathcal{F}$ on $X$.
\begin{lemma}
\label{lemma-largest-quasi-coherent-subsheaf}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module.
Let $\mathcal{G} \subset \mathcal{F}$ be a $\mathcal{O}_X$-submodule.
There exists a unique quasi-coherent $\mathcal{O}_X$-submodule
$\mathcal{G}' \subset \mathcal{G}$ with the following property:
For every quasi-coherent $\mathcal{O}_X$-module $\mathcal{H}$ the map
$$
\Hom_{\mathcal{O}_X}(\mathcal{H}, \mathcal{G}')
\longrightarrow
\Hom_{\mathcal{O}_X}(\mathcal{H}, \mathcal{G})
$$
is bijective. In particular $\mathcal{G}'$ is the largest quasi-coherent
$\mathcal{O}_X$-submodule of $\mathcal{F}$ contained in $\mathcal{G}$.
\end{lemma}
\begin{proof}
Let $\mathcal{G}_a$, $a \in A$ be the set of quasi-coherent
$\mathcal{O}_X$-submodules contained in $\mathcal{G}$.
Then the image $\mathcal{G}'$ of
$$
\bigoplus\nolimits_{a \in A} \mathcal{G}_a \longrightarrow \mathcal{F}
$$
is quasi-coherent as the image of a map of quasi-coherent sheaves
on $X$ is quasi-coherent and since a direct sum of quasi-coherent sheaves
is quasi-coherent, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-properties-quasi-coherent}.
The module $\mathcal{G}'$ is contained in $\mathcal{G}$. Hence this is the
largest quasi-coherent $\mathcal{O}_X$-module contained in $\mathcal{G}$.
\medskip\noindent
To prove the formula, let $\mathcal{H}$ be a quasi-coherent
$\mathcal{O}_X$-module and let $\alpha : \mathcal{H} \to \mathcal{G}$
be an $\mathcal{O}_X$-module map. The image of the composition
$\mathcal{H} \to \mathcal{G} \to \mathcal{F}$ is quasi-coherent
as the image of a map of quasi-coherent sheaves. Hence it is contained
in $\mathcal{G}'$. Hence $\alpha$ factors through $\mathcal{G}'$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-i-upper-shriek}
Let $S$ be a scheme.
Let $i : Z \to X$ be a closed immersion of algebraic spaces over $S$.
There is a functor\footnote{This is likely nonstandard notation.}
$i^! : \QCoh(\mathcal{O}_X) \to \QCoh(\mathcal{O}_Z)$
which is a right adjoint to $i_*$. (Compare
Modules, Lemma \ref{modules-lemma-i-star-right-adjoint}.)
\end{lemma}
\begin{proof}
Given quasi-coherent $\mathcal{O}_X$-module $\mathcal{G}$ we consider
the subsheaf $\mathcal{H}_Z(\mathcal{G})$ of $\mathcal{G}$ of local sections
annihilated by $\mathcal{I}$. By
Lemma \ref{lemma-largest-quasi-coherent-subsheaf}
there is a canonical largest quasi-coherent $\mathcal{O}_X$-submodule
$\mathcal{H}_Z(\mathcal{G})'$. By construction we have
$$
\Hom_{\mathcal{O}_X}(i_*\mathcal{F}, \mathcal{H}_Z(\mathcal{G})')
=
\Hom_{\mathcal{O}_X}(i_*\mathcal{F}, \mathcal{G})
$$
for any quasi-coherent $\mathcal{O}_Z$-module $\mathcal{F}$.
Hence we can set $i^!\mathcal{G} = i^*(\mathcal{H}_Z(\mathcal{G})')$.
Details omitted.
\end{proof}
\noindent
Using the $1$-to-$1$ corresponding between quasi-coherent sheaves
of ideals and closed subspaces (see
Lemma \ref{lemma-closed-immersion-ideals})
we can define scheme theoretic intersections and unions
of closed subschemes.
\begin{definition}
\label{definition-scheme-theoretic-intersection-union}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $Z, Y \subset X$ be closed subspaces
corresponding to quasi-coherent ideal sheaves
$\mathcal{I}, \mathcal{J} \subset \mathcal{O}_X$.
The {\it scheme theoretic intersection} of $Z$ and $Y$
is the closed subspace of $X$ cut out by $\mathcal{I} + \mathcal{J}$.
Then {\it scheme theoretic union} of $Z$ and $Y$
is the closed subspace of $X$ cut out by
$\mathcal{I} \cap \mathcal{J}$.
\end{definition}
\noindent
It is clear that formation of scheme theoretic intersection
commutes with \'etale localization and the same is true for
scheme theoretic union.
\begin{lemma}
\label{lemma-scheme-theoretic-intersection}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $Z, Y \subset X$ be closed subspaces.
Let $Z \cap Y$ be the scheme theoretic intersection of $Z$ and $Y$.
Then $Z \cap Y \to Z$ and $Z \cap Y \to Y$ are closed immersions
and
$$
\xymatrix{
Z \cap Y \ar[r] \ar[d] & Z \ar[d] \\
Y \ar[r] & X
}
$$
is a cartesian diagram of algebraic spaces over $S$, i.e.,
$Z \cap Y = Z \times_X Y$.
\end{lemma}
\begin{proof}
The morphisms $Z \cap Y \to Z$ and $Z \cap Y \to Y$ are closed immersions
by Lemma \ref{lemma-closed-immersion-ideals}.
Since formation of the scheme theoretic intersection commutes
with \'etale localization we conclude the diagram is cartesian
by the case of schemes. See
Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-intersection}.
\end{proof}
\begin{lemma}
\label{lemma-scheme-theoretic-union}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $Y, Z \subset X$ be closed subspaces.
Let $Y \cup Z$ be the scheme theoretic union of $Y$ and $Z$.
Let $Y \cap Z$ be the scheme theoretic intersection of $Y$ and $Z$.
Then $Y \to Y \cup Z$ and $Z \to Y \cup Z$ are closed immersions,
there is a short exact sequence
$$
0 \to \mathcal{O}_{Y \cup Z} \to \mathcal{O}_Y \times \mathcal{O}_Z
\to \mathcal{O}_{Y \cap Z} \to 0
$$
of $\mathcal{O}_Z$-modules, and the diagram
$$
\xymatrix{
Y \cap Z \ar[r] \ar[d] & Y \ar[d] \\
Z \ar[r] & Y \cup Z
}
$$
is cocartesian in the category of algebraic spaces over $S$, i.e.,
$Y \cup Z = Y \amalg_{Y \cap Z} Z$.
\end{lemma}
\begin{proof}
The morphisms $Y \to Y \cup Z$ and $Z \to Y \cup Z$ are closed immersions
by Lemma \ref{lemma-closed-immersion-ideals}. In the short exact sequence
we use the equivalence of Lemma \ref{lemma-i-star-equivalence} to think of
quasi-coherent modules on closed subspaces of $X$ as quasi-coherent modules
on $X$. For the first map in the sequence we use the canonical maps
$\mathcal{O}_{Y \cup Z} \to \mathcal{O}_Y$ and
$\mathcal{O}_{Y \cup Z} \to \mathcal{O}_Z$
and for the second map we use the canonical map
$\mathcal{O}_Y \to \mathcal{O}_{Y \cap Z}$ and
the negative of the canonical map
$\mathcal{O}_Z \to \mathcal{O}_{Y \cap Z}$. Then to check
exactness we may work \'etale locally and deduce exactness
from the case of schemes
(Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-union}).
\medskip\noindent
To show the diagram is cocartesian, suppose we are given an algebraic space
$T$ over $S$ and morphisms $f : Y \to T$, $g : Z \to T$ agreeing as morphisms
$Y \cap Z \to T$. Goal: Show there exists a unique morphism
$h : Y \cup Z \to T$ agreeing with $f$ and $g$.
To construct $h$ we may work \'etale locally on $Y \cup Z$
(as $Y \cup Z$ is an \'etale sheaf being an algebraic space).
Hence we may assume that $X$ is a scheme.
In this case we know that $Y \cup Z$ is the pushout
of $Y$ and $Z$ along $Y \cap Z$ in the category of schemes
by Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-union}.
Choose a scheme $T'$ and a surjective \'etale morphism $T' \to T$.
Set $Y' = T' \times_{T, f} Y$ and $Z' = T' \times_{T, g} Z$.
Then $Y'$ and $Z'$ are schemes and we have a canonical isomorphism
$\varphi : Y' \times_Y (Y \cap Z) \to Z' \times_Z (Y \cap Z)$
of schemes. By More on Morphisms, Lemma
\ref{more-morphisms-lemma-pushout-along-closed-immersions}
the pushout $W' = Y' \amalg_{Y' \times_Y (Y \cap Z), \varphi} Z'$
exists in the category of schemes.
The morphism $W' \to Y \cup Z$ is \'etale by
More on Morphisms, Lemma
\ref{more-morphisms-lemma-pushout-along-closed-immersions-properties-above}.
It is surjective as $Y' \to Y$ and $Z' \to Z$ are surjective.
The morphisms $f' : Y' \to T'$ and $g' : Z' \to T'$
glue to a unique morphism of schemes $h' : W' \to T'$.
By uniqueness the composition $W' \to T' \to T$
descends to the desired morphism $h : Y \cup Z \to T$.
Some details omitted.
\end{proof}
\section{Supports of modules}
\label{section-support}
\noindent
In this section we collect some elementary results on supports of
quasi-coherent modules on algebraic spaces. Let $X$ be an algebraic
space. The support of an abelian sheaf on $X_\etale$
has been defined in Properties of Spaces, Section
\ref{spaces-properties-section-support}.
We use the same definition for supports of modules.
The following lemma tells us this agrees with the notion
as defined for quasi-coherent modules on schemes.
\begin{lemma}
\label{lemma-support-covering}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module.
Let $U$ be a scheme and let $\varphi : U \to X$ be an \'etale morphism.
Then
$$
\text{Supp}(\varphi^*\mathcal{F}) = |\varphi|^{-1}(\text{Supp}(\mathcal{F}))
$$
where the left hand side is the support of $\varphi^*\mathcal{F}$ as a
quasi-coherent module on the scheme $U$.
\end{lemma}
\begin{proof}
Let $u\in U$ be a (usual) point and let $\overline{x}$ be a
geometric point lying over $u$. By
Properties of Spaces, Lemma \ref{spaces-properties-lemma-stalk-quasi-coherent}
we have
$(\varphi^*\mathcal{F})_u \otimes_{\mathcal{O}_{U, u}}
\mathcal{O}_{X, \overline{x}} = \mathcal{F}_{\overline{x}}$.
Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \overline{x}}$
is the strict henselization by
Properties of Spaces, Lemma
\ref{spaces-properties-lemma-describe-etale-local-ring}
we see that it is faithfully flat (see
More on Algebra, Lemma
\ref{more-algebra-lemma-dumb-properties-henselization}).
Thus we see that $(\varphi^*\mathcal{F})_u = 0$ if and only if
$\mathcal{F}_{\overline{x}} = 0$. This proves the lemma.
\end{proof}
\noindent
For finite type quasi-coherent modules the support is closed,
can be checked on fibres, and commutes with base change.
\begin{lemma}
\label{lemma-support-finite-type}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_X$-module.
Then
\begin{enumerate}
\item The support of $\mathcal{F}$ is closed.
\item For a geometric point $\overline{x}$ lying over $x \in |X|$ we have
$$
x \in \text{Supp}(\mathcal{F})
\Leftrightarrow
\mathcal{F}_{\overline{x}} \not = 0
\Leftrightarrow
\mathcal{F}_{\overline{x}} \otimes_{\mathcal{O}_{X, \overline{x}}}
\kappa(\overline{x}) \not = 0.
$$
\item For any morphism of algebraic spaces $f : Y \to X$ the pullback
$f^*\mathcal{F}$ is of finite type as well and we have
$\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a scheme $U$ and a surjective \'etale morphism $\varphi : U \to X$.
By Lemma \ref{lemma-support-covering} the inverse image of the support of
$\mathcal{F}$ is the support of $\varphi^*\mathcal{F}$ which is closed by
Morphisms, Lemma \ref{morphisms-lemma-support-finite-type}.
Thus (1) follows from the definition of the topology on $|X|$.
\medskip\noindent
The first equivalence in (2) is the definition of support.
The second equivalence follows from Nakayama's lemma, see
Algebra, Lemma \ref{algebra-lemma-NAK}.
\medskip\noindent
Let $f : Y \to X$ be as in (3). Note that $f^*\mathcal{F}$ is of finite type
by Properties of Spaces, Section
\ref{spaces-properties-section-properties-modules}.
For the final assertion, let $\overline{y}$ be a geometric point of $Y$
mapping to the geometric point $\overline{x}$ on $X$. Recall that
$$
(f^*\mathcal{F})_{\overline{y}} =
\mathcal{F}_{\overline{x}} \otimes_{\mathcal{O}_{X, \overline{x}}}
\mathcal{O}_{Y, \overline{y}},
$$
see Properties of Spaces, Lemma
\ref{spaces-properties-lemma-stalk-pullback-quasi-coherent}.
Hence $(f^*\mathcal{F})_{\overline{y}} \otimes \kappa(\overline{y})$
is nonzero if and only if
$\mathcal{F}_{\overline{x}} \otimes \kappa(\overline{x})$ is nonzero.
By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only
if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of
assertion (3).
\end{proof}
\noindent
Our next task is to show that the scheme theoretic support
of a finite type quasi-coherent module (see
Morphisms, Definition \ref{morphisms-definition-scheme-theoretic-support})
also makes sense for finite type quasi-coherent modules on
algebraic spaces.
\begin{lemma}
\label{lemma-scheme-theoretic-support}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_X$-module.
There exists a smallest closed subspace $i : Z \to X$ such that there
exists a quasi-coherent $\mathcal{O}_Z$-module $\mathcal{G}$ with
$i_*\mathcal{G} \cong \mathcal{F}$. Moreover:
\begin{enumerate}
\item If $U$ is a scheme and $\varphi : U \to X$ is an \'etale morphism
then $Z \times_X U$ is the scheme theoretic support of $\varphi^*\mathcal{F}$.
\item The quasi-coherent sheaf $\mathcal{G}$ is unique up to unique
isomorphism.
\item The quasi-coherent sheaf $\mathcal{G}$ is of finite type.
\item The support of $\mathcal{G}$ and of $\mathcal{F}$ is $|Z|$.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a scheme $U$ and a surjective \'etale morphism $\varphi : U \to X$.
Let $R = U \times_X U$ with projections $s, t : R \to U$.
Let $i' : Z' \to U$ be the scheme theoretic support of $\varphi^*\mathcal{F}$
and let $\mathcal{G}'$ be the (unique up to unique isomorphism)
finite type quasi-coherent $\mathcal{O}_{Z'}$-module
with $i'_*\mathcal{G}' = \varphi^*\mathcal{F}$, see
Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-support}.
As $s^*\varphi^*\mathcal{F} = t^*\varphi^*\mathcal{F}$ we see that
$R' = s^{-1}Z' = t^{-1}Z'$ as closed subschemes of $R$ by
Morphisms, Lemma \ref{morphisms-lemma-flat-pullback-support}.
Thus we may apply Properties of Spaces, Lemma
\ref{spaces-properties-lemma-subspaces-presentation}
to find a closed subspace $i : Z \to X$ whose pullback to $U$ is $Z'$.
Writing $s', t' : R' \to Z'$ the projections and
$j' : R' \to R$ the given closed immersion, we see that
$$
j'_* (s')^*\mathcal{G}' = s^* i'_*\mathcal{G}' =
s^*\varphi^*\mathcal{F} = t^*\varphi^*\mathcal{F} =
t^*i'_*\mathcal{G}' = j'_*(t')^*\mathcal{G}'
$$
(the first and the last equality by Cohomology of Schemes,
Lemma \ref{coherent-lemma-flat-base-change-cohomology}).
Hence the uniqueness of
Morphisms, Lemma \ref{morphisms-lemma-flat-pullback-support}
applied to $R' \to R$ gives an isomorphism
$\alpha : (t')^*\mathcal{G}' \to (s')^*\mathcal{G}'$
compatible with the canonical isomorphism
$t^*\varphi^*\mathcal{F} = s^*\varphi^*\mathcal{F}$
via $j'_*$. Clearly $\alpha$ satisfies the cocycle condition, hence
we may apply
Properties of Spaces, Proposition
\ref{spaces-properties-proposition-quasi-coherent}
to obtain a quasi-coherent module $\mathcal{G}$ on $Z$ whose restriction
to $Z'$ is $\mathcal{G}'$ compatible with $\alpha$.
Again using the equivalence of the proposition mentioned above
(this time for $X$) we conclude that $i_*\mathcal{G} \cong \mathcal{F}$.
\medskip\noindent
This proves existence. The other properties of the lemma follow
by comparing with the result for schemes using
Lemma \ref{lemma-support-covering}.
Detailed proofs omitted.
\end{proof}
\begin{definition}
\label{definition-scheme-theoretic-support}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_X$-module.
The {\it scheme theoretic support of $\mathcal{F}$} is the closed subspace
$Z \subset X$ constructed in Lemma \ref{lemma-scheme-theoretic-support}.
\end{definition}
\noindent
In this situation we often think of $\mathcal{F}$ as a quasi-coherent
sheaf of finite type on $Z$ (via the equivalence of categories of
Lemma \ref{lemma-i-star-equivalence}).
\section{Scheme theoretic image}
\label{section-scheme-theoretic-image}
\noindent
Caution: Some of the material in this section is ultra-general and
behaves differently from what you might expect.
\begin{lemma}
\label{lemma-scheme-theoretic-image}
\begin{slogan}
The scheme-theoretic image of a morphism of algebraic spaces exists.
\end{slogan}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. There exists a closed subspace $Z \subset Y$ such that $f$ factors
through $Z$ and such that for any other closed subspace $Z' \subset Y$
such that $f$ factors through $Z'$ we have $Z \subset Z'$.
\end{lemma}
\begin{proof}
Let $\mathcal{I} = \Ker(\mathcal{O}_Y \to f_*\mathcal{O}_X)$.
If $\mathcal{I}$ is quasi-coherent then we just take $Z$ to be the
closed subscheme determined by $\mathcal{I}$, see
Lemma \ref{lemma-closed-immersion-ideals}.
In general the lemma requires us to show that there exists
a largest quasi-coherent sheaf of ideals $\mathcal{I}'$ contained in
$\mathcal{I}$.
This follows from Lemma \ref{lemma-largest-quasi-coherent-subsheaf}.
\end{proof}
\noindent
Suppose that in the situation of Lemma \ref{lemma-scheme-theoretic-image}
above $X$ and $Y$ are representable. Then the closed subspace $Z \subset Y$
found in the lemma agrees with the closed subscheme $Z \subset Y$ found in
Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-image}.
The reason is that closed subspaces (or subschemes) are in a inclusion
reversing correspondence with quasi-coherent ideal sheaves on
$X_\etale$ and $X$. As the category of quasi-coherent modules
on $X_\etale$ and $X$ are the same
(Properties of Spaces, Section \ref{spaces-properties-section-quasi-coherent})
we conclude. Thus the following definition agrees with the earlier
definition for morphisms of schemes.
\begin{definition}
\label{definition-scheme-theoretic-image}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. The {\it scheme theoretic image} of $f$ is the smallest closed
subspace $Z \subset Y$ through which $f$
factors, see Lemma \ref{lemma-scheme-theoretic-image} above.
\end{definition}
\noindent
We often just denote $f : X \to Z$ the factorization of $f$.
If the morphism $f$ is not quasi-compact, then (in general) the
construction of the scheme theoretic image does not commute with
restriction to open subspaces of $Y$.
\begin{lemma}
\label{lemma-quasi-compact-scheme-theoretic-image}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Let $Z \subset Y$ be the scheme theoretic image of $f$.
If $f$ is quasi-compact then
\begin{enumerate}
\item the sheaf of ideals
$\mathcal{I} = \Ker(\mathcal{O}_Y \to f_*\mathcal{O}_X)$
is quasi-coherent,
\item the scheme theoretic image $Z$ is the closed subspace
corresponding to $\mathcal{I}$,
\item for any \'etale morphism $V \to Y$ the scheme theoretic image of
$X \times_Y V \to V$ is equal to $Z \times_Y V$, and
\item the image $|f|(|X|) \subset |Z|$ is a dense subset of $|Z|$.
\end{enumerate}
\end{lemma}
\begin{proof}
To prove (3) it suffices to prove (1) and (2) since the
formation of $\mathcal{I}$ commutes with \'etale localization.
If (1) holds then in the proof of Lemma \ref{lemma-scheme-theoretic-image}
we showed (2). Let us prove that $\mathcal{I}$ is quasi-coherent.
Since the property of being quasi-coherent is \'etale local we may
assume $Y$ is an affine scheme. As $f$ is quasi-compact,
we can find an affine scheme $U$ and a surjective \'etale morphism
$U \to X$. Denote $f'$ the composition $U \to X \to Y$.
Then $f_*\mathcal{O}_X$ is a subsheaf of $f'_*\mathcal{O}_U$,
and hence $\mathcal{I} = \Ker(\mathcal{O}_Y \to \mathcal{O}_{X'})$.
By Lemma \ref{lemma-pushforward}
the sheaf $f'_*\mathcal{O}_U$ is quasi-coherent on $Y$. Hence $\mathcal{I}$
is quasi-coherent as a kernel of a map between coherent modules.
Finally, part (4) follows from parts (1), (2), and (3) as the ideal
$\mathcal{I}$ will be the unit ideal in any point of $|Y|$ which is
not contained in the closure of $|f|(|X|)$.
\end{proof}
\begin{lemma}
\label{lemma-scheme-theoretic-image-reduced}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Assume $X$ is reduced. Then
\begin{enumerate}
\item the scheme theoretic image $Z$ of $f$ is the reduced induced algebraic
space structure on $\overline{|f|(|X|)}$, and
\item for any \'etale morphism $V \to Y$ the scheme theoretic image of
$X \times_Y V \to V$ is equal to $Z \times_Y V$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) is true because the reduced induced algebraic space structure on
$\overline{|f|(|X|)}$ is the smallest closed subspace
of $Y$ through which $f$ factors, see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-map-into-reduction}.
Part (2) follows from (1), the fact that $|V| \to |Y|$ is open, and the
fact that being reduced is preserved under \'etale localization.
\end{proof}
\begin{lemma}
\label{lemma-reach-points-scheme-theoretic-image}
Let $S$ be a scheme.
Let $f : X \to Y$ be a quasi-compact morphism of algebraic spaces over $S$.
Let $Z$ be the scheme theoretic image of $f$.
Let $z \in |Z|$. There exists a valuation ring $A$ with
fraction field $K$ and a commutative diagram
$$
\xymatrix{
\Spec(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\
\Spec(A) \ar[r] & Z \ar[r] & Y
}
$$
such that the closed point of $\Spec(A)$ maps to $z$.
\end{lemma}
\begin{proof}
Choose an affine scheme $V$ with a point $z' \in V$
and an \'etale morphism $V \to Y$ mapping $z'$ to $z$.
Let $Z' \subset V$ be the scheme theoretic image of $X \times_Y V \to V$.
By Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} we have
$Z' = Z \times_Y V$. Thus $z' \in Z'$.
Since $f$ is quasi-compact and $V$ is affine we see that
$X \times_Y V$ is quasi-compact. Hence
there exists an affine scheme $W$ and a surjective \'etale
morphism $W \to X \times_Y V$. Then $Z' \subset V$ is also the
scheme theoretic image of $W \to V$.
By Morphisms, Lemma \ref{morphisms-lemma-reach-points-scheme-theoretic-image}
we can choose a diagram
$$
\xymatrix{
\Spec(K) \ar[r] \ar[d] &
W \ar[r] \ar[d] &
X \times_Y V \ar[d] \ar[r] &
X \ar[d] \\
\Spec(A) \ar[r] &
Z' \ar[r] &
V \ar[r] &
Y
}
$$
such that the closed point of $\Spec(A)$ maps to $z'$.
Composing with $Z' \to Z$ and $W \to X \times_Y V \to X$
we obtain a solution.
\end{proof}
\begin{lemma}
\label{lemma-factor-factor}
Let $S$ be a scheme. Let
$$
\xymatrix{
X_1 \ar[d] \ar[r]_{f_1} & Y_1 \ar[d] \\
X_2 \ar[r]^{f_2} & Y_2
}
$$
be a commutative diagram of algebraic spaces over $S$.
Let $Z_i \subset Y_i$, $i = 1, 2$ be
the scheme theoretic image of $f_i$. Then the morphism
$Y_1 \to Y_2$ induces a morphism $Z_1 \to Z_2$ and a
commutative diagram
$$
\xymatrix{
X_1 \ar[r] \ar[d] & Z_1 \ar[d] \ar[r] & Y_1 \ar[d] \\
X_2 \ar[r] & Z_2 \ar[r] & Y_2
}
$$
\end{lemma}
\begin{proof}
The scheme theoretic inverse image of $Z_2$ in $Y_1$
is a closed subspace of $Y_1$ through
which $f_1$ factors. Hence $Z_1$ is contained in this.
This proves the lemma.
\end{proof}
\begin{lemma}
\label{lemma-scheme-theoretic-image-of-partial-section}
Let $S$ be a scheme.
Let $f : X \to Y$ be a separated morphism of algebraic spaces over $S$.
Let $V \subset Y$ be an open subspace such that $V \to Y$ is quasi-compact.
Let $s : V \to X$ be a morphism such that $f \circ s = \text{id}_V$.
Let $Y'$ be the scheme theoretic image of $s$.
Then $Y' \to Y$ is an isomorphism over $V$.
\end{lemma}
\begin{proof}
By Lemma \ref{lemma-quasi-compact-permanence}
the morphism $s : V \to X$ is quasi-compact.
Hence the construction of the scheme theoretic image $Y'$
of $s$ commutes with restriction to opens by
Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}.
In particular, we see that $Y' \cap f^{-1}(V)$ is the
scheme theoretic image of a section of the separated
morphism $f^{-1}(V) \to V$. Since a section of a separated
morphism is a closed immersion
(Lemma \ref{lemma-section-immersion}),
we conclude that
$Y' \cap f^{-1}(V) \to V$ is an isomorphism as desired.
\end{proof}
\section{Scheme theoretic closure and density}
\label{section-scheme-theoretic-closure}
\noindent
This section is the analogue of
Morphisms, Section \ref{morphisms-section-scheme-theoretic-closure}.
\begin{lemma}
\label{lemma-scheme-theoretically-dense-representable}
Let $S$ be a scheme. Let $W \subset S$ be a scheme theoretically
dense open subscheme
(Morphisms, Definition \ref{morphisms-definition-scheme-theoretically-dense}).
Let $f : X \to S$ be a morphism of schemes which is flat, locally of
finite presentation, and locally quasi-finite.
Then $f^{-1}(W)$ is scheme theoretically dense in $X$.
\end{lemma}
\begin{proof}
We will use the characterization of Morphisms, Lemma
\ref{morphisms-lemma-characterize-scheme-theoretically-dense}.
Assume $V \subset X$ is an open and $g \in \Gamma(V, \mathcal{O}_V)$
is a function which restricts to zero on $f^{-1}(W) \cap V$.
We have to show that $g = 0$. Assume $g \not = 0$ to get a
contradiction. By
More on Morphisms, Lemma \ref{more-morphisms-lemma-go-down-with-annihilators}
we may shrink $V$, find an open $U \subset S$ fitting into a
commutative diagram
$$
\xymatrix{
V \ar[r] \ar[d]_\pi & X \ar[d]^f \\
U \ar[r] & S,
}
$$
a quasi-coherent subsheaf $\mathcal{F} \subset \mathcal{O}_U$, an integer
$r > 0$, and an injective $\mathcal{O}_U$-module map
$\mathcal{F}^{\oplus r} \to \pi_*\mathcal{O}_V$
whose image contains $g|_V$. Say
$(g_1, \ldots, g_r) \in \Gamma(U, \mathcal{F}^{\oplus r})$ maps to $g$.
Then we see that $g_i|_{W \cap U} = 0$ because $g|_{f^{-1}W \cap V} = 0$.
Hence $g_i = 0$ because $\mathcal{F} \subset \mathcal{O}_U$ and
$W$ is scheme theoretically dense in $S$.
This implies $g = 0$ which is the desired contradiction.
\end{proof}
\begin{lemma}
\label{lemma-scheme-theoretically-dense}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $U \subset X$ be an open subspace.
The following are equivalent
\begin{enumerate}
\item for every \'etale morphism $\varphi : V \to X$ (of algebraic spaces)
the scheme theoretic closure of $\varphi^{-1}(U)$ in $V$ is equal to $V$,
\item there exists a scheme $V$ and a surjective \'etale morphism
$\varphi : V \to X$ such that the scheme theoretic closure of
$\varphi^{-1}(U)$ in $V$ is equal to $V$,
\end{enumerate}
\end{lemma}
\begin{proof}
Observe that if $V \to V'$ is a morphism of algebraic spaces \'etale
over $X$, and $Z \subset V$, resp.\ $Z' \subset V'$ is the scheme theoretic
closure of $U \times_X V$, resp.\ $U \times_X V'$ in $V$, resp.\ $V'$,
then $Z$ maps into $Z'$. Thus if $V \to V'$ is surjective and \'etale
then $Z = V$ implies $Z' = V'$. Next, note that an \'etale morphism is
flat, locally of finite presentation, and locally quasi-finite
(see Morphisms, Section \ref{morphisms-section-etale}).
Thus Lemma \ref{lemma-scheme-theoretically-dense-representable}
implies that if $V$ and $V'$ are schemes, then $Z' = V'$ implies
$Z = V$. A formal argument using that every algebraic space has an
\'etale covering by a scheme shows that (1) and (2) are equivalent.
\end{proof}
\noindent
It follows from
Lemma \ref{lemma-scheme-theoretically-dense}
that the following definition is compatible with the definition
in the case of schemes.
\begin{definition}
\label{definition-scheme-theoretically-dense}
Let $S$ be a scheme.
Let $X$ be an algebraic space over $S$.
Let $U \subset X$ be an open subspace.
\begin{enumerate}
\item The scheme theoretic image of the morphism $U \to X$
is called the {\it scheme theoretic closure of $U$ in $X$}.
\item We say $U$ is {\it scheme theoretically dense in $X$}
if the equivalent conditions of
Lemma \ref{lemma-scheme-theoretically-dense} are satisfied.
\end{enumerate}
\end{definition}
\noindent
With this definition it is {\bf not} the case that $U$ is scheme
theoretically dense in $X$ if and only if the scheme theoretic closure
of $U$ is $X$. This is somewhat inelegant. But with suitable
finiteness conditions we will see that it does hold.
\begin{lemma}
\label{lemma-scheme-theoretically-dense-quasi-compact}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
Let $U \subset X$ be an open subspace.
If $U \to X$ is quasi-compact, then $U$
is scheme theoretically dense in $X$ if and only if the scheme theoretic
closure of $U$ in $X$ is $X$.
\end{lemma}
\begin{proof}
Follows from Lemma \ref{lemma-quasi-compact-scheme-theoretic-image} part (3).
\end{proof}
\begin{lemma}
\label{lemma-characterize-scheme-theoretically-dense}
Let $S$ be a scheme.
Let $j : U \to X$ be an open immersion of algebraic spaces over $S$.
Then $U$ is scheme theoretically dense in $X$ if and only if
$\mathcal{O}_X \to j_*\mathcal{O}_U$ is injective.
\end{lemma}
\begin{proof}
If $\mathcal{O}_X \to j_*\mathcal{O}_U$ is injective,
then the same is true when restricted to any
algebraic space $V$ \'etale over $X$.
Hence the scheme theoretic closure of $U \times_X V$ in $V$
is equal to $V$, see proof of
Lemma \ref{lemma-scheme-theoretic-image}.
Conversely, assume the scheme theoretic
closure of $U \times_X V$ is equal to $V$ for all $V$ \'etale over $X$.
Suppose that $\mathcal{O}_X \to j_*\mathcal{O}_U$ is not injective.
Then we can find an affine, say $V = \Spec(A)$, \'etale over $X$
and a nonzero element $f \in A$ such that $f$ maps to zero in
$\Gamma(V \times_X U, \mathcal{O})$. In this case the scheme theoretic
closure of $V \times_X U$ in $V$ is clearly contained in $\Spec(A/(f))$
a contradiction.
\end{proof}
\begin{lemma}
\label{lemma-intersection-scheme-theoretically-dense}
Let $S$ be a scheme. Let $X$ be an algebraic space over $S$.
If $U$, $V$ are scheme theoretically dense
open subspaces of $X$, then so is $U \cap V$.
\end{lemma}
\begin{proof}
Let $W \to X$ be any \'etale morphism. Consider the map
$\mathcal{O}(W) \to \mathcal{O}(W \times_X V)
\to \mathcal{O}(W \times_X (V \cap U))$.
By Lemma \ref{lemma-characterize-scheme-theoretically-dense}
both maps are injective. Hence the composite is injective.
Hence by Lemma \ref{lemma-characterize-scheme-theoretically-dense}
$U \cap V$ is scheme theoretically dense in $X$.
\end{proof}
\begin{lemma}
\label{lemma-quasi-compact-immersion}
Let $S$ be a scheme. Let $h : Z \to X$ be an immersion of algebraic spaces
over $S$. Assume either $Z \to X$ is quasi-compact or $Z$ is reduced.
Let $\overline{Z} \subset X$ be the scheme theoretic image of $h$.
Then the morphism $Z \to \overline{Z}$ is an open immersion
which identifies $Z$ with a scheme theoretically dense open
subspace of $\overline{Z}$. Moreover, $Z$ is topologically
dense in $\overline{Z}$.
\end{lemma}
\begin{proof}
In both cases the formation of $\overline{Z}$ commutes with
\'etale localization, see
Lemmas \ref{lemma-quasi-compact-scheme-theoretic-image} and
\ref{lemma-scheme-theoretic-image-reduced}.
Hence this lemma follows from the case of schemes, see
Morphisms, Lemma \ref{morphisms-lemma-quasi-compact-immersion}.
\end{proof}
\begin{lemma}
\label{lemma-equality-of-morphisms}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $f, g : X \to Y$ be morphisms of algebraic spaces over $B$.
Let $U \subset X$ be an open subspace such that
$f|_U = g|_U$. If the scheme theoretic closure of $U$
in $X$ is $X$ and $Y \to B$ is separated, then $f = g$.
\end{lemma}
\begin{proof}
As $Y \to B$ is separated the fibre product
$Y \times_{\Delta, Y \times_B Y, (f, g)} X$ is a closed subspace $Z \subset X$.
As $f|_U = g|_U$ we see that $U \subset Z$. Hence $Z = X$ as $U$ is assumed
scheme theoretically dense in $X$.
\end{proof}
\section{Dominant morphisms}
\label{section-dominant}
\noindent
We copy the definition of a dominant morphism of schemes to get the
notion of a dominant morphism of algebraic spaces. We caution the
reader that this definition is not well behaved unless the morphism
is quasi-compact and the algebraic spaces satisfy some separation
axioms.
\begin{definition}
\label{definition-dominant}
Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$ is
called {\it dominant} if the image of $|f| : |X| \to |Y|$ is dense in $|Y|$.
\end{definition}
\section{Universally injective morphisms}
\label{section-universally-injective}
\noindent
We have already defined in Section \ref{section-representable}
what it means for a representable morphism of algebraic spaces
to be universally injective. For a field $K$ over $S$ (recall this means that
we are given a structure morphism $\Spec(K) \to S$) and an
algebraic space $X$ over $S$ we write
$X(K) = \Mor_S(\Spec(K), X)$. We first translate the
condition for representable morphisms into a condition on the functor
of points.
\begin{lemma}
\label{lemma-universally-injective-representable}
Let $S$ be a scheme. Let $f : X \to Y$ be a representable
morphism of algebraic spaces over $S$. Then $f$ is universally injective
(in the sense of Section \ref{section-representable})
if and only if for all fields $K$ the map $X(K) \to Y(K)$ is injective.
\end{lemma}
\begin{proof}
We are going to use
Morphisms, Lemma \ref{morphisms-lemma-universally-injective}
without further mention.
Suppose that $f$ is universally injective. Then for any field $K$ and any
morphism $\Spec(K) \to Y$ the morphism of schemes
$\Spec(K) \times_Y X \to \Spec(K)$ is universally injective.
Hence there exists at most one section of the morphism
$\Spec(K) \times_Y X \to \Spec(K)$. Hence the map
$X(K) \to Y(K)$ is injective. Conversely, suppose that for every field $K$
the map $X(K) \to Y(K)$ is injective. Let $T \to Y$ be a morphism from a
scheme into $Y$, and consider the base change $f_T : T \times_Y X \to T$.
For any field $K$ we have
$$
(T \times_Y X)(K) = T(K) \times_{Y(K)} X(K)
$$
by definition of the fibre product, and hence the injectivity of
$X(K) \to Y(K)$ guarantees the injectivity of
$(T \times_Y X)(K) \to T(K)$ which means that $f_T$ is universally injective
as desired.
\end{proof}
\noindent
Next, we translate the property that the transformation between field valued
points is injective into something more geometric.
\begin{lemma}
\label{lemma-universally-injective}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
The following are equivalent:
\begin{enumerate}
\item the map $X(K) \to Y(K)$ is injective for every field $K$ over $S$
\item for every morphism $Y' \to Y$ of algebraic spaces over $S$
the induced map $|Y' \times_Y X| \to |Y'|$ is injective, and
\item the diagonal morphism $X \to X \times_Y X$ is surjective.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1). Let $g : Y' \to Y$ be a morphism of algebraic
spaces, and denote $f' : Y' \times_Y X \to Y'$ the base change of $f$.
Let $K_i$, $i = 1, 2$ be fields and let
$\varphi_i : \Spec(K_i) \to Y' \times_Y X$ be morphisms
such that $f' \circ \varphi_1$ and $f' \circ \varphi_2$ define the
same element of $|Y'|$. By definition this means there exists a
field $\Omega$ and embeddings $\alpha_i : K_i \subset \Omega$ such that
the two morphisms
$f' \circ \varphi_i \circ \alpha_i : \Spec(\Omega) \to Y'$ are equal.
Here is the corresponding commutative diagram
$$
\xymatrix{
\Spec(\Omega) \ar@/_5ex/[ddrr] \ar[rd]^{\alpha_1} \ar[r]_{\alpha_2} &
\Spec(K_2) \ar[rd]^{\varphi_2} \\
& \Spec(K_1) \ar[r]^{\varphi_1} &
Y' \times_Y X \ar[d]^{f'} \ar[r]^{g'} &
X \ar[d]^f \\
& & Y' \ar[r]^g & Y.
}
$$
In particular the compositions $g \circ f' \circ \varphi_i \circ \alpha_i$
are equal. By assumption (1) this implies that the morphism
$g' \circ \varphi_i \circ \alpha_i$ are equal, where $g' : Y' \times_Y X \to X$
is the projection. By the universal property of the fibre product we conclude
that the morphisms
$\varphi_i \circ \alpha_i : \Spec(\Omega) \to Y' \times_Y X$ are
equal. In other words $\varphi_1$ and $\varphi_2$ define the same point
of $Y' \times_Y X$. We conclude that (2) holds.
\medskip\noindent
Assume (2). Let $K$ be a field over $S$, and let $a, b : \Spec(K) \to X$
be two morphisms such that $f \circ a = f \circ b$. Denote
$c : \Spec(K) \to Y$ the common value. By assumption
$|\Spec(K) \times_{c, Y} X| \to |\Spec(K)|$ is injective.
This means there exists a field $\Omega$ and embeddings
$\alpha_i : K \to \Omega$ such that
$$
\xymatrix{
\Spec(\Omega) \ar[r]_{\alpha_1} \ar[d]_{\alpha_2} &
\Spec(K) \ar[d]^a \\
\Spec(K) \ar[r]^-b &
\Spec(K) \times_{c, Y} X
}
$$
is commutative. Composing with the projection to $\Spec(K)$
we see that $\alpha_1 = \alpha_2$. Denote the common value $\alpha$.
Then we see that $\{\alpha : \Spec(\Omega) \to \Spec(K)\}$
is a fpqc covering of $\Spec(K)$ such that the two morphisms
$a, b$ become equal on the members of the covering. By
Properties of Spaces, Proposition
\ref{spaces-properties-proposition-sheaf-fpqc}
we conclude that $a = b$. We conclude that (1) holds.
\medskip\noindent
Assume (3). Let $x, x' \in |X|$ be a pair of points such that
$f(x) = f(x')$ in $|Y|$. By
Properties of Spaces, Lemma \ref{spaces-properties-lemma-points-cartesian}
we see there exists a $x'' \in |X \times_Y X|$ whose projections
are $x$ and $x'$. By assumption and
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-characterize-surjective}
there exists a $x''' \in |X|$ with $\Delta_{X/Y}(x''') = x''$.