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\input{preamble}
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\begin{document}
\title{More on Morphisms of Spaces}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
In this chapter we continue our study of properties of morphisms of algebraic
spaces. A fundamental reference is \cite{Kn}.
\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{Radicial morphisms}
\label{section-radicial}
\noindent
It turns out that a radicial morphism is not the same thing as a
universally injective morphism, contrary to what happens with
morphisms of schemes. In fact it is a bit stronger.
\begin{definition}
\label{definition-radicial}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. We say $f$ is {\it radicial} if for any morphism
$\Spec(K) \to Y$ where $K$ is a field the reduction
$(\Spec(K) \times_Y X)_{red}$ is either empty or
representable by the spectrum of a purely inseparable field extension of $K$.
\end{definition}
\begin{lemma}
\label{lemma-radicial-implies-universally-injective}
A radicial morphism of algebraic spaces is universally injective.
\end{lemma}
\begin{proof}
Let $S$ be a scheme. Let $f : X \to Y$ be a radicial
morphism of algebraic spaces over $S$.
It is clear from the definition that given a morphism
$\Spec(K) \to Y$ there is at most one lift of this morphism
to a morphism into $X$. Hence we conclude that $f$ is universally
injective by
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-universally-injective}.
\end{proof}
\begin{example}
\label{example-universally-injective-not-radicial}
It is no longer true that universally injective is equivalent to radicial.
For example the morphism
$$
X = [\Spec(\overline{\mathbf{Q}})/
\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})]
\longrightarrow
S = \Spec(\mathbf{Q})
$$
of
Spaces, Example \ref{spaces-example-Qbar}
is universally injective, but is not radicial in the sense above.
\end{example}
\noindent
Nonetheless it is often the case that the reverse implication holds.
\begin{lemma}
\label{lemma-when-universally-injective-radicial}
Let $S$ be a scheme. Let $f : X \to Y$ be a universally injective
morphism of algebraic spaces over $S$.
\begin{enumerate}
\item If $f$ is decent then $f$ is radicial.
\item If $f$ is quasi-separated then $f$ is radicial.
\item If $f$ is locally separated then $f$ is radicial.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $\mathcal{P}$ be a property of morphisms of algebraic spaces
which is stable under base change and composition and holds for
closed immersions. Assume $f : X \to Y$ has $\mathcal{P}$ and
is universally injective. Then, in the situation of
Definition \ref{definition-radicial}
the morphism $(\Spec(K) \times_Y X)_{red} \to \Spec(K)$
is universally injective and has $\mathcal{P}$. This reduces the
problem of proving
$$
\mathcal{P} + \text{universally injective}
\Rightarrow
\text{radicial}
$$
to the problem of proving that any nonempty reduced algebraic space $X$
over field whose structure morphism $X \to \Spec(K)$ is universally
injective and $\mathcal{P}$ is representable by the spectrum of a field.
Namely, then $X \to \Spec(K)$ will be a morphism of schemes and
we conclude by the equivalence of radicial and universally injective for
morphisms of schemes, see
Morphisms, Lemma \ref{morphisms-lemma-universally-injective}.
\medskip\noindent
Let us prove (1). Assume $f$ is decent and universally injective. By
Decent Spaces,
Lemmas \ref{decent-spaces-lemma-base-change-relative-conditions},
\ref{decent-spaces-lemma-composition-relative-conditions}, and
\ref{decent-spaces-lemma-properties-trivial-implications}
(to see that an immersion is decent) we see that the discussion in
the first paragraph applies.
Let $X$ be a nonempty decent reduced algebraic space
universally injective over a field $K$. In particular we see that $|X|$
is a singleton. By
Decent Spaces, Lemma \ref{decent-spaces-lemma-when-field}
we conclude that $X \cong \Spec(L)$ for some extension
$K \subset L$ as desired.
\medskip\noindent
A quasi-separated morphism is decent, see
Decent Spaces,
Lemma \ref{decent-spaces-lemma-properties-trivial-implications}.
Hence (1) implies (2).
\medskip\noindent
Let us prove (3).
Recall that the separation axioms are stable under base change
and composition and that closed immersions are separated, see
Morphisms of Spaces,
Lemmas \ref{spaces-morphisms-lemma-base-change-separated},
\ref{spaces-morphisms-lemma-composition-separated}, and
\ref{spaces-morphisms-lemma-immersions-monomorphisms}.
Thus the discussion in the first paragraph of the proof applies.
Let $X$ be a reduced algebraic space universally injective and
locally separated over a field $K$.
In particular $|X|$ is a singleton hence $X$ is quasi-compact, see
Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-space}.
We can find a surjective \'etale morphism $U \to X$ with $U$ affine, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover}.
Consider the morphism of schemes
$$
j :
U \times_X U
\longrightarrow
U \times_{\Spec(K)} U
$$
As $X \to \Spec(K)$ is universally injective $j$ is surjective,
and as $X \to \Spec(K)$ is locally separated $j$ is an immersion.
A surjective immersion is a closed immersion, see
Schemes, Lemma \ref{schemes-lemma-immersion-when-closed}.
Hence $R = U \times_X U$ is affine as a closed subscheme of an affine scheme.
In particular $R$ is quasi-compact.
It follows that $X = U/R$ is quasi-separated, and the result follows from (2).
\end{proof}
\begin{remark}
\label{remark-weakly-radicial}
Let $X \to Y$ be a morphism of algebraic spaces.
For some applications (of radicial morphisms)
it is enough to require that for every
$\Spec(K) \to Y$ where $K$ is a field
\begin{enumerate}
\item the space $|\Spec(K) \times_Y X|$ is a singleton,
\item there exists a monomorphism
$\Spec(L) \to \Spec(K) \times_Y X$, and
\item $K \subset L$ is purely inseparable.
\end{enumerate}
If needed later we will may call such a morphism {\it weakly radicial}.
For example if $X \to Y$ is a surjective weakly radicial morphism
then $X(k) \to Y(k)$ is surjective for every algebraically closed field $k$.
Note that the base change
$X_{\overline{\mathbf{Q}}} \to \Spec(\overline{\mathbf{Q}})$
of the morphism in
Example \ref{example-universally-injective-not-radicial}
is weakly radicial, but not radicial. The analogue of
Lemma \ref{lemma-when-universally-injective-radicial}
is that if $X \to Y$ has property ($\beta$) and is universally
injective, then it is weakly radicial (proof omitted).
\end{remark}
\begin{lemma}
\label{lemma-check-universally-injective}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic
spaces over $S$. Assume
\begin{enumerate}
\item $f$ is locally of finite type,
\item for every \'etale morphism $V \to Y$ the map $|X \times_Y V| \to |V|$
is injective.
\end{enumerate}
Then $f$ is universally injective.
\end{lemma}
\begin{proof}
The question is \'etale local on $Y$ by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-universally-injective-local}.
Hence we may assume that $Y$ is a scheme.
Then $Y$ is in particular decent and by Decent Spaces, Lemma
\ref{decent-spaces-lemma-conditions-on-point-in-fibre-and-qf}
we see that $f$ is locally quasi-finite.
Let $y \in Y$ be a point and let $X_y$ be the scheme theoretic
fibre. Assume $X_y$ is not empty. By Spaces over Fields, Lemma
\ref{spaces-over-fields-lemma-locally-quasi-finite-over-field}
we see that $X_y$ is a scheme which is locally quasi-finite over
$\kappa(y)$. Since $|X_y| \subset |X|$ is the fibre of $|X| \to |Y|$
over $y$ we see that $X_y$ has a unique point $x$. The same is true
for $X_y \times_{\Spec(\kappa(y))} \Spec(k)$ for any
finite separable extension $\kappa(y) \subset k$
because we can realize $k$ as the residue field at a point
lying over $y$ in an \'etale scheme over $Y$, see
see More on Morphisms, Lemma
\ref{more-morphisms-lemma-realize-prescribed-residue-field-extension-etale}.
Thus $X_y$ is geometrically connected, see
Varieties, Lemma \ref{varieties-lemma-characterize-geometrically-disconnected}.
This implies that the finite extension $\kappa(y) \subset \kappa(x)$
is purely inseparable.
\medskip\noindent
We conclude (in the case that $Y$ is a scheme)
that for every $y \in Y$ either the fibre $X_y$ is empty,
or $(X_y)_{red} = \Spec(\kappa(x))$ with
$\kappa(y) \subset \kappa(x)$ purely inseparable.
Hence $f$ is radicial (some details omitted), whence universally injective by
Lemma \ref{lemma-radicial-implies-universally-injective}.
\end{proof}
\section{Monomorphisms}
\label{section-monomorphisms}
\noindent
This section is the continuation of
Morphisms of Spaces, Section \ref{spaces-morphisms-section-monomorphisms}.
We would like to know whether or not every monomorphism of algebraic
spaces is representable. If you can prove this is true or have a
counterexample, please email
\href{mailto:stacks.project@gmail.com}{stacks.project@gmail.com}.
For the moment this is known in the following cases
\begin{enumerate}
\item for monomorphisms which are locally of finite type
(more generally any separated, locally quasi-finite morphism
is representable by Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-locally-quasi-finite-separated-representable}
and a monomorphism which is locally of finite type is
locally quasi-finite by Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite}),
\item if the target is a disjoint union of spectra of zero dimensional
local rings (Decent Spaces, Lemma
\ref{decent-spaces-lemma-monomorphism-toward-disjoint-union-dim-0-rings}), and
\item for flat monomorphisms (see below).
\end{enumerate}
\begin{lemma}[David Rydh]
\label{lemma-flat-case}
A flat monomorphism of algebraic spaces is representable by schemes.
\end{lemma}
\begin{proof}
Let $f : X \to Y$ be a flat morphism of algebraic spaces.
To prove $f$ is representable, we have to show
$X \times_Y V$ is a scheme for every scheme $V$ mapping to $Y$.
Since being a scheme is local (Properties of Spaces,
Lemma \ref{spaces-properties-lemma-subscheme}), we may
assume $V$ is affine. Thus we may assume $Y = \Spec(B)$
is an affine scheme. Next, we can assume that $X$ is quasi-compact
by replacing $X$ by a quasi-compact open. The space $X$ is
separated as $X \to X \times_{\Spec(B)} X$ is an isomorphism.
Applying Limits of Spaces, Lemma \ref{spaces-limits-lemma-enough-local}
we reduce to the case where $B$ is local, $X \to \Spec(B)$ is a
flat monomorphism, and
there exists a point $x \in X$ mapping to the closed point of $\Spec(B)$.
Then $X \to \Spec(B)$ is surjective as generalizations
lift along flat morphisms of separated algebraic spaces, see
Decent Spaces, Lemma \ref{decent-spaces-lemma-generalizations-lift-flat}.
Hence we see that $\{X \to \Spec(B)\}$ is an fpqc cover.
Then $X \to \Spec(B)$ is a morphism which becomes an isomorphism
after base change by $X \to \Spec(B)$. Hence it is an isomorphism by
fpqc descent, see Descent on Spaces, Lemma
\ref{spaces-descent-lemma-descending-property-isomorphism}.
\end{proof}
\noindent
The following is (in some sense) a variant of the lemma above.
\begin{lemma}
\label{lemma-ui-case}
Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact monomorphism
of algebraic spaces $f : X \to Y$ such that for every $T \to X$ the map
$$
\mathcal{O}_T \to f_{T,*}\mathcal{O}_{X \times_Y T}
$$
is injective. Then $f$ is an isomorphism (and hence representable by schemes).
\end{lemma}
\begin{proof}
The question is \'etale local on $Y$, hence we may assume $Y = \Spec(A)$
is affine. Then $X$ is quasi-compact and we may choose an affine scheme
$U = \Spec(B)$ and a surjective \'etale morphism $U \to X$
(Properties of Spaces, Lemma
\ref{spaces-properties-lemma-quasi-compact-affine-cover}).
Note that $U \times_X U = \Spec(B \otimes_A B)$. Hence the category of
quasi-coherent $\mathcal{O}_X$-modules is equivalent to the
category $DD_{B/A}$ of descent data on modules for $A \to B$.
See Properties of Spaces, Proposition
\ref{spaces-properties-proposition-quasi-coherent},
Descent, Definition \ref{descent-definition-descent-datum-modules}, and
Descent, Subsection \ref{descent-subsection-descent-modules-morphisms}.
On the other hand,
$$
A \to B
$$
is a universally injective ring map. Namely, given an
$A$-module $M$ we see that $A \oplus M \to B \otimes_A (A \oplus M)$
is injective by the assumption of the lemma. Hence
$DD_{B/A}$ is equivalent to the category of $A$-modules by
Descent, Theorem \ref{descent-theorem-descent}. Thus pullback along
$f : X \to \Spec(A)$ determines an equivalence of categories of
quasi-coherent modules. In particular $f^*$ is exact on
quasi-coherent modules and we see that $f$ is flat
(small detail omitted). Moreover, it is clear that $f$ is surjective
(for example because $\Spec(B) \to \Spec(A)$ is surjective).
Hence we see that $\{X \to \Spec(A)\}$ is an fpqc cover.
Then $X \to \Spec(A)$ is a morphism which becomes an isomorphism
after base change by $X \to \Spec(A)$. Hence it is an isomorphism by
fpqc descent, see Descent on Spaces, Lemma
\ref{spaces-descent-lemma-descending-property-isomorphism}.
\end{proof}
\begin{lemma}
\label{lemma-flat-surjective-monomorphism}
A quasi-compact flat surjective monomorphism of algebraic spaces
is an isomorphism.
\end{lemma}
\begin{proof}
Such a morphism satisfies the assumptions of Lemma \ref{lemma-ui-case}.
\end{proof}
\section{Conormal sheaf of an immersion}
\label{section-conormal-sheaf}
\noindent
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 corresponding
quasi-coherent sheaf of ideals, see
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}.
Consider the short exact sequence
$$
0 \to \mathcal{I}^2 \to \mathcal{I} \to \mathcal{I}/\mathcal{I}^2 \to 0
$$
of quasi-coherent sheaves on $X$. Since the sheaf $\mathcal{I}/\mathcal{I}^2$
is annihilated by $\mathcal{I}$ it corresponds to a sheaf on $Z$ by
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-i-star-equivalence}.
This quasi-coherent $\mathcal{O}_Z$-module is the
{\it conormal sheaf of $Z$ in $X$} and is often denoted
$\mathcal{I}/\mathcal{I}^2$ by the abuse of notation mentioned in
Morphisms of Spaces,
Section \ref{spaces-morphisms-section-closed-immersions-quasi-coherent}.
\medskip\noindent
In case $i : Z \to X$ is a (locally closed) immersion we define the
conormal sheaf of $i$ as the conormal sheaf of the closed
immersion $i : Z \to X \setminus \partial Z$, see
Morphisms of Spaces, Remark \ref{spaces-morphisms-remark-immersion}.
It is often denoted
$\mathcal{I}/\mathcal{I}^2$ where $\mathcal{I}$ is the ideal sheaf
of the closed immersion $i : Z \to X \setminus \partial Z$.
\begin{definition}
\label{definition-conormal-sheaf}
Let $i : Z \to X$ be an immersion. The {\it conormal sheaf
$\mathcal{C}_{Z/X}$ of $Z$ in $X$} or the {\it conormal sheaf of $i$}
is the quasi-coherent $\mathcal{O}_Z$-module $\mathcal{I}/\mathcal{I}^2$
described above.
\end{definition}
\noindent
In \cite[IV Definition 16.1.2]{EGA} this sheaf is denoted
$\mathcal{N}_{Z/X}$. We will not follow this convention since we would
like to reserve the notation $\mathcal{N}_{Z/X}$
for the {\it normal sheaf of the immersion}. It is defined as
$$
\mathcal{N}_{Z/X} =
\SheafHom_{\mathcal{O}_Z}(\mathcal{C}_{Z/X}, \mathcal{O}_Z) =
\SheafHom_{\mathcal{O}_Z}(\mathcal{I}/\mathcal{I}^2, \mathcal{O}_Z)
$$
provided the conormal sheaf is of finite presentation (otherwise the
normal sheaf may not even be quasi-coherent). We will come back to the
normal sheaf later (insert future reference here).
\begin{lemma}
\label{lemma-etale-conormal}
Let $S$ be a scheme. Let $i : Z \to X$ be an immersion.
Let $\varphi : U \to X$ be an \'etale morphism where $U$ is a scheme.
Set $Z_U = U \times_X Z$ which is a locally closed subscheme of $U$.
Then
$$
\mathcal{C}_{Z/X}|_{Z_U} = \mathcal{C}_{Z_U/U}
$$
canonically and functorially in $U$.
\end{lemma}
\begin{proof}
Let $T \subset X$ be a closed subspace such that $i$ defines a closed
immersion into $X \setminus T$.
Let $\mathcal{I}$ be the quasi-coherent sheaf of ideals on
$X \setminus T$ defining $Z$. Then the lemma just states that
$\mathcal{I}|_{U \setminus \varphi^{-1}(T)}$ is the sheaf of ideals of
the immersion $Z_U \to U \setminus \varphi^{-1}(T)$.
This is clear from the construction of $\mathcal{I}$ in
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}.
\end{proof}
\begin{lemma}
\label{lemma-conormal-functorial}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\
Z' \ar[r]^{i'} & X'
}
$$
be a commutative diagram of algebraic spaces over $S$.
Assume $i$, $i'$ immersions. There is a canonical map
of $\mathcal{O}_Z$-modules
$$
f^*\mathcal{C}_{Z'/X'}
\longrightarrow
\mathcal{C}_{Z/X}
$$
\end{lemma}
\begin{proof}
First find open subspaces $U' \subset X'$ and $U \subset X$ such that
$g(U) \subset U'$ and such that $i(Z) \subset U$ and $i(Z') \subset U'$
are closed (proof existence omitted). Replacing $X$ by $U$ and $X'$ by
$U'$ we may assume that $i$ and $i'$ are closed immersions.
Let $\mathcal{I}' \subset \mathcal{O}_{X'}$ and
$\mathcal{I} \subset \mathcal{O}_X$ be the quasi-coherent sheaves of
ideals associated to $i'$ and $i$, see
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}.
Consider the composition
$$
g^{-1}\mathcal{I}' \to g^{-1}\mathcal{O}_{X'}
\xrightarrow{g^\sharp} \mathcal{O}_X \to
\mathcal{O}_X/\mathcal{I} = i_*\mathcal{O}_Z
$$
Since $g(i(Z)) \subset Z'$ we conclude this composition is zero (see
statement on factorizations in
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}).
Thus we obtain a commutative diagram
$$
\xymatrix{
0 \ar[r] &
\mathcal{I} \ar[r] &
\mathcal{O}_X \ar[r] &
i_*\mathcal{O}_Z \ar[r] &
0 \\
0 \ar[r] &
g^{-1}\mathcal{I}' \ar[r] \ar[u] &
g^{-1}\mathcal{O}_{X'} \ar[r] \ar[u] &
g^{-1}i'_*\mathcal{O}_{Z'} \ar[r] \ar[u] &
0
}
$$
The lower row is exact since $g^{-1}$ is an exact functor.
By exactness we also see that
$(g^{-1}\mathcal{I}')^2 = g^{-1}((\mathcal{I}')^2)$.
Hence the diagram induces a map
$g^{-1}(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{I}/\mathcal{I}^2$.
Pulling back (using $i^{-1}$ for example) to $Z$ we obtain
$i^{-1}g^{-1}(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{C}_{Z/X}$.
Since $i^{-1}g^{-1} = f^{-1}(i')^{-1}$ this gives a map
$f^{-1}\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}$, which induces
the desired map.
\end{proof}
\begin{lemma}
\label{lemma-conormal-functorial-more}
Let $S$ be a scheme. The conormal sheaf of
Definition \ref{definition-conormal-sheaf}, and its functoriality of
Lemma \ref{lemma-conormal-functorial}
satisfy the following properties:
\begin{enumerate}
\item If $Z \to X$ is an immersion of schemes over $S$, then the conormal
sheaf agrees with the one from
Morphisms, Definition \ref{morphisms-definition-conormal-sheaf}.
\item If in
Lemma \ref{lemma-conormal-functorial}
all the spaces are schemes, then the map
$f^*\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}$ is the same
as the one constructed in
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial}.
\item Given a commutative diagram
$$
\xymatrix{
Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\
Z' \ar[r]^{i'} \ar[d]_{f'} & X' \ar[d]^{g'} \\
Z'' \ar[r]^{i''} & X''
}
$$
then the map $(f' \circ f)^*\mathcal{C}_{Z''/X''} \to \mathcal{C}_{Z/X}$
is the same as the composition of
$f^*\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}$
with the pullback by $f$ of
$(f')^*\mathcal{C}_{Z''/X''} \to \mathcal{C}_{Z'/X'}$
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Note that Part (1) is a special case of
Lemma \ref{lemma-etale-conormal}.
\end{proof}
\begin{lemma}
\label{lemma-conormal-functorial-flat}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\
Z' \ar[r]^{i'} & X'
}
$$
be a fibre product diagram of algebraic spaces over $S$. Assume
$i$, $i'$ immersions. Then the canonical map
$f^*\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}$ of
Lemma \ref{lemma-conormal-functorial}
is surjective. If $g$ is flat, then it is an isomorphism.
\end{lemma}
\begin{proof}
Choose a commutative diagram
$$
\xymatrix{
U \ar[r] \ar[d] & X \ar[d] \\
U' \ar[r] & X'
}
$$
where $U$, $U'$ are schemes and the horizontal arrows are surjective
and \'etale, see
Spaces, Lemma \ref{spaces-lemma-lift-morphism-presentations}.
Then using
Lemmas \ref{lemma-etale-conormal} and \ref{lemma-conormal-functorial-more}
we see that the question reduces to the case of a morphism of schemes.
In the schemes case this is
Morphisms, Lemma \ref{morphisms-lemma-conormal-functorial-flat}.
\end{proof}
\begin{lemma}
\label{lemma-transitivity-conormal}
Let $S$ be a scheme.
Let $Z \to Y \to X$ be immersions of algebraic spaces.
Then there is a canonical exact sequence
$$
i^*\mathcal{C}_{Y/X} \to
\mathcal{C}_{Z/X} \to
\mathcal{C}_{Z/Y} \to 0
$$
where the maps come from
Lemma \ref{lemma-conormal-functorial}
and $i : Z \to Y$ is the first morphism.
\end{lemma}
\begin{proof}
Let $U$ be a scheme and let $U \to X$ be a surjective \'etale morphism. Via
Lemmas \ref{lemma-etale-conormal} and \ref{lemma-conormal-functorial-more}
the exactness of the sequence translates immediately into the
exactness of the corresponding sequence for the immersions of schemes
$Z \times_X U \to Y \times_X U \to U$. Hence the lemma follows from
Morphisms, Lemma \ref{morphisms-lemma-transitivity-conormal}.
\end{proof}
\section{The normal cone of an immersion}
\label{section-normal-cone}
\noindent
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
corresponding quasi-coherent sheaf of ideals, see
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-closed-immersion-ideals}.
Consider the quasi-coherent sheaf of graded $\mathcal{O}_X$-algebras
$\bigoplus_{n \geq 0} \mathcal{I}^n/\mathcal{I}^{n + 1}$.
Since the sheaves $\mathcal{I}^n/\mathcal{I}^{n + 1}$
are each annihilated by $\mathcal{I}$ this graded algebra
corresponds to a quasi-coherent sheaf of graded $\mathcal{O}_Z$-algebras by
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-i-star-equivalence}.
This quasi-coherent graded $\mathcal{O}_Z$-algebra is called the
{\it conormal algebra of $Z$ in $X$} and is often simply denoted
$\bigoplus_{n \geq 0} \mathcal{I}^n/\mathcal{I}^{n + 1}$
by the abuse of notation mentioned in
Morphisms of Spaces, Section
\ref{spaces-morphisms-section-closed-immersions-quasi-coherent}.
\medskip\noindent
In case $i : Z \to X$ is a (locally closed) immersion we define the conormal
algebra of $i$ as the conormal algebra of the closed immersion
$i : Z \to X \setminus \partial Z$, see Morphisms of Spaces, Remark
\ref{spaces-morphisms-remark-immersion}.
It is often denoted
$\bigoplus_{n \geq 0} \mathcal{I}^n/\mathcal{I}^{n + 1}$
where $\mathcal{I}$ is the ideal sheaf
of the closed immersion $i : Z \to X \setminus \partial Z$.
\begin{definition}
\label{definition-conormal-algebra}
Let $i : Z \to X$ be an immersion. The {\it conormal algebra
$\mathcal{C}_{Z/X, *}$ of $Z$ in $X$} or the {\it conormal algebra of $i$}
is the quasi-coherent sheaf of graded $\mathcal{O}_Z$-algebras
$\bigoplus_{n \geq 0} \mathcal{I}^n/\mathcal{I}^{n + 1}$ described above.
\end{definition}
\noindent
Thus $\mathcal{C}_{Z/X, 1} = \mathcal{C}_{Z/X}$ is the conormal sheaf
of the immersion. Also $\mathcal{C}_{Z/X, 0} = \mathcal{O}_Z$ and
$\mathcal{C}_{Z/X, n}$ is a quasi-coherent $\mathcal{O}_Z$-module
characterized by the property
\begin{equation}
\label{equation-conormal-in-degree-n}
i_*\mathcal{C}_{Z/X, n} = \mathcal{I}^n/\mathcal{I}^{n + 1}
\end{equation}
where $i : Z \to X \setminus \partial Z$ and $\mathcal{I}$ is the ideal
sheaf of $i$ as above. Finally, note that there is a canonical surjective map
\begin{equation}
\label{equation-conormal-algebra-quotient}
\text{Sym}^*(\mathcal{C}_{Z/X}) \longrightarrow \mathcal{C}_{Z/X, *}
\end{equation}
of quasi-coherent graded $\mathcal{O}_Z$-algebras which is an isomorphism
in degrees $0$ and $1$.
\begin{lemma}
\label{lemma-etale-conormal-algebra}
Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic spaces
over $S$. Let $\varphi : U \to X$ be an \'etale morphism where $U$ is a
scheme. Set $Z_U = U \times_X Z$ which is a locally closed subscheme of $U$.
Then
$$
\mathcal{C}_{Z/X, *}|_{Z_U} = \mathcal{C}_{Z_U/U, *}
$$
canonically and functorially in $U$.
\end{lemma}
\begin{proof}
Let $T \subset X$ be a closed subspace such that $i$ defines a closed
immersion into $X \setminus T$. Let $\mathcal{I}$ be the quasi-coherent
sheaf of ideals on $X \setminus T$ defining $Z$. Then the lemma follows
from the fact that
$\mathcal{I}|_{U \setminus \varphi^{-1}(T)}$ is the sheaf of ideals of
the immersion $Z_U \to U \setminus \varphi^{-1}(T)$.
This is clear from the construction of $\mathcal{I}$ in
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-closed-immersion-ideals}.
\end{proof}
\begin{lemma}
\label{lemma-conormal-algebra-functorial}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\
Z' \ar[r]^{i'} & X'
}
$$
be a commutative diagram of algebraic spaces over $S$.
Assume $i$, $i'$ immersions. There is a canonical map
of graded $\mathcal{O}_Z$-algebras
$$
f^*\mathcal{C}_{Z'/X', *}
\longrightarrow
\mathcal{C}_{Z/X, *}
$$
\end{lemma}
\begin{proof}
First find open subspaces $U' \subset X'$ and $U \subset X$ such that
$g(U) \subset U'$ and such that $i(Z) \subset U$ and $i(Z') \subset U'$
are closed (proof existence omitted). Replacing $X$ by $U$ and $X'$ by
$U'$ we may assume that $i$ and $i'$ are closed immersions.
Let $\mathcal{I}' \subset \mathcal{O}_{X'}$ and
$\mathcal{I} \subset \mathcal{O}_X$ be the quasi-coherent sheaves of
ideals associated to $i'$ and $i$, see
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}.
Consider the composition
$$
g^{-1}\mathcal{I}' \to g^{-1}\mathcal{O}_{X'}
\xrightarrow{g^\sharp} \mathcal{O}_X \to
\mathcal{O}_X/\mathcal{I} = i_*\mathcal{O}_Z
$$
Since $g(i(Z)) \subset Z'$ we conclude this composition is zero (see
statement on factorizations in
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}).
Thus we obtain a commutative diagram
$$
\xymatrix{
0 \ar[r] &
\mathcal{I} \ar[r] &
\mathcal{O}_X \ar[r] &
i_*\mathcal{O}_Z \ar[r] &
0 \\
0 \ar[r] &
g^{-1}\mathcal{I}' \ar[r] \ar[u] &
g^{-1}\mathcal{O}_{X'} \ar[r] \ar[u] &
g^{-1}i'_*\mathcal{O}_{Z'} \ar[r] \ar[u] &
0
}
$$
The lower row is exact since $g^{-1}$ is an exact functor.
By exactness we also see that
$(g^{-1}\mathcal{I}')^n = g^{-1}((\mathcal{I}')^n)$ for all $n \geq 1$.
Hence the diagram induces a map
$g^{-1}((\mathcal{I}')^n/(\mathcal{I}')^{n + 1}) \to
\mathcal{I}^n/\mathcal{I}^{n + 1}$.
Pulling back (using $i^{-1}$ for example) to $Z$ we obtain
$i^{-1}g^{-1}((\mathcal{I}')^n/(\mathcal{I}')^{n + 1}) \to
\mathcal{C}_{Z/X, n}$.
Since $i^{-1}g^{-1} = f^{-1}(i')^{-1}$ this gives maps
$f^{-1}\mathcal{C}_{Z'/X', n} \to \mathcal{C}_{Z/X, n}$, which induce
the desired map.
\end{proof}
\begin{lemma}
\label{lemma-conormal-algebra-functorial-flat}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_i \ar[d]_f & X \ar[d]^g \\
Z' \ar[r]^{i'} & X'
}
$$
be a cartesian square of algebraic spaces over $S$ with
$i$, $i'$ immersions. Then the canonical map
$f^*\mathcal{C}_{Z'/X', *} \to \mathcal{C}_{Z/X, *}$ of
Lemma \ref{lemma-conormal-algebra-functorial}
is surjective. If $g$ is flat, then it is an isomorphism.
\end{lemma}
\begin{proof}
We may check the statement after \'etale localizing $X'$.
In this case we may assume $X' \to X$ is a morphism of schemes,
hence $Z$ and $Z'$ are schemes and the result follows from
the case of schemes, see
Divisors, Lemma \ref{divisors-lemma-conormal-algebra-functorial-flat}.
\end{proof}
\noindent
We use the same conventions for cones and vector bundles over
algebraic spaces as we do for schemes (where we use
the conventions of EGA), see
Constructions, Sections \ref{constructions-section-cone} and
\ref{constructions-section-vector-bundle}.
In particular, a vector bundle is a very general gadget
(and not locally isomorphic to an affine space bundle).
\begin{definition}
\label{definition-normal-cone}
Let $S$ be a scheme. Let $i : Z \to X$ be an immersion of algebraic spaces
over $S$. The {\it normal cone $C_ZX$} of $Z$ in $X$ is
$$
C_ZX = \underline{\Spec}_Z(\mathcal{C}_{Z/X, *})
$$
see Morphisms of Spaces,
Definition \ref{spaces-morphisms-definition-relative-spec}. The
{\it normal bundle} of $Z$ in $X$ is the vector bundle
$$
N_ZX = \underline{\Spec}_Z(\text{Sym}(\mathcal{C}_{Z/X}))
$$
\end{definition}
\noindent
Thus $C_ZX \to Z$ is a cone over $Z$ and $N_ZX \to Z$ is a vector bundle
over $Z$. Moreover, the canonical surjection
(\ref{equation-conormal-algebra-quotient}) of graded algebras
defines a canonical closed immersion
\begin{equation}
\label{equation-normal-cone-in-normal-bundle}
C_ZX \longrightarrow N_ZX
\end{equation}
of cones over $Z$.
\section{Sheaf of differentials of a morphism}
\label{section-sheaf-differentials}
\noindent
We suggest the reader take a look at the corresponding section
in the chapter on commutative algebra
(Algebra, Section \ref{algebra-section-differentials}),
the corresponding section in the chapter on morphism of schemes
(Morphisms, Section \ref{morphisms-section-sheaf-differentials})
as well as
Modules on Sites, Section \ref{sites-modules-section-differentials}.
We first show that the notion of sheaf of differentials for a
morphism of schemes agrees with the corresponding morphism of
small \'etale (ringed) sites.
\medskip\noindent
To clearly state the following lemma we temporarily go back to
denoting $\mathcal{F}^a$ the sheaf of $\mathcal{O}_{X_\etale}$-modules
associated to a quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$
on the scheme $X$, see
Descent, Definition \ref{descent-definition-structure-sheaf}.
\begin{lemma}
\label{lemma-match-modules-differentials}
Let $f : X \to Y$ be a morphism of schemes. Let
$f_{small} : X_\etale \to Y_\etale$ be the associated
morphism of small \'etale sites, see
Descent, Remark \ref{descent-remark-change-topologies-ringed}.
Then there is a canonical isomorphism
$$
(\Omega_{X/Y})^a = \Omega_{X_\etale/Y_\etale}
$$
compatible with universal derivations. Here the first module
is the sheaf on $X_\etale$ associated
to the quasi-coherent $\mathcal{O}_X$-module $\Omega_{X/Y}$, see
Morphisms, Definition \ref{morphisms-definition-sheaf-differentials},
and the second module is the one from
Modules on Sites,
Definition \ref{sites-modules-definition-module-differentials}.
\end{lemma}
\begin{proof}
Let $h : U \to X$ be an \'etale morphism. In this case the natural map
$h^*\Omega_{X/Y} \to \Omega_{U/Y}$ is an isomorphism, see
More on Morphisms,
Lemma \ref{more-morphisms-lemma-sheaf-differentials-etale-localization}.
This means that there is a natural $\mathcal{O}_{Y_\etale}$-derivation
$$
\text{d}^a : \mathcal{O}_{X_\etale} \longrightarrow (\Omega_{X/Y})^a
$$
since we have just seen that the value of $(\Omega_{X/Y})^a$ on any object
$U$ of $X_\etale$ is canonically identified with
$\Gamma(U, \Omega_{U/Y})$. By the universal property of
$\text{d}_{X/Y} :
\mathcal{O}_{X_\etale}
\to
\Omega_{X_\etale/Y_\etale}$
there is a unique $\mathcal{O}_{X_\etale}$-linear map
$c : \Omega_{X_\etale/Y_\etale} \to (\Omega_{X/Y})^a$
such that
$\text{d}^a = c \circ \text{d}_{X/Y}$.
\medskip\noindent
Conversely, suppose that $\mathcal{F}$ is an
$\mathcal{O}_{X_\etale}$-module
and $D : \mathcal{O}_{X_\etale} \to \mathcal{F}$ is a
$\mathcal{O}_{Y_\etale}$-derivation. Then we can simply restrict
$D$ to the small Zariski site $X_{Zar}$ of $X$. Since sheaves on $X_{Zar}$
agree with sheaves on $X$, see
Descent, Remark \ref{descent-remark-Zariski-site-space},
we see that $D|_{X_{Zar}} : \mathcal{O}_X \to \mathcal{F}|_{X_{Zar}}$
is just a ``usual'' $Y$-derivation. Hence we obtain a map
$\psi : \Omega_{X/Y} \longrightarrow \mathcal{F}|_{X_{Zar}}$
such that $D|_{X_{Zar}} = \psi \circ \text{d}$. In particular, if we
apply this with $\mathcal{F} = \Omega_{X_\etale/Y_\etale}$
we obtain a map
$$
c' :
\Omega_{X/Y}
\longrightarrow
\Omega_{X_\etale/Y_\etale}|_{X_{Zar}}
$$
Consider the morphism of ringed sites
$\text{id}_{small, \etale, Zar} : X_\etale \to X_{Zar}$
discussed in
Descent, Remark \ref{descent-remark-change-topologies-ringed} and
Lemma \ref{descent-lemma-compare-sites}.
Since the restriction functor $\mathcal{F} \mapsto \mathcal{F}|_{X_{Zar}}$
is equal to $\text{id}_{small, \etale, Zar, *}$, since
$\text{id}_{small, \etale, Zar}^*$ is left adjoint to
$\text{id}_{small, \etale, Zar, *}$ and since
$(\Omega_{X/Y})^a = \text{id}_{small, \etale, Zar}^*\Omega_{X/Y}$
we see that $c'$ is adjoint to a map
$$
c'' :
(\Omega_{X/Y})^a
\longrightarrow
\Omega_{X_\etale/Y_\etale}.
$$
We claim that $c''$ and $c'$ are mutually inverse.
This claim finishes the proof of the lemma.
To see this it is enough to show that $c''(\text{d}(f)) = \text{d}_{X/Y}(f)$
and $c(\text{d}_{X/Y}(f)) = \text{d}(f)$ if $f$ is a local section of
$\mathcal{O}_X$ over an open of $X$. We omit the verification.
\end{proof}
\noindent
This clears the way for the following definition. For an alternative, see
Remark \ref{remark-alternative}.
\begin{definition}
\label{definition-sheaf-differentials}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. The {\it sheaf of differentials $\Omega_{X/Y}$ of $X$ over $Y$}
is sheaf of differentials
(Modules on Sites,
Definition \ref{sites-modules-definition-sheaf-differentials})
for the morphism of ringed topoi
$$
(f_{small}, f^\sharp) :
(X_\etale, \mathcal{O}_X)
\to
(Y_\etale, \mathcal{O}_Y)
$$
of
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-morphism-ringed-topoi}.
The {\it universal $Y$-derivation} will be denoted
$\text{d}_{X/Y} : \mathcal{O}_X \to \Omega_{X/Y}$.
\end{definition}
\noindent
By
Lemma \ref{lemma-match-modules-differentials}
this does not conflict with the already existing
notion in case $X$ and $Y$ are representable. From now on, if $X$ and $Y$
are representable, we no longer distinguish between the sheaf of differentials
defined above and the one defined in
Morphisms, Definition \ref{morphisms-definition-sheaf-differentials}.
We want to relate this to the usual modules of differentials for
morphisms of schemes. Here is the key lemma.
\begin{lemma}
\label{lemma-localize-differentials}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Consider any commutative diagram
$$
\xymatrix{
U \ar[d]_a \ar[r]_\psi & V \ar[d]^b \\
X \ar[r]^f & Y
}
$$
where the vertical arrows are \'etale morphisms of algebraic spaces. Then
$$
\Omega_{X/Y}|_{U_\etale} = \Omega_{U/V}
$$
In particular, if $U$, $V$ are schemes, then this is equal to the usual
sheaf of differentials of the morphism of schemes $U \to V$.
\end{lemma}
\begin{proof}
By
Properties of Spaces, Lemma \ref{spaces-properties-lemma-etale-morphism-topoi}
and Equation (\ref{spaces-properties-equation-restrict})
we may think of the restriction of a sheaf on $X_\etale$ to
$U_\etale$ as the pullback by $a_{small}$. Similarly for $b$. By
Modules on Sites, Lemma \ref{sites-modules-lemma-localize-differentials}
we have
$$
\Omega_{X/Y}|_{U_\etale} =
\Omega_{\mathcal{O}_{U_\etale}/
a_{small}^{-1}f_{small}^{-1}\mathcal{O}_{Y_\etale}}
$$
Since $a_{small}^{-1}f_{small}^{-1}\mathcal{O}_{Y_\etale}
= \psi_{small}^{-1}b_{small}^{-1}\mathcal{O}_{Y_\etale}
= \psi_{small}^{-1}\mathcal{O}_{V_\etale}$ we see that the lemma holds.
\end{proof}
\begin{lemma}
\label{lemma-module-differentials-quasi-coherent}
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces
over $S$. Then $\Omega_{X/Y}$ is a quasi-coherent $\mathcal{O}_X$-module.
\end{lemma}
\begin{proof}
Choose a diagram as in
Lemma \ref{lemma-localize-differentials}
with $a$ and $b$ surjective and $U$ and $V$ schemes.
Then we see that $\Omega_{X/Y}|_U = \Omega_{U/V}$ which is
quasi-coherent (for example by
Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}).
Hence we conclude that $\Omega_{X/Y}$ is quasi-coherent by
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-characterize-quasi-coherent}.
\end{proof}
\begin{remark}
\label{remark-alternative}
Now that we know that $\Omega_{X/Y}$ is quasi-coherent we can attempt
to construct it in another manner. For example we can use the result of
Properties of Spaces,
Section \ref{spaces-properties-section-quasi-coherent-presentation}
to construct the sheaf of differentials by glueing.
For example if $Y$ is a scheme and if $U \to X$ is a surjective \'etale morphism
from a scheme towards $X$, then we see that $\Omega_{U/Y}$ is
a quasi-coherent $\mathcal{O}_U$-module, and since $s, t : R \to U$
are \'etale we get an isomorphism
$$
\alpha : s^*\Omega_{U/Y} \to \Omega_{R/Y} \to t^*\Omega_{U/Y}
$$
by using
Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials-smooth}.
You check that this satisfies the cocycle condition and you're done.
If $Y$ is not a scheme, then you define $\Omega_{U/Y}$ as the cokernel
of the map $(U \to Y)^*\Omega_{Y/S} \to \Omega_{U/S}$, and proceed as
before. This two step process is a little bit ugly. Another possibility
is to glue the sheaves $\Omega_{U/V}$ for any diagram as in
Lemma \ref{lemma-localize-differentials}
but this is not very elegant either. Both approaches will work however, and
will give a slightly more elementary construction of the sheaf of
differentials.
\end{remark}
\begin{lemma}
\label{lemma-functoriality-differentials}
Let $S$ be a scheme. Let
$$
\xymatrix{
X' \ar[d] \ar[r]_f & X \ar[d] \\
Y' \ar[r] & Y
}
$$
be a commutative diagram of algebraic spaces. The map
$f^\sharp : \mathcal{O}_X \to f_*\mathcal{O}_{X'}$ composed with the map
$f_*\text{d}_{X'/Y'} : f_*\mathcal{O}_{X'} \to f_*\Omega_{X'/Y'}$ is a
$Y$-derivation. Hence we obtain a canonical map of $\mathcal{O}_X$-modules
$\Omega_{X/Y} \to f_*\Omega_{X'/Y'}$, and by
adjointness of $f_*$ and $f^*$ a
canonical $\mathcal{O}_{X'}$-module homomorphism
$$
c_f : f^*\Omega_{X/Y} \longrightarrow \Omega_{X'/Y'}.
$$
It is uniquely characterized by the property that
$f^*\text{d}_{X/Y}(t)$ mapsto $\text{d}_{X'/Y'}(f^* t)$
for any local section $t$ of $\mathcal{O}_X$.
\end{lemma}
\begin{proof}
This is a special case of
Modules on Sites, Lemma
\ref{sites-modules-lemma-functoriality-differentials-ringed-topoi}.
\end{proof}
\begin{lemma}
\label{lemma-check-functoriality-differentials}
Let $S$ be a scheme. Let
$$
\xymatrix{
X'' \ar[d] \ar[r]_g & X' \ar[d] \ar[r]_f & X \ar[d] \\
Y'' \ar[r] & Y' \ar[r] & Y
}
$$
be a commutative diagram of algebraic spaces over $S$. Then we have
$$
c_{f \circ g} = c_g \circ g^* c_f
$$
as maps $(f \circ g)^*\Omega_{X/Y} \to \Omega_{X''/Y''}$.
\end{lemma}
\begin{proof}
Omitted. Hint: Use the characterization of $c_f, c_g, c_{f \circ g}$
in terms of the effect these maps have on local sections.
\end{proof}
\begin{lemma}
\label{lemma-triangle-differentials}
Let $S$ be a scheme.
Let $f : X \to Y$, $g : Y \to B$ be morphisms of algebraic spaces over $S$.
Then there is a canonical exact sequence
$$
f^*\Omega_{Y/B} \to \Omega_{X/B} \to \Omega_{X/Y} \to 0
$$
where the maps come from applications of
Lemma \ref{lemma-functoriality-differentials}.
\end{lemma}
\begin{proof}
Follows from the schemes version, see
Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials},
of this result via \'etale localization, see
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-immersion-differentials}
Let $S$ be a scheme. If $X \to Y$ is an immersion
of algebraic spaces over $S$ then $\Omega_{X/S}$ is zero.
\end{lemma}
\begin{proof}
Follows from the schemes version, see
Morphisms, Lemma \ref{morphisms-lemma-immersion-differentials},
of this result via \'etale localization, see
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-differentials-relative-immersion}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $i : Z \to X$ be an immersion of algebraic spaces over $B$.
There is a canonical exact sequence
$$
\mathcal{C}_{Z/X} \to i^*\Omega_{X/B} \to \Omega_{Z/B} \to 0
$$
where the first arrow is induced by $\text{d}_{X/B}$
and the second arrow comes from
Lemma \ref{lemma-functoriality-differentials}.
\end{lemma}
\begin{proof}
This is the algebraic spaces version of
Morphisms, Lemma \ref{morphisms-lemma-differentials-relative-immersion}
and will be a consequence of that lemma by
\'etale localization, see
Lemmas \ref{lemma-localize-differentials} and
\ref{lemma-etale-conormal}.
However, we should make sure we can define the first arrow globally.
Hence we explain the meaning of ``induced by $\text{d}_{X/B}$'' here.
Namely, we may assume that $i$ is a closed immersion after replacing $X$
by an open subspace. Let $\mathcal{I} \subset \mathcal{O}_X$
be the quasi-coherent sheaf of ideals corresponding to $Z \subset X$.
Then $\text{d}_{X/S} : \mathcal{I} \to \Omega_{X/S}$
maps the subsheaf $\mathcal{I}^2 \subset \mathcal{I}$ to
$\mathcal{I}\Omega_{X/S}$. Hence it induces a map
$\mathcal{I}/\mathcal{I}^2 \to \Omega_{X/S}/\mathcal{I}\Omega_{X/S}$
which is $\mathcal{O}_X/\mathcal{I}$-linear.
By
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-i-star-equivalence}
this corresponds to a map $\mathcal{C}_{Z/X} \to i^*\Omega_{X/S}$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-differentials-relative-immersion-section}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $i : Z \to X$ be an immersion of algebraic spaces over $B$, and
assume $i$ (\'etale locally) has a left inverse. Then the canonical
sequence
$$
0 \to \mathcal{C}_{Z/X} \to i^*\Omega_{X/B} \to \Omega_{Z/B} \to 0
$$
of
Lemma \ref{lemma-differentials-relative-immersion}
is (\'etale locally) split exact.
\end{lemma}
\begin{proof}
Clarification: we claim that if $g : X \to Z$ is a left inverse of $i$
over $B$, then $i^*c_g$ is a right inverse of the map
$i^*\Omega_{X/B} \to \Omega_{Z/B}$.
Having said this, the result follows from the corresponding result for
morphisms of schemes by \'etale localization, see
Lemmas \ref{lemma-localize-differentials} and
\ref{lemma-etale-conormal}.
\end{proof}
\begin{lemma}
\label{lemma-base-change-differentials}
Let $S$ be a scheme.
Let $X \to Y$ be a morphism of algebraic spaces over $S$.
Let $g : Y' \to Y$ be a morphism of algebraic spaces over $S$.
Let $X' = X_{Y'}$ be the base change of $X$.
Denote $g' : X' \to X$ the projection.
Then the map
$$
(g')^*\Omega_{X/Y} \to \Omega_{X'/Y'}
$$
of
Lemma \ref{lemma-functoriality-differentials}
is an isomorphism.
\end{lemma}
\begin{proof}
Follows from the schemes version, see
Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials}
and \'etale localization, see
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-differential-product}
Let $S$ be a scheme.
Let $f : X \to B$ and $g : Y \to B$ be morphisms of algebraic spaces
over $S$ with the same target.
Let $p : X \times_B Y \to X$ and $q : X \times_B Y \to Y$ be the
projection morphisms. The maps from
Lemma \ref{lemma-functoriality-differentials}
$$
p^*\Omega_{X/B} \oplus q^*\Omega_{Y/B}
\longrightarrow
\Omega_{X \times_B Y/B}
$$
give an isomorphism.
\end{lemma}
\begin{proof}
Follows from the schemes version, see
Morphisms, Lemma \ref{morphisms-lemma-differential-product}
and \'etale localization, see
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-finite-type-differentials}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
If $f$ is locally of finite type, then $\Omega_{X/Y}$ is
a finite type $\mathcal{O}_X$-module.
\end{lemma}
\begin{proof}
Follows from the schemes version, see
Morphisms, Lemma \ref{morphisms-lemma-finite-type-differentials}
and \'etale localization, see
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-finite-presentation-differentials}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
If $f$ is locally of finite presentation, then $\Omega_{X/Y}$ is
an $\mathcal{O}_X$-module of finite presentation.
\end{lemma}
\begin{proof}
Follows from the schemes version, see
Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-differentials}
and \'etale localization, see
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-smooth-omega-finite-locally-free}
Let $S$ be a scheme.
Let $f : X \to Y$ be a smooth morphism of algebraic spaces over $S$.
Then the module of differentials $\Omega_{X/Y}$
is finite locally free.
\end{lemma}
\begin{proof}
The statement is \'etale local on $X$ and $Y$ by
Lemma \ref{lemma-localize-differentials}.
Hence this follows from the case of schemes, see
Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}.
\end{proof}
\section{Topological invariance of the \'etale site}
\label{section-topological-invariance}
\noindent
We show that the site $X_{spaces, \etale}$ is a ``topological
invariant''. It then follows that $X_\etale$, which
consists of the representable objects in $X_{spaces, \etale}$,
is a topological invariant too, see Lemma \ref{lemma-topological-invariance}.
\begin{theorem}
\label{theorem-topological-invariance}
Let $S$ be a scheme.
Let $f : X \to Y$ be a morphism of algebraic spaces over $S$.
Assume $f$ is integral, universally injective and surjective.
The functor
$$
V \longmapsto V_X = X \times_Y V
$$
defines an equivalence of categories
$Y_{spaces, \etale} \to X_{spaces, \etale}$.
\end{theorem}
\begin{proof}
The morphism $f$ is representable and a universal homeomorphism, see
Morphisms of Spaces,
Section \ref{spaces-morphisms-section-universal-homeomorphisms}.
\medskip\noindent
We first prove that the functor is faithful.
Suppose that $V', V$ are objects of $Y_{spaces, \etale}$ and
that $a, b : V' \to V$ are distinct morphisms over $Y$.
Since $V', V$ are \'etale over $Y$ the equalizer
$$
E = V' \times_{(a, b), V \times_Y V, \Delta_{V/Y}} V
$$
of $a, b$ is \'etale over $Y$ also. Hence $E \to V'$ is an \'etale monomorphism
(i.e., an open immersion) which is an isomorphism if and only if it is
surjective. Since $X \to Y$ is a universal homeomorphism we see that this
is the case if and only if $E_X = V'_X$, i.e., if and only if $a_X = b_X$.
\medskip\noindent
Next, we prove that the functor is fully faithful.
Suppose that $V', V$ are objects of $Y_{spaces, \etale}$ and
that $c : V'_X \to V_X$ is a morphism over $X$. We want to construct
a morphism $a : V' \to V$ over $Y$ such that $a_X = c$.
Let $a' : V'' \to V'$ be a surjective \'etale morphism such that $V''$ is
a separated algebraic space. If we can construct a morphism
$a'' : V'' \to V$ such that $a''_X = c \circ a'_X$, then the two compositions
$$
V'' \times_{V'} V'' \xrightarrow{\text{pr}_i} V'' \xrightarrow{a''} V
$$
will be equal by the faithfulness of the functor proved in the first
paragraph. Hence $a''$ will factor through a unique morphism
$a : V' \to V$ as $V'$ is (as a sheaf) the quotient of $V''$ by
the equivalence relation $V'' \times_{V'} V''$. Hence we may assume that
$V'$ is separated. In this case the graph
$$
\Gamma_c \subset (V' \times_Y V)_X
$$
is open and closed (details omitted). Since $X \to Y$ is a universal
homeomorphism, there exists an open and closed subspace
$\Gamma \subset V' \times_Y V$ such that $\Gamma_X = \Gamma_c$.
The projection $\Gamma \to V'$ is an \'etale morphism whose base
change to $X$ is an isomorphism. Hence $\Gamma \to V'$ is \'etale,
universally injective, and surjective, so an isomorphism by
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-etale-universally-injective-open}.
Thus $\Gamma$ is the graph of a morphism $a : V' \to V$ as desired.
\medskip\noindent
Finally, we prove that the functor is essentially surjective.
Suppose that $U$ is an object of $X_{spaces, \etale}$.
We have to find an object $V$ of $Y_{spaces, \etale}$
such that $V_X \cong U$. Let $U' \to U$ be a surjective \'etale morphism
such that $U' \cong V'_X$ and $U' \times_U U' \cong V''_X$
for some objects $V'', V'$ of $Y_{spaces, \etale}$.
Then by fully faithfulness of the functor we obtain morphisms
$s, t : V'' \to V'$ with $t_X = \text{pr}_0$ and $s_X = \text{pr}_1$
as morphisms $U' \times_U U' \to U'$. Using that
$(\text{pr}_0, \text{pr}_1) : U' \times_U U' \to U' \times_S U'$
is an \'etale equivalence relation, and that $U' \to V'$ and
$U' \times_U U' \to V''$ are universally injective and surjective
we deduce that
$(t, s) : V'' \to V' \times_S V'$ is an \'etale equivalence relation.
Then the quotient $V = V'/V''$ (see
Spaces, Theorem \ref{spaces-theorem-presentation})
is an algebraic space $V$ over $Y$. There is a morphism
$V' \to V$ such that $V'' = V' \times_V V'$. Thus we obtain a morphism
$V \to Y$ (see
Descent on Spaces, Lemma
\ref{spaces-descent-lemma-fpqc-universal-effective-epimorphisms}).
On base change to $X$ we see that we have a morphism $U' \to V_X$
and a compatible isomorphism $U' \times_{V_X} U' = U' \times_U U'$, which
implies that $V_X \cong U$ (by the lemma just cited once more).
\medskip\noindent
Pick a scheme $W$ and a surjective \'etale morphism $W \to Y$.
Pick a scheme $U'$ and a surjective \'etale morphism $U' \to U \times_X W_X$.
Note that $U'$ and $U' \times_U U'$ are schemes \'etale over $X$ whose
structure morphism to $X$ factors through the scheme $W_X$.
Hence by
\'Etale Cohomology,
Theorem \ref{etale-cohomology-theorem-topological-invariance}
there exist schemes $V', V''$ \'etale over $W$ whose base change to
$W_X$ is isomorphic to respectively $U'$ and $U' \times_U U'$.
This finishes the proof.
\end{proof}
\begin{lemma}
\label{lemma-topological-invariance}
With assumption and notation as in
Theorem \ref{theorem-topological-invariance}
the equivalence of categories
$Y_{spaces, \etale} \to X_{spaces, \etale}$
restricts to an equivalence of categories
$Y_\etale \to X_\etale$.
\end{lemma}
\begin{proof}
This is just the statement that given an object
$V \in Y_{spaces, \etale}$ we have $V$ is a scheme if and
only if $V \times_Y X$ is a scheme. Since $V \times_Y X \to V$
is integral, universally injective, and surjective (as a base
change of $X \to Y$) this
follows from Limits of Spaces, Lemma
\ref{spaces-limits-lemma-integral-universally-bijective-scheme}.
\end{proof}
\begin{remark}
\label{remark-topological-invariance-etale-site}
\begin{reference}
Email by Lenny Taelman dated May 1, 2016.
\end{reference}
A universal homeomorphism of algebraic spaces need not be representable, see
Morphisms of Spaces,
Example \ref{spaces-morphisms-example-universal-homeomorphism}.
In fact Theorem \ref{theorem-topological-invariance} does
not hold for universal homeomorphisms. To see this, let $k$ be an
algebraically closed field of characteristic $0$ and let
$$
\mathbf{A}^1 \to X \to \mathbf{A}^1
$$
be as in Morphisms of Spaces,
Example \ref{spaces-morphisms-example-universal-homeomorphism}.
Recall that the first morphism is \'etale and identifies
$t$ with $-t$ for $t \in \mathbf{A}^1_k \setminus \{0\}$
and that the second morphism is our universal homeomorphism.
Since $\mathbf{A}^1_k$ has no
nontrivial connected finite \'etale coverings
(because $k$ is algebraically closed of characteristic zero; details omitted),
it suffices to construct a nontrivial connected finite \'etale covering
$Y \to X$. To do this, let $Y$ be the affine line
with zero doubled
(Schemes, Example \ref{schemes-example-affine-space-zero-doubled}).
Then $Y = Y_1 \cup Y_2$ with $Y_i = \mathbf{A}^1_k$ glued
along $\mathbf{A}^1_k \setminus \{0\}$.
To define the morphism $Y \to X$ we use the morphisms
$$
Y_1 \xrightarrow{1} \mathbf{A}^1_k \to X
\quad\text{and}\quad
Y_2 \xrightarrow{-1} \mathbf{A}^1_k \to X.
$$
These glue over $Y_1 \cap Y_2$ by the construction of $X$ and
hence define a morphism $Y \to X$. In fact, we claim that
$$
\xymatrix{
Y \ar[d] & Y_1 \amalg Y_2 \ar[l] \ar[d] \\
X & \mathbf{A}^1_k \ar[l]
}
$$
is a cartesian square. We omit the details; you can use for example
Groupoids, Lemma \ref{groupoids-lemma-criterion-fibre-product}.
Since $\mathbf{A}^1_k \to X$ is \'etale and
surjective, this proves that $Y \to X$
is finite \'etale of degree $2$ which gives the desired example.
\medskip\noindent
More simply, you can argue as follows. The scheme $Y$ has a free
action of the group $G = \{+1, -1\}$ where $-1$ acts by swapping
$Y_1$ and $Y_2$ and changing the sign of the coordinate. Then
$X = Y/G$ (see Spaces, Definition \ref{spaces-definition-quotient})
and hence $Y \to X$ is finite \'etale. You can also show directly
that there exists a universal homeomorphism $X \to \mathbf{A}^1_k$
by using $t \mapsto t^2$ on affine spaces. In fact, this $X$ is
the same as the $X$ above.
\end{remark}
\section{Thickenings}
\label{section-thickenings}
\noindent
The following terminology may not be completely standard, but it is convenient.
\begin{definition}
\label{definition-thickening}
Thickenings. Let $S$ be a scheme.
\begin{enumerate}
\item We say an algebraic space $X'$ is a {\it thickening} of an algebraic
space $X$ if $X$ is a closed subspace of $X'$ and the associated topological
spaces are equal.
\item We say $X'$ is a {\it first order thickening} of $X$ if
$X$ is a closed subspace of $X'$ and the quasi-coherent sheaf of ideals
$\mathcal{I} \subset \mathcal{O}_{X'}$ defining $X$ has square zero.
\item Given two thickenings $X \subset X'$ and $Y \subset Y'$ a
{\it morphism of thickenings} is a morphism $f' : X' \to Y'$ such that
$f(X) \subset Y$, i.e., such that $f'|_X$ factors through the closed
subspace $Y$. In this situation we set $f = f'|_X : X \to Y$ and we say
that $(f, f') : (X \subset X') \to (Y \subset Y')$ is a morphism of
thickenings.
\item Let $B$ be an algebraic space. We similarly define
{\it thickenings over $B$}, and
{\it morphisms of thickenings over $B$}. This means that the spaces
$X, X', Y, Y'$ above are algebraic spaces endowed with a structure
morphism to $B$, and that the morphisms
$X \to X'$, $Y \to Y'$ and $f' : X' \to Y'$ are morphisms over $B$.
\end{enumerate}
\end{definition}
\noindent
The fundamental equivalence.
Note that if $X \subset X'$ is a thickening, then $X \to X'$
is integral and universally bijective. This implies that
\begin{equation}
\label{equation-equivalence-etale-spaces}
X_{spaces, \etale} = X'_{spaces, \etale}
\end{equation}
via the pullback functor, see
Theorem \ref{theorem-topological-invariance}.
Hence we may think of $\mathcal{O}_{X'}$ as a sheaf on
$X_{spaces, \etale}$. Thus a canonical equivalence
of locally ringed topoi
\begin{equation}
\label{equation-fundamental-equivalence}
(\Sh(X'_{spaces, \etale}), \mathcal{O}_{X'})
\cong
(\Sh(X_{spaces, \etale}), \mathcal{O}_{X'})
\end{equation}
Below we will frequently combine this with the fully faithfulness result of
Properties of Spaces, Theorem \ref{spaces-properties-theorem-fully-faithful}.
For example the closed immersion $i_X : X \to X'$ corresponds
to the surjective map $i_X^\sharp : \mathcal{O}_{X'} \to \mathcal{O}_X$.
\medskip\noindent
Let $S$ be a scheme, and let $B$ be an algebraic space over $S$.
Let $(f, f') : (X \subset X') \to (Y \subset Y')$ be a morphism of
thickenings over $B$. Note that the diagram of continuous functors
$$
\xymatrix{
X_{spaces, \etale} &
Y_{spaces, \etale} \ar[l] \\
X'_{spaces, \etale} \ar[u] &
Y'_{spaces, \etale} \ar[u] \ar[l]
}
$$
is commutative and the vertical arrows are equivalences. Hence
$f_{spaces, \etale}$, $f_{small}$,
$f'_{spaces, \etale}$, and $f'_{small}$
all define the same morphism of topoi. Thus we may think of
$$
(f')^\sharp :
f_{spaces, \etale}^{-1}\mathcal{O}_{Y'}
\longrightarrow
\mathcal{O}_{X'}
$$
as a map of sheaves of $\mathcal{O}_B$-algebras fitting into the commutative
diagram
$$
\xymatrix{
f_{spaces, \etale}^{-1}\mathcal{O}_Y \ar[r]_-{f^\sharp} \ar[r] &
\mathcal{O}_X \\
f_{spaces, \etale}^{-1}\mathcal{O}_{Y'} \ar[r]^-{(f')^\sharp}
\ar[u]^{i_Y^\sharp} &
\mathcal{O}_{X'} \ar[u]_{i_X^\sharp}
}
$$
Here $i_X : X \to X'$ and $i_Y : Y \to Y'$ are the names of the given
closed immersions.
\begin{lemma}
\label{lemma-first-order-thickening-maps}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $X \subset X'$ and $Y \subset Y'$ be thickenings
of algebraic spaces over $B$. Let $f : X \to Y$ be a morphism of algebraic
spaces over $B$. Given any map of $\mathcal{O}_B$-algebras
$$
\alpha : f_{spaces, \etale}^{-1}\mathcal{O}_{Y'} \to \mathcal{O}_{X'}
$$
such that
$$
\xymatrix{
f_{spaces, \etale}^{-1}\mathcal{O}_Y \ar[r]_-{f^\sharp} \ar[r] &
\mathcal{O}_X \\
f_{spaces, \etale}^{-1}\mathcal{O}_{Y'} \ar[r]^-\alpha
\ar[u]^{i_Y^\sharp} &
\mathcal{O}_{X'} \ar[u]_{i_X^\sharp}
}
$$
commutes, there exists a unique morphism of $(f, f')$ of
thickenings over $B$ such that $\alpha = (f')^\sharp$.
\end{lemma}
\begin{proof}
To find $f'$, by
Properties of Spaces, Theorem \ref{spaces-properties-theorem-fully-faithful},
all we have to do is show that the morphism of ringed topoi
$$
(f_{spaces, \etale}, \alpha) :
(\Sh(X_{spaces, \etale}), \mathcal{O}_{X'})
\longrightarrow
(\Sh(Y_{spaces, \etale}), \mathcal{O}_{Y'})
$$
is a morphism of locally ringed topoi. This follows directly
from the definition of morphisms of locally ringed topoi
(Modules on Sites,
Definition \ref{sites-modules-definition-morphism-locally-ringed-topoi}),
the fact that $(f, f^\sharp)$ is a morphism of locally ringed topoi
(Properties of Spaces,
Lemma \ref{spaces-properties-lemma-morphism-locally-ringed}),
that $\alpha$ fits into the given commutative diagram, and
the fact that the kernels of $i_X^\sharp$ and $i_Y^\sharp$ are
locally nilpotent. Finally, the fact that $f' \circ i_X = i_Y \circ f$
follows from the commutativity of the diagram and another application of
Properties of Spaces, Theorem \ref{spaces-properties-theorem-fully-faithful}.
We omit the verification that $f'$ is a morphism over $B$.
\end{proof}
\begin{lemma}
\label{lemma-open-subspace-thickening}
Let $S$ be a scheme. Let $X \subset X'$ be a thickening
of algebraic spaces over $S$. For any open subspace $U \subset X$ there
exists a unique open subspace $U' \subset X'$ such that
$U = X \times_{X'} U'$.
\end{lemma}
\begin{proof}
Let $U' \to X'$ be the object of $X'_{spaces, \etale}$
corresponding to the object $U \to X$ of $X_{spaces, \etale}$
via (\ref{equation-equivalence-etale-spaces}). The morphism
$U' \to X'$ is \'etale and universally injective, hence an open immersion, see
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-etale-universally-injective-open}.
\end{proof}
\noindent
Finite order thickenings. Let $i_X : X \to X'$ be a thickening of
algebraic spaces. Any local section of the kernel
$\mathcal{I} = \Ker(i_X^\sharp) \subset \mathcal{O}_{X'}$ is
locally nilpotent.
Let us say that $X \subset X'$ is a {\it finite order thickening}
if the ideal sheaf $\mathcal{I}$ is ``globally'' nilpotent, i.e.,
if there exists an $n \geq 0$ such that $\mathcal{I}^{n + 1} = 0$.
Technically the class of finite order thickenings $X \subset X'$
is much easier to handle than the general case.
Namely, in this case we have a filtration
$$
0 \subset \mathcal{I}^n \subset \mathcal{I}^{n - 1} \subset
\ldots \subset \mathcal{I} \subset \mathcal{O}_{X'}
$$
and we see that $X'$ is filtered by closed subspaces
$$
X = X_0 \subset X_1 \subset \ldots \subset X_{n - 1} \subset X_{n + 1} = X'
$$
such that each pair $X_i \subset X_{i + 1}$ is a first order thickening
over $B$. Using simple induction arguments many results proved for first order
thickenings can be rephrased as results on finite order thickenings.
\begin{lemma}
\label{lemma-first-order-thickening-surjective}
Let $S$ be a scheme. Let $X \subset X'$ be a thickening
of algebraic spaces over $S$. Let $U$ be an affine object of
$X_{spaces, \etale}$. Then
$$
\Gamma(U, \mathcal{O}_{X'}) \to \Gamma(U, \mathcal{O}_X)
$$
is surjective where we think of $\mathcal{O}_{X'}$ as a sheaf on
$X_{spaces, \etale}$ via (\ref{equation-fundamental-equivalence}).
\end{lemma}
\begin{proof}
Let $U' \to X'$ be the \'etale morphism of algebraic spaces such that
$U = X \times_{X'} U'$, see Theorem \ref{theorem-topological-invariance}.
By Limits of Spaces, Lemma \ref{spaces-limits-lemma-affine} we see
that $U'$ is an affine scheme. Hence
$\Gamma(U, \mathcal{O}_{X'}) = \Gamma(U', \mathcal{O}_{U'}) \to
\Gamma(U, \mathcal{O}_U)$
is surjective as $U \to U'$ is a closed immersion of affine schemes.
Below we give a direct proof for finite order thickenings
which is the case most used in practice.
\end{proof}
\begin{proof}[Proof for finite order thickenings]
We may assume that $X \subset X'$ is a first order thickening by the
principle explained above. Denote $\mathcal{I}$ the kernel of the surjection
$\mathcal{O}_{X'} \to \mathcal{O}_X$. As $\mathcal{I}$ is a quasi-coherent
$\mathcal{O}_{X'}$-module and since $\mathcal{I}^2 = 0$ by the definition
of a first order thickening we may apply
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-i-star-equivalence}
to see that $\mathcal{I}$ is a quasi-coherent $\mathcal{O}_X$-module.
Hence the lemma follows from the long exact cohomology sequence
associated to the short exact sequence
$$
0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0
$$
and the fact that $H^1_\etale(U, \mathcal{I}) = 0$ as
$\mathcal{I}$ is quasi-coherent, see
Descent, Proposition \ref{descent-proposition-same-cohomology-quasi-coherent}
and Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}.
\end{proof}
\begin{lemma}
\label{lemma-thickening-scheme}
Let $S$ be a scheme. Let $X \subset X'$ be a thickening of algebraic spaces
over $S$. If $X$ is (representable by) a scheme, then so is $X'$.
\end{lemma}
\begin{proof}
Note that $X'_{red} = X_{red}$. Hence if $X$ is a scheme, then
$X'_{red}$ is a scheme. Thus the result follows from
Limits of Spaces, Lemma
\ref{spaces-limits-lemma-reduction-scheme}.
Below we give a direct proof for finite order thickenings which is
the case most often used in practice.
\end{proof}
\begin{proof}[Proof for finite order thickenings]
It suffices to prove this when $X'$ is a first order thickening of $X$. By
Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme}
there is a largest open subspace of $X'$ which is a scheme. Thus we have
to show that every point $x$ of $|X'| = |X|$ is contained in an open subspace of
$X'$ which is a scheme. Using
Lemma \ref{lemma-open-subspace-thickening}
we may replace $X \subset X'$ by $U \subset U'$ with $x \in U$ and $U$
an affine scheme. Hence we may assume that $X$ is affine.
Thus we reduce to the case discussed in the next paragraph.
\medskip\noindent
Assume $X \subset X'$ is a first order thickening where $X$ is an affine
scheme. Set $A = \Gamma(X, \mathcal{O}_X)$ and
$A' = \Gamma(X', \mathcal{O}_{X'})$. By
Lemma \ref{lemma-first-order-thickening-surjective}
the map $A \to A'$ is surjective. The kernel $I$ is an ideal of square zero. By
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-morphism-to-affine-scheme}
we obtain a canonical morphism $f : X' \to \Spec(A')$ which fits
into the following commutative diagram
$$
\xymatrix{
X \ar@{=}[d] \ar[r] & X' \ar[d]^f \\
\Spec(A) \ar[r] & \Spec(A')
}
$$
Because the horizontal arrows are thickenings it is clear that $f$ is
universally injective and surjective. Hence it suffices to show that
$f$ is \'etale, since then
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-etale-universally-injective-open}
will imply that $f$ is an isomorphism.
\medskip\noindent
To prove that $f$ is \'etale choose an affine scheme $U'$ and an
\'etale morphism $U' \to X'$. It suffices to show that
$U' \to X' \to \Spec(A')$ is \'etale, see
Properties of Spaces, Definition \ref{spaces-properties-definition-etale}.
Write $U' = \Spec(B')$. Set $U = X \times_{X'} U'$. Since $U$
is a closed subspace of $U'$, it is a closed subscheme, hence
$U = \Spec(B)$ with $B' \to B$ surjective. Denote
$J = \Ker(B' \to B)$ and note that $J = \Gamma(U, \mathcal{I})$
where $\mathcal{I} = \Ker(\mathcal{O}_{X'} \to \mathcal{O}_X)$
on $X_{spaces, \etale}$ as in the proof of
Lemma \ref{lemma-first-order-thickening-surjective}.
The morphism $U' \to X' \to \Spec(A')$ induces a commutative
diagram
$$
\xymatrix{
0 \ar[r] &
J \ar[r] &
B' \ar[r] &
B \ar[r] & 0 \\
0 \ar[r] &
I \ar[r] \ar[u] &
A' \ar[r] \ar[u] &
A \ar[r] \ar[u] & 0
}
$$
Now, since $\mathcal{I}$ is a quasi-coherent $\mathcal{O}_X$-module
we have $\mathcal{I} = (\widetilde I)^a$, see
Descent, Definition \ref{descent-definition-structure-sheaf}
for notation and
Descent, Proposition \ref{descent-proposition-equivalence-quasi-coherent}
for why this is true. Hence we see that $J = I \otimes_A B$.
Finally, note that $A \to B$ is \'etale as $U \to X$ is \'etale as
the base change of the \'etale morphism $U' \to X'$.
We conclude that $A' \to B'$ is \'etale by
Algebra, Lemma \ref{algebra-lemma-lift-etale-infinitesimal}.
\end{proof}
\begin{lemma}
\label{lemma-thickening-equivalence}
Let $S$ be a scheme. Let $X \subset X'$ be a thickening of algebraic spaces
over $S$. The functor
$$
V' \longmapsto V = X \times_{X'} V'
$$
defines an equivalence of categories
$X'_\etale \to X_\etale$.
\end{lemma}
\begin{proof}
The functor $V' \mapsto V$ defines an equivalence of categories
$X'_{spaces, \etale} \to X_{spaces, \etale}$, see
Theorem \ref{theorem-topological-invariance}.
Thus it suffices to show that $V$ is a scheme if and only if $V'$ is
a scheme. This is the content of
Lemma \ref{lemma-thickening-scheme}.
\end{proof}
\noindent
First order thickening are described as follows.
\begin{lemma}
\label{lemma-first-order-thickening}
Let $S$ be a scheme.
Let $f : X \to B$ be a morphism of algebraic spaces over $S$.
Consider a short exact sequence
$$
0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_X \to 0
$$
of sheaves on $X_\etale$ where $\mathcal{A}$ is a sheaf of
$f^{-1}\mathcal{O}_B$-algebras, $\mathcal{A} \to \mathcal{O}_X$ is a surjection
of sheaves of $f^{-1}\mathcal{O}_B$-algebras, and $\mathcal{I}$ is its kernel.
If
\begin{enumerate}
\item $\mathcal{I}$ is an ideal of square zero in $\mathcal{A}$, and
\item $\mathcal{I}$ is quasi-coherent as an $\mathcal{O}_X$-module
\end{enumerate}
then there exists a first order thickening
$X \subset X'$ over $B$ and an isomorphism
$\mathcal{O}_{X'} \to \mathcal{A}$ of $f^{-1}\mathcal{O}_B$-algebras
compatible with the surjections to $\mathcal{O}_X$.
\end{lemma}
\begin{proof}
In this proof we redo some of the arguments used in the
proofs of
Lemmas \ref{lemma-first-order-thickening-surjective} and
\ref{lemma-thickening-scheme}.
We first handle the case $B = S = \Spec(\mathbf{Z})$.
Let $U$ be an affine scheme, and let $U \to X$ be \'etale.
Then
$$
0 \to \mathcal{I}(U) \to \mathcal{A}(U) \to \mathcal{O}_X(U) \to 0
$$
is exact as $H^1(U_\etale, \mathcal{I}) = 0$ as
$\mathcal{I}$ is quasi-coherent, see
Descent, Proposition \ref{descent-proposition-same-cohomology-quasi-coherent}
and Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}.
If $V \to U$ is a morphism of affine objects of $X_{spaces, \etale}$
then
$$
\mathcal{I}(V) = \mathcal{I}(U) \otimes_{\mathcal{O}_X(U)} \mathcal{O}_X(V)
$$
since $\mathcal{I}$ is a quasi-coherent $\mathcal{O}_X$-module, see
Descent, Proposition \ref{descent-proposition-equivalence-quasi-coherent}.
Hence $\mathcal{A}(U) \to \mathcal{A}(V)$ is an
\'etale ring map, see
Algebra, Lemma \ref{algebra-lemma-lift-etale-infinitesimal}.
Hence we see that
$$
U \longmapsto U' = \Spec(\mathcal{A}(U))
$$
is a functor from $X_{affine, \etale}$ to the category of affine
schemes and \'etale morphisms. In fact, we claim that this functor can
be extended to a functor $U \mapsto U'$ on all of $X_\etale$.
To see this, if $U$ is an object of $X_\etale$, note that
$$
0 \to \mathcal{I}|_{U_{Zar}} \to \mathcal{A}|_{U_{Zar}} \to
\mathcal{O}_X|_{U_{Zar}} \to 0
$$
and $\mathcal{I}|_{U_{Zar}}$ is a quasi-coherent sheaf on $U$, see
Descent,
Proposition \ref{descent-proposition-equivalence-quasi-coherent-functorial}.
Hence by
More on Morphisms, Lemma \ref{more-morphisms-lemma-first-order-thickening}
we obtain a first order thickening $U \subset U'$ of schemes such that
$\mathcal{O}_{U'}$ is isomorphic to $\mathcal{A}|_{U_{Zar}}$. It is clear that
this construction is compatible with the construction for affines above.
\medskip\noindent
Choose a presentation $X = U/R$, see
Spaces, Definition \ref{spaces-definition-presentation}
so that $s, t : R \to U$ define an \'etale equivalence relation.
Applying the functor above we obtain an \'etale equivalence
relation $s', t' : R' \to U'$ in schemes. Consider the algebraic space
$X' = U'/R'$ (see
Spaces, Theorem \ref{spaces-theorem-presentation}).
The morphism $X = U/R \to U'/R' = X'$ is a first order thickening.
Consider $\mathcal{O}_{X'}$ viewed as a sheaf on $X_\etale$.
By construction we have an isomorphism
$$
\gamma :
\mathcal{O}_{X'}|_{U_\etale}
\longrightarrow
\mathcal{A}|_{U_\etale}
$$
such that $s^{-1}\gamma$ agrees with $t^{-1}\gamma$ on $R_\etale$.
Hence by
Properties of Spaces, Lemma \ref{spaces-properties-lemma-descent-sheaf}
this implies that $\gamma$ comes from a unique isomorphism
$\mathcal{O}_{X'} \to \mathcal{A}$ as desired.
\medskip\noindent
To handle the case of a general base algebraic space $B$, we first
construct $X'$ as an algebraic space over $\mathbf{Z}$ as above.
Then we use the isomorphism $\mathcal{O}_{X'} \to \mathcal{A}$ to
define $f^{-1}\mathcal{O}_B \to \mathcal{O}_{X'}$. According to
Lemma \ref{lemma-first-order-thickening-maps}
this defines a morphism $X' \to B$ compatible with the given morphism
$X \to B$ and we are done.
\end{proof}
\begin{lemma}
\label{lemma-base-change-thickening}
Let $S$ be a scheme. Let $Y \subset Y'$ be a thickening of algebraic spaces
over $S$. Let $X' \to Y'$ be a morphism and set $X = Y \times_{Y'} X'$.
Then $(X \subset X') \to (Y \subset Y')$
is a morphism of thickenings. If $Y \subset Y'$ is a first
(resp.\ finite order) thickening, then $X \subset X'$ is a first
(resp.\ finite order) thickening.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-composition-thickening}
Let $S$ be a scheme. If $X \subset X'$ and $X' \subset X''$ are
thickenings of algebraic spaces over $S$, then so is $X \subset X''$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-descending-property-thickening}
The property of being a thickening is fpqc local.
Similarly for first order thickenings.
\end{lemma}
\begin{proof}
The statement means the following: Let $S$ be a scheme and let
$X \to X'$ be a morphism of algebraic spaces over $S$.
Let $\{g_i : X'_i \to X'\}$ be an fpqc covering of algebraic spaces
such that the base change $X_i \to X'_i$ is a thickening for all $i$.
Then $X \to X'$ is a thickening. Since the morphisms $g_i$ are jointly
surjective we conclude that $X \to X'$ is surjective. By
Descent on Spaces, Lemma
\ref{spaces-descent-lemma-descending-property-closed-immersion}
we conclude that $X \to X'$ is a closed immersion.
Thus $X \to X'$ is a thickening. We omit the proof in the
case of first order thickenings.
\end{proof}
\section{Morphisms of thickenings}
\label{section-morphisms-thickenings}
\noindent
If $(f, f') : (X \subset X') \to (Y \subset Y')$ is a morphism
of thickenings of algebraic spaces, then often properties of the morphism
$f$ are inherited by $f'$. There are several variants.
\begin{lemma}
\label{lemma-thicken-property-morphisms}
Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \subset Y')$
be a morphism of thickenings of algebraic spaces over $S$. Then
\begin{enumerate}
\item $f$ is an affine morphism if and only if $f'$ is an affine morphism,
\item $f$ is a surjective morphism if and only if $f'$ is a surjective morphism,
\item $f$ is quasi-compact if and only if $f'$ quasi-compact,
\item $f$ is universally closed if and only if $f'$ is universally closed,
\item $f$ is integral if and only if $f'$ is integral,
\item $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,
\item $f$ is universally injective if and only if $f'$ is universally injective,
\item $f$ is universally open if and only if $f'$ is universally open,
\item $f$ is representable if and only if $f'$ is representable, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
Observe that $Y \to Y'$ and $X \to X'$ are integral and
universal homeomorphisms. This immediately implies parts
(2), (3), (4), (7), and (8).
Part (1) follows from
Limits of Spaces, Proposition \ref{spaces-limits-proposition-affine}
which tells us that there is a 1-to-1 correspondence between
affine schemes \'etale over $X$ and $X'$ and between affine schemes
\'etale over $Y$ and $Y'$.
Part (5) follows from (1) and (4) by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-integral-universally-closed}.
Finally, note that
$$
X \times_Y X = X \times_{Y'} X \to X \times_{Y'} X' \to X' \times_{Y'} X'
$$
is a thickening (the two arrows are thickenings by
Lemma \ref{lemma-base-change-thickening}).
Hence applying (3) and (4) to the morphism
$(X \subset X') \to (X \times_Y X \to X' \times_{Y'} X')$
we obtain (6). Finally, part (9) follows from the fact that an
algebraic space thickening of a scheme is again a scheme, see
Lemma \ref{lemma-thickening-scheme}.
\end{proof}
\begin{lemma}
\label{lemma-thicken-property-morphisms-cartesian}
Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \subset Y')$ be a
morphism of thickenings of algebraic spaces over $S$ such that
$X = Y \times_{Y'} X'$. If $X \subset X'$ is a finite order thickening, then
\begin{enumerate}
\item $f$ is a closed immersion if and only if $f'$ is a closed immersion,
\item $f$ is locally of finite type if and only if $f'$ is
locally of finite type,
\item $f$ is locally quasi-finite if and only if $f'$ is locally
quasi-finite,
\item $f$ is locally of finite type of relative dimension $d$ if and
only if $f'$ is locally of finite type of relative dimension $d$,
\item $\Omega_{X/Y} = 0$ if and only if $\Omega_{X'/Y'} = 0$,
\item $f$ is unramified if and only if $f'$ is unramified,
\item $f$ is proper if and only if $f'$ is proper,
\item $f$ is a finite morphism if and only if $f'$ is an finite morphism,
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\item $f$ is an immersion if and only if $f'$ is an immersion, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a scheme $V'$ and a surjective \'etale morphism $V' \to Y'$.
Choose a scheme $U'$ and a surjective \'etale morphism
$U' \to X' \times_{Y'} V'$. Set $V = Y \times_{Y'} V'$ and
$U = X \times_{X'} U'$. Then for \'etale local properties of morphisms
we can reduce to the morphism of thickenings of schemes
$(U \subset U') \to (V \subset V')$ and apply More on Morphisms, Lemma
\ref{more-morphisms-lemma-thicken-property-morphisms-cartesian}.
This proves (2), (3), (4), (5), and (6).
\medskip\noindent
The properties of morphisms in (1), (7), (8), (9), (10) are stable
under base change, hence if $f'$ has property $\mathcal{P}$, then so
does $f$. See
Spaces, Lemma \ref{spaces-lemma-base-change-immersions},
and
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-base-change-proper},
\ref{spaces-morphisms-lemma-base-change-integral}, and
\ref{spaces-morphisms-lemma-base-change-monomorphism}.
\medskip\noindent
The interesting direction in (1), (7), (8), (9), (10) is to assume
that $f$ has the property and deduce that $f'$ has it too.
By induction on the order of the thickening we may
assume that $Y \subset Y'$ is a first order thickening, see
discussion on finite order thickenings above.
\medskip\noindent
Proof of (1). Choose a scheme $V'$ and a surjective \'etale morphism
$V' \to Y'$. Set $V = Y \times_{Y'} V'$, $U' = X' \times_{Y'} V'$
and $U = X \times_Y V$. Then $U \to V$ is a closed immersion, which
implies that $U$ is a scheme, which in turn implies that $U'$ is
a scheme (Lemma \ref{lemma-thickening-scheme}). Thus we can apply
the lemma in the case of schemes
(More on Morphisms, Lemma
\ref{more-morphisms-lemma-thicken-property-morphisms-cartesian})
to $(U \subset U') \to (V \subset V')$ to conclude.
\medskip\noindent
Proof of (7). Follows by combining (2) with
results of Lemma \ref{lemma-thicken-property-morphisms}
and the fact that proper equals quasi-compact $+$
separated $+$ locally of finite type $+$ universally closed.
\medskip\noindent
Proof of (8). Follows by combining (2) with
results of Lemma \ref{lemma-thicken-property-morphisms}
and using the fact that finite equals integral $+$ locally
of finite type (Morphisms, Lemma \ref{morphisms-lemma-finite-integral}).
\medskip\noindent
Proof of (9). As $f$ is a monomorphism we have $X = X \times_Y X$.
We may apply the results proved so far to the morphism
of thickenings $(X \subset X') \to (X \times_Y X \subset X' \times_{Y'} X')$.
We conclude $X' \to X' \times_{Y'} X'$ is a closed immersion by (1).
In fact, it is a first order thickening as the ideal defining the
closed immersion $X' \to X' \times_{Y'} X'$ is contained in the pullback
of the ideal $\mathcal{I} \subset \mathcal{O}_{Y'}$ cutting out $Y$ in $Y'$.
Indeed, $X = X \times_Y X = (X' \times_{Y'} X') \times_{Y'} Y$ is contained
in $X'$. The conormal sheaf of the closed immersion
$\Delta : X' \to X' \times_{Y'} X'$ is equal to $\Omega_{X'/Y'}$
(this is the analogue of
Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}
for algebraic spaces and follows either by \'etale localization
or by combining
Lemmas \ref{lemma-differentials-relative-immersion-section} and
\ref{lemma-differential-product}; some details omitted).
Thus it suffices to show that $\Omega_{X'/Y'} = 0$ which follows from (5)
and the corresponding statement for $X/Y$.
\medskip\noindent
Proof of (10). If $f : X \to Y$ is an immersion, then it factors as
$X \to V \to Y$ where $V \to Y$ is an open subspace and $X \to V$ is a
closed immersion, see
Morphisms of Spaces, Remark \ref{spaces-morphisms-remark-immersion}.
Let $V' \subset Y'$ be the open subspace whose
underlying topological space $|V'|$ is the same as $|V| \subset |Y| = |Y'|$.
Then $X' \to Y'$ factors through $V'$ and we conclude that $X' \to V'$
is a closed immersion by part (1). This finishes the proof.
\end{proof}
\noindent
The following lemma is a variant on the preceding one. Rather than assume
that the thickenings involved are finite order (which allows us to transfer
the property of being locally of finite type from $f$ to $f'$),
we instead take as given that each of $f$ and $f'$ is locally of
finite type.
\begin{lemma}
\label{lemma-properties-that-extend-over-thickenings}
Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \to Y')$ be a
morphism of thickenings of algebraic spaces over $S$. Assume $f$ and $f'$
are locally of finite type and $X = Y \times_{Y'} X'$. Then
\begin{enumerate}
\item $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,
\item $f$ is finite if and only if $f'$ is finite,
\item $f$ is a closed immersion if and only if $f'$ is a closed immersion,
\item $\Omega_{X/Y} = 0$ if and only if $\Omega_{X'/Y'} = 0$,
\item $f$ is unramified if and only if $f'$ is unramified,
\item $f$ is a monomorphism if and only if $f'$ is a monomorphism,
\item $f$ is an immersion if and only if $f'$ is an immersion,
\item $f$ is proper if and only if $f'$ is proper, and
\item add more here.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a scheme $V'$ and a surjective \'etale morphism $V' \to Y'$.
Choose a scheme $U'$ and a surjective \'etale morphism
$U' \to X' \times_{Y'} V'$. Set $V = Y \times_{Y'} V'$ and
$U = X \times_{X'} U'$. Then for \'etale local properties of morphisms
we can reduce to the morphism of thickenings of schemes
$(U \subset U') \to (V \subset V')$ and apply
More on Morphisms, Lemma
\ref{more-morphisms-lemma-properties-that-extend-over-thickenings}.
This proves (1), (4), and (5).
\medskip\noindent
The properties in (2), (3), (6), (7), and (8) are stable
under base change, hence if $f'$ has property $\mathcal{P}$, then so
does $f$. See Spaces, Lemma \ref{spaces-lemma-base-change-immersions},
and
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-base-change-proper},
\ref{spaces-morphisms-lemma-base-change-integral}, and
\ref{spaces-morphisms-lemma-base-change-monomorphism}.
Hence in each case we need only to prove that if $f$ has
the desired property, so does $f'$.
\medskip\noindent
Case (2) follows from case (5) of Lemma \ref{lemma-thicken-property-morphisms}
and the fact that the finite morphisms are precisely
the integral morphisms that are locally of finite type
(Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-integral}).
\medskip\noindent
Case (3). This follows immediately from
Limits of Spaces, Lemma
\ref{spaces-limits-lemma-check-closed-infinitesimally}.
\medskip\noindent
Proof of (6). As $f$ is a monomorphism we have $X = X \times_Y X$.
We may apply the results proved so far to the morphism of thickenings
$(X \subset X') \to (X \times_Y X \subset X' \times_{Y'} X')$.
We conclude $\Delta_{X'/Y'} : X' \to X' \times_{Y'} X'$
is a closed immersion by (3). In fact $\Delta_{X'/Y'}$ induces a bijection
$|X'| \to |X' \times_{Y'} X'|$, hence $\Delta_{X'/Y'}$ is a thickening.
On the other hand $\Delta_{X'/Y'}$ is locally of finite presentation by
Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-diagonal-morphism-finite-type}.
In other words, $\Delta_{X'/Y'}(X')$ is cut out by
a quasi-coherent sheaf of ideals
$\mathcal{J} \subset \mathcal{O}_{X' \times_{Y'} X'}$ of finite type.
Since $\Omega_{X'/Y'} = 0$ by (5) we see that
the conormal sheaf of $X' \to X' \times_{Y'} X'$ is zero.
(The conormal sheaf of the closed immersion $\Delta_{X'/Y'}$ is equal to
$\Omega_{X'/Y'}$; this is the analogue of
Morphisms, Lemma \ref{morphisms-lemma-differentials-diagonal}
for algebraic spaces and follows either by \'etale localization
or by combining
Lemmas \ref{lemma-differentials-relative-immersion-section} and
\ref{lemma-differential-product}; some details omitted.)
In other words, $\mathcal{J}/\mathcal{J}^2 = 0$.
This implies $\Delta_{X'/Y'}$ is an isomorphism, for example
by Algebra, Lemma \ref{algebra-lemma-ideal-is-squared-union-connected}.
\medskip\noindent
Proof of (7). If $f : X \to Y$ is an immersion, then it factors as
$X \to V \to Y$ where $V \to Y$ is an open subspace and $X \to V$ is a
closed immersion, see
Morphisms of Spaces, Remark \ref{spaces-morphisms-remark-immersion}.
Let $V' \subset Y'$ be the open subspace whose
underlying topological space $|V'|$ is the same as $|V| \subset |Y| = |Y'|$.
Then $X' \to Y'$ factors through $V'$ and we conclude that $X' \to V'$
is a closed immersion by part (3).
\medskip\noindent
Case (8) follows from Lemma \ref{lemma-thicken-property-morphisms}
and the definition of proper morphisms as being the quasi-compact,
universally closed, and separated morphisms that are locally of finite type.
\end{proof}
\section{First order infinitesimal neighbourhood}
\label{section-first-order-infinitesimal-neighbourhood}
\noindent
A natural construction of first order thickenings is the following.
Suppose that $i : Z \to X$ be an immersion of algebraic spaces. Choose an
open subspace $U \subset X$ such that $i$ identifies $Z$ with a closed
subspace $Z \subset U$ (see
Morphisms of Spaces, Remark \ref{spaces-morphisms-remark-immersion}).
Let $\mathcal{I} \subset \mathcal{O}_U$ be the
quasi-coherent sheaf of ideals defining $Z$ in $U$, see
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}.
Then we can consider
the closed subspace $Z' \subset U$ defined by the quasi-coherent sheaf
of ideals $\mathcal{I}^2$.
\begin{definition}
\label{definition-first-order-infinitesimal-neighbourhood}
Let $i : Z \to X$ be an immersion of algebraic spaces. The
{\it first order infinitesimal neighbourhood} of $Z$ in $X$ is
the first order thickening $Z \subset Z'$ over $X$ described above.
\end{definition}
\noindent
This thickening has the following universal property (which will assuage
any fears that the construction above depends on the choice of the open
$U$).
\begin{lemma}
\label{lemma-first-order-infinitesimal-neighbourhood}
Let $i : Z \to X$ be an immersion of algebraic spaces.
The first order infinitesimal neighbourhood $Z'$ of $Z$ in $X$
has the following universal property:
Given any commutative diagram
$$
\xymatrix{
Z \ar[d]_i & T \ar[l]^a \ar[d] \\
X & T' \ar[l]_b
}
$$
where $T \subset T'$ is a first order thickening over $X$, there exists
a unique morphism $(a', a) : (T \subset T') \to (Z \subset Z')$ of
thickenings over $X$.
\end{lemma}
\begin{proof}
Let $U \subset X$ be the open subspace used in the construction of $Z'$,
i.e., an open such that $Z$ is identified with a closed subspace of $U$
cut out by the quasi-coherent sheaf of ideals $\mathcal{I}$.
Since $|T| = |T'|$ we see that $|b|(|T'|) \subset |U|$. Hence we can
think of $b$ as a morphism into $U$, see
Properties of Spaces,
Lemma \ref{spaces-properties-lemma-factor-through-open-subspace}.
Let $\mathcal{J} \subset \mathcal{O}_{T'}$
be the square zero quasi-coherent sheaf of ideals cutting out $T$.
By the commutativity of the diagram we have $b|_T = i \circ a$ where
$i : Z \to U$ is the closed immersion. We conclude that
$b^\sharp(b^{-1}\mathcal{I}) \subset \mathcal{J}$ by
Morphisms of Spaces,
Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}.
As $T'$ is a first order thickening of $T$ we see that $\mathcal{J}^2 = 0$
hence $b^\sharp(b^{-1}(\mathcal{I}^2)) = 0$. By
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-ideals}
this implies that $b$ factors through $Z'$. Letting $a' : T' \to Z'$
be this factorization we win.
\end{proof}
\begin{lemma}
\label{lemma-infinitesimal-neighbourhood-conormal}
Let $i : Z \to X$ be an immersion of algebraic spaces.
Let $Z \subset Z'$ be the first order infinitesimal neighbourhood
of $Z$ in $X$. Then the diagram
$$
\xymatrix{
Z \ar[r] \ar[d] & Z' \ar[d] \\
Z \ar[r] & X
}
$$
induces a map of conormal sheaves
$\mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Z'}$ by
Lemma \ref{lemma-conormal-functorial}.
This map is an isomorphism.
\end{lemma}
\begin{proof}
This is clear from the construction of $Z'$ above.
\end{proof}
\section{Formally smooth, \'etale, unramified transformations}
\label{section-formally-smooth-etale-unramified}
\noindent
Recall that a ring map $R \to A$ is called
{\it formally smooth}, resp.\ {\it formally \'etale},
resp.\ {\it formally unramified}
(see Algebra, Definition \ref{algebra-definition-formally-smooth},
resp.\ Definition \ref{algebra-definition-formally-etale},
resp.\ Definition \ref{algebra-definition-formally-unramified})
if for every commutative solid diagram
$$
\xymatrix{
A \ar[r] \ar@{-->}[rd] & B/I \\
R \ar[r] \ar[u] & B \ar[u]
}
$$
where $I \subset B$ is an ideal of square zero, there
exists a, resp.\ exists a unique, resp.\ exists at most one dotted
arrow which makes the diagram commute. This motivates
the following analogue for morphisms of algebraic spaces, and more
generally functors.
\begin{definition}
\label{definition-formally-smooth-etale-unramified}
Let $S$ be a scheme.
Let $a : F \to G$ be a transformation of functors
$F, G : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Consider commutative solid diagrams of the form
$$
\xymatrix{
F \ar[d]_a & T \ar[d]^i \ar[l] \\
G & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T$ and $T'$ are affine schemes and $i$ is a closed immersion
defined by an ideal of square zero.
\begin{enumerate}
\item We say $a$ is {\it formally smooth} if given any solid
diagram as above there exists a dotted arrow making the diagram
commute\footnote{This is just one possible definition that one can
make here. Another slightly weaker condition would be to require that
the dotted arrow exists fppf locally on $T'$. This weaker notion
has in some sense better formal properties.}.
\item We say $a$ is {\it formally \'etale} if given any solid
diagram as above there exists exactly one dotted arrow making the diagram
commute.
\item We say $a$ is {\it formally unramified} if given any solid
diagram as above there exists at most one dotted arrow making the diagram
commute.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-formally-etale-is-combination}
Let $S$ be a scheme.
Let $a : F \to G$ be a transformation of functors
$F, G : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Then $a$ is formally \'etale if and only if $a$ is both formally
smooth and formally unramified.
\end{lemma}
\begin{proof}
Formal from the definition.
\end{proof}
\begin{lemma}
\label{lemma-composition-formally-smooth-etale-unramified}
Composition.
\begin{enumerate}
\item A composition of formally smooth transformations of functors is formally
smooth.
\item A composition of formally \'etale transformations of functors is formally
\'etale.
\item A composition of formally unramified transformations of functors is
formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-base-change-formally-smooth-etale-unramified}
Let $S$ be a scheme contained in $\Sch_{fppf}$.
Let $F, G, H : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Let $a : F \to G$, $b : H \to G$ be transformations of functors.
Consider the fibre product diagram
$$
\xymatrix{
H \times_{b, G, a} F \ar[r]_-{b'} \ar[d]_{a'} & F \ar[d]^a \\
H \ar[r]^b & G
}
$$
\begin{enumerate}
\item If $a$ is formally smooth, then the base change $a'$ is
formally smooth.
\item If $a$ is formally \'etale, then the base change $a'$ is
formally \'etale.
\item If $a$ is formally unramified, then the base change $a'$ is
formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-representable-property-formally-property}
Let $S$ be a scheme.
Let $F, G : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Let $a : F \to G$ be a representable transformation of functors.
\begin{enumerate}
\item If $a$ is smooth then $a$ is formally smooth.
\item If $a$ is \'etale, then $a$ is formally \'etale.
\item If $a$ is unramified, then $a$ is formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
Consider a solid commutative diagram
$$
\xymatrix{
F \ar[d]_a & T \ar[d]^i \ar[l] \\
G & T' \ar[l] \ar@{-->}[lu]
}
$$
as in
Definition \ref{definition-formally-smooth-etale-unramified}.
Then $F \times_G T'$ is a scheme smooth (resp.\ \'etale, resp.\ unramified)
over $T'$. Hence by
More on Morphisms, Lemma \ref{more-morphisms-lemma-smooth-formally-smooth}
(resp.\ Lemma \ref{more-morphisms-lemma-etale-formally-etale},
resp.\ Lemma \ref{more-morphisms-lemma-unramified-formally-unramified})
we can fill in (resp.\ uniquely fill in, resp.\ fill in in at most
one way) the dotted arrow in the diagram
$$
\xymatrix{
F \times_G T' \ar[d] & T \ar[d]^i \ar[l] \\
T' & T' \ar[l] \ar@{-->}[lu]
}
$$
an hence we also obtain the corresponding assertion in the first diagram.
\end{proof}
\begin{lemma}
\label{lemma-etale-on-top}
Let $S$ be a scheme contained in $\Sch_{fppf}$.
Let $F, G, H : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Let $a : F \to G$, $b : G \to H$ be transformations of functors.
Assume that $a$ is representable, surjective, and \'etale.
\begin{enumerate}
\item If $b$ is formally smooth, then $b \circ a$ is formally smooth.
\item If $b$ is formally \'etale, then $b \circ a$ is formally \'etale.
\item If $b$ is formally unramified, then $b \circ a$ is formally unramified.
\end{enumerate}
Conversely, consider a solid commutative diagram
$$
\xymatrix{
G \ar[d]_b & T \ar[d]^i \ar[l] \\
H & T' \ar[l] \ar@{-->}[lu]
}
$$
with $T'$ an affine scheme over $S$
and $i : T \to T'$ a closed immersion defined by an ideal of square zero.
\begin{enumerate}
\item[(4)] If $b \circ a$ is formally smooth, then for every $t \in T$
there exists an \'etale morphism of affines $U' \to T'$ and a morphism
$U' \to G$ such that
$$
\xymatrix{
G \ar[d]_b & T \ar[l] & T \times_{T'} U' \ar[d] \ar[l]\\
H & T' \ar[l] & U' \ar[llu] \ar[l]
}
$$
commutes and $t$ is in the image of $U' \to T'$.
\item[(5)] If $b \circ a$ is formally unramified, then there exists at most
one dotted arrow in the diagram above, i.e., $b$ is formally unramified.
\item[(6)] If $b \circ a$ is formally \'etale, then there exists exactly one
dotted arrow in the diagram above, i.e., $b$ is formally \'etale.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $b$ is formally smooth (resp.\ formally \'etale,
resp.\ formally unramified). Since an \'etale morphism is both
smooth and unramified we see that $a$ is representable and smooth
(resp.\ \'etale, resp. unramified). Hence parts (1), (2) and (3)
follow from a combination of
Lemma \ref{lemma-representable-property-formally-property}
and
Lemma \ref{lemma-composition-formally-smooth-etale-unramified}.
\medskip\noindent
Assume that $b \circ a$ is formally smooth. Consider a diagram
as in the statement of the lemma. Let $W = F \times_G T$.
By assumption $W$ is a scheme surjective \'etale over $T$. By
\'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence}
there exists a scheme $W'$ \'etale over $T'$ such that $W = T \times_{T'} W'$.
Choose an affine open subscheme $U' \subset W'$ such that $t$ is in
the image of $U' \to T'$. Because $b \circ a$ is formally
smooth we see that the exist morphisms $U' \to F$ such that
$$
\xymatrix{
F \ar[d]_{b \circ a} & W \ar[l] & T \times_{T'} U' \ar[d] \ar[l]\\
H & T' \ar[l] & U' \ar[llu] \ar[l]
}
$$
commutes. Taking the composition $U' \to F \to G$ gives a
map as in part (5) of the lemma.
\medskip\noindent
Assume that $f, g : T' \to G$ are two dotted arrows fitting into the
diagram of the lemma. Let $W = F \times_G T$.
By assumption $W$ is a scheme surjective \'etale over $T$. By
\'Etale Morphisms, Theorem \ref{etale-theorem-remarkable-equivalence}
there exists a scheme $W'$ \'etale over $T'$ such that $W = T \times_{T'} W'$.
Since $a$ is formally \'etale the compositions
$$
W' \to T' \xrightarrow{f} G
\quad\text{and}\quad
W' \to T' \xrightarrow{g} G
$$
lift to morphisms $f', g' : W' \to F$ (lift on affine opens and glue by
uniqueness). Now if $b \circ a : F \to H$ is formally unramified, then
$f' = g'$ and hence $f = g$ as $W' \to T'$ is an \'etale covering. This proves
part (6) of the lemma.
\medskip\noindent
Assume that $b \circ a$ is formally \'etale. Then by part (4) we
can \'etale locally on $T'$ find a dotted arrow fitting into the diagram
and by part (5) this dotted arrow is unique. Hence we may glue the
local solutions to get assertion (6). Some details omitted.
\end{proof}
\begin{remark}
\label{remark-tempting}
It is tempting to think that in the situation of
Lemma \ref{lemma-etale-on-top}
we have
``$b$ formally smooth'' $\Leftrightarrow$ ``$b \circ a$ formally smooth''.
However, this is likely not true in general.
\end{remark}
\begin{lemma}
\label{lemma-formally-permanence}
Let $S$ be a scheme.
Let $F, G, H : (\Sch/S)_{fppf}^{opp} \to \textit{Sets}$.
Let $a : F \to G$, $b : G \to H$ be transformations of functors.
Assume $b$ is formally unramified.
\begin{enumerate}
\item If $b \circ a$ is formally unramified then $a$ is formally unramified.
\item If $b \circ a$ is formally \'etale then $a$ is formally \'etale.
\item If $b \circ a$ is formally smooth then $a$ is formally smooth.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $T \subset T'$ be a closed immersion of affine schemes defined by an ideal
of square zero. Let $g' : T' \to G$ and $f : T \to F$ be given such that
$g'|_T = a \circ f$. Because $b$ is formally unramified, there is a one
to one correspondence between
$$
\{f' : T' \to F \mid f = f'|_T\text{ and }a \circ f' = g'\}
$$
and
$$
\{f' : T' \to F \mid f = f'|_T\text{ and }b \circ a \circ f' = b \circ g'\}.
$$
From this the lemma follows formally.
\end{proof}
\section{Formally unramified morphisms}
\label{section-formally-unramified}
\noindent
In this section we work out what it means that a morphism of algebraic spaces
is formally unramified.
\begin{definition}
\label{definition-formally-unramified}
Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$
is said to be {\it formally unramified} if it is formally unramified as a
transformation of functors as in
Definition \ref{definition-formally-smooth-etale-unramified}.
\end{definition}
\noindent
We will not restate the results proved in the more general setting of
formally unramified transformations of functors in
Section \ref{section-formally-smooth-etale-unramified}.
It turns out we can characterize this property in terms of vanishing of the
module of relative differentials, see
Lemma \ref{lemma-characterize-formally-unramified}.
\begin{lemma}
\label{lemma-formally-unramified}
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 formally unramified,
\item for every diagram
$$
\xymatrix{
U \ar[d] \ar[r]_\psi & V \ar[d] \\
X \ar[r]^f & Y
}
$$
where $U$ and $V$ are schemes and the vertical arrows are \'etale
the morphism of schemes $\psi$ is formally unramified (as in
More on Morphisms,
Definition \ref{more-morphisms-definition-formally-unramified}), and
\item for one such diagram with surjective vertical arrows the morphism
$\psi$ is formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $f$ is formally unramified. By
Lemma \ref{lemma-representable-property-formally-property}
the morphisms $U \to X$ and $V \to Y$ are formally unramified. Thus by
Lemma \ref{lemma-composition-formally-smooth-etale-unramified}
the composition $U \to Y$ is formally unramified. Then it follows from
Lemma \ref{lemma-formally-permanence}
that $U \to V$ is formally unramified. Thus (1) implies (2). And (2)
implies (3) trivially
\medskip\noindent
Assume given a diagram as in (3). By
Lemma \ref{lemma-representable-property-formally-property}
the morphism $V \to Y$ is formally unramified. Thus by
Lemma \ref{lemma-composition-formally-smooth-etale-unramified}
the composition $U \to Y$ is formally unramified. Then it follows from
Lemma \ref{lemma-etale-on-top}
that $X \to Y$ is formally unramified, i.e., (1) holds.
\end{proof}
\begin{lemma}
\label{lemma-formally-unramified-not-affine}
Let $S$ be a scheme.
If $f : X \to Y$ is a formally unramified morphism of algebraic spaces
over $S$, then given any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l] \\
S & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of algebraic spaces
over $S$ there exists at most one dotted arrow making the diagram commute.
In other words, in
Definition \ref{definition-formally-unramified}
the condition that $T$ be an affine scheme may be dropped.
\end{lemma}
\begin{proof}
This is true because there exists a surjective \'etale morphism
$U' \to T'$ where $U'$ is a disjoint union of affine schemes (see
Properties of Spaces, Lemma
\ref{spaces-properties-lemma-cover-by-union-affines})
and a morphism $T' \to X$ is determined by its restriction to $U'$.
\end{proof}
\begin{lemma}
\label{lemma-composition-formally-unramified}
A composition of formally unramified morphisms is formally unramified.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-base-change-formally-unramified}
A base change of a formally unramified morphism is formally unramified.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-characterize-formally-unramified}
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 formally unramified, and
\item $\Omega_{X/Y} = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
This is a combination of
Lemma \ref{lemma-formally-unramified},
More on Morphisms,
Lemma \ref{more-morphisms-lemma-formally-unramified-differentials},
and
Lemma \ref{lemma-localize-differentials}.
\end{proof}
\begin{lemma}
\label{lemma-unramified-formally-unramified}
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 morphism $f$ is unramified,
\item the morphism $f$ is locally of finite type and $\Omega_{X/Y} = 0$, and
\item the morphism $f$ is locally of finite type and formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
Choose a diagram
$$
\xymatrix{
U \ar[d] \ar[r]_\psi & V \ar[d] \\
X \ar[r]^f & Y
}
$$
where $U$ and $V$ are schemes and the vertical arrows are \'etale and
surjective. Then we see
\begin{align*}
f\text{ unramified}
& \Leftrightarrow
\psi\text{ unramified} \\
& \Leftrightarrow
\psi\text{ locally finite type and }\Omega_{U/V} = 0 \\
& \Leftrightarrow
f\text{ locally finite type and }\Omega_{X/Y} = 0 \\
& \Leftrightarrow
f\text{ locally finite type and formally unramified}
\end{align*}
Here we have used
Morphisms, Lemma \ref{morphisms-lemma-unramified-omega-zero} and
Lemma \ref{lemma-characterize-formally-unramified}.
\end{proof}
\begin{lemma}
\label{lemma-universally-injective-unramified}
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 unramified and a monomorphism,
\item $f$ is unramified and universally injective,
\item $f$ is locally of finite type and a monomorphism,
\item $f$ is universally injective, locally of finite type, and
formally unramified.
\end{enumerate}
Moreover, in this case $f$ is also representable, separated, and
locally quasi-finite.
\end{lemma}
\begin{proof}
We have seen in
Lemma \ref{lemma-unramified-formally-unramified}
that being formally unramified and locally of finite type is the same thing
as being unramified.
Hence (4) is equivalent to (2).
A monomorphism is certainly formally unramified hence (3) implies (4).
It is clear that (1) implies (3). Finally, if (2) holds, then
$\Delta : X \to X \times_Y X$ is both an open immersion
(Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-diagonal-unramified-morphism})
and surjective
(Morphisms of Spaces, Lemma
\ref{spaces-morphisms-lemma-universally-injective})
hence an isomorphism, i.e., $f$ is a monomorphism. In this way we see that
(2) implies (1).
Finally, we see that $f$ is representable, separated, and locally
quasi-finite by
Morphisms of Spaces, Lemmas
\ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite} and
\ref{spaces-morphisms-lemma-locally-quasi-finite-separated-representable}.
\end{proof}
\begin{lemma}
\label{lemma-characterize-closed-immersion}
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,
\item $f$ is universally closed, unramified, and a monomorphism,
\item $f$ is universally closed, unramified, and universally injective,
\item $f$ is universally closed, locally of finite type, and a monomorphism,
\item $f$ is universally closed, universally injective, locally of
finite type, and formally unramified.
\end{enumerate}
\end{lemma}
\begin{proof}
The equivalence of (2) -- (5) follows immediately from
Lemma \ref{lemma-universally-injective-unramified}.
Moreover, if (2) -- (5) are satisfied then $f$ is representable.
Similarly, if (1) is satisfied then $f$ is representable.
Hence the result follows from the case of schemes, see
\'Etale Morphisms, Lemma \ref{etale-lemma-characterize-closed-immersion}.
\end{proof}
\section{Universal first order thickenings}
\label{section-universal-thickening}
\noindent
Let $S$ be a scheme.
Let $h : Z \to X$ be a morphism of algebraic spaces over $S$.
A {\it universal first order thickening} of $Z$ over $X$ is a
first order thickening $Z \subset Z'$ over $X$ such that given
any first order thickening $T \subset T'$
over $X$ and a solid commutative diagram
\begin{equation}
\label{equation-universal-first-order-thickening}
\vcenter{
\xymatrix{
& Z \ar[ld] & & T \ar[rd] \ar[ll]^a \\
Z' \ar[rrd] & & & & T' \ar@{..>}[llll]_{a'} \ar[lld]^b \\
& & X
}
}
\end{equation}
there exists a unique dotted arrow making the diagram commute.
Note that in this situation $(a, a') : (T \subset T') \to (Z \subset Z')$
is a morphism of thickenings over $X$. Thus if a universal first order
thickening exists, then it is unique up to unique isomorphism.
In general a universal first order thickening
does not exist, but if $h$ is formally unramified then it does.
Before we prove this, let us show that a universal first order thickening
in the category of schemes is a universal first order thickening in the
category of algebraic spaces.
\begin{lemma}
\label{lemma-check-universal-first-order-thickening}
Let $S$ be a scheme.
Let $h : Z \to X$ be a morphism of algebraic spaces over $S$.
Let $Z \subset Z'$ be a first order thickening over $X$.
The following are equivalent
\begin{enumerate}
\item $Z \subset Z'$ is a universal first order thickening,
\item for any diagram (\ref{equation-universal-first-order-thickening})
with $T'$ a scheme a unique dotted arrow exists making the diagram commute, and
\item for any diagram (\ref{equation-universal-first-order-thickening})
with $T'$ an affine scheme a unique dotted arrow exists making the
diagram commute.
\end{enumerate}
\end{lemma}
\begin{proof}
The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) are formal.
Assume (3) a assume given an arbitrary diagram
(\ref{equation-universal-first-order-thickening}).
Choose a presentation $T' = U'/R'$, see
Spaces, Definition \ref{spaces-definition-presentation}.
We may assume that $U' = \coprod U'_i$ is a disjoint union
of affines, so $R' = U' \times_{T'} U' = \coprod_{i, j} U'_i \times_T' U'_j$.
For each pair $(i, j)$ choose an affine open covering
$U'_i \times_T' U'_j = \bigcup_k R'_{ijk}$. Denote $U_i, R_{ijk}$
the fibre products with $T$ over $T'$. Then each
$U_i \subset U'_i$ and $R_{ijk} \subset R'_{ijk}$
is a first order thickening of affine schemes.
Denote $a_i : U_i \to Z$, resp.\ $a_{ijk} : R_{ijk} \to Z$
the composition of $a : T \to Z$ with the morphism
$U_i \to T$, resp.\ $R_{ijk} \to T$.
By (3) applied to $a_i : U_i \to Z$
we obtain unique morphisms $a'_i : U'_i \to Z'$.
By (3) applied to $a_{ijk}$ we see that the two compositions
$R'_{ijk} \to R'_i \to Z'$ and $R'_{ijk} \to R'_j \to Z'$
are equal. Hence $a' = \coprod a'_i : U' = \coprod U'_i \to Z'$
descends to the quotient sheaf $T' = U'/R'$ and we win.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-over-formally-etale}
Let $S$ be a scheme.
Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$.
If $Z \subset Z'$ is a universal first order thickening of
$Z$ over $Y$ and $Y \to X$ is formally \'etale, then $Z \subset Z'$ is
a universal first order thickening of $Z$ over $X$.
\end{lemma}
\begin{proof}
This is formal. Namely, by
Lemma \ref{lemma-check-universal-first-order-thickening}
it suffices to consider solid commutative diagrams
(\ref{equation-universal-first-order-thickening})
with $T'$ an affine scheme. The composition
$T \to Z \to Y$ lifts uniquely to $T' \to Y$ as $Y \to X$ is
assumed formally \'etale. Hence the fact that
$Z \subset Z'$ is a universal first order thickening over $Y$
produces the desired morphism $a' : T' \to Z'$.
\end{proof}
\begin{lemma}
\label{lemma-etale-morphism-of-universal-thickenings}
Let $S$ be a scheme.
Let $Z \to Y \to X$ be morphisms of algebraic spaces over $S$.
Assume $Z \to Y$ is \'etale.
\begin{enumerate}
\item If $Y \subset Y'$ is a universal first order thickening of
$Y$ over $X$, then the unique \'etale morphism $Z' \to Y'$ such
that $Z = Y \times_{Y'} Z'$ (see
Theorem \ref{theorem-topological-invariance})
is a universal first order thickening of $Z$ over $X$.
\item If $Z \to Y$ is surjective and
$(Z \subset Z') \to (Y \subset Y')$ is an \'etale morphism
of first order thickenings over $X$ and $Z'$ is a universal first
order thickening of $Z$ over $X$, then $Y'$ is a universal first
order thickening of $Y$ over $X$.
\end{enumerate}
\end{lemma}
\begin{proof}
Proof of (1). By
Lemma \ref{lemma-check-universal-first-order-thickening}
it suffices to consider solid commutative diagrams
(\ref{equation-universal-first-order-thickening})
with $T'$ an affine scheme. The composition
$T \to Z \to Y$ lifts uniquely to $T' \to Y'$ as $Y'$ is
the universal first order thickening. Then the fact that
$Z' \to Y'$ is \'etale implies (see
Lemma \ref{lemma-representable-property-formally-property})
that $T' \to Y'$ lifts to the
desired morphism $a' : T' \to Z'$.
\medskip\noindent
Proof of (2). Let $T \subset T'$ be a first order thickening over
$X$ and let $a : T \to Y$ be a morphism. Set $W = T \times_Y Z$
and denote $c : W \to Z$ the projection
Let $W' \to T'$ be the unique \'etale morphism such that
$W = T \times_{T'} W'$, see
Theorem \ref{theorem-topological-invariance}.
Note that $W' \to T'$ is surjective as $Z \to Y$ is surjective.
By assumption we obtain a unique morphism $c' : W' \to Z'$
over $X$ restricting to $c$ on $W$. By uniqueness the two restrictions
of $c'$ to $W' \times_{T'} W'$ are equal (as the two restrictions of
$c$ to $W \times_T W$ are equal). Hence $c'$ descends to a unique
morphism $a' : T' \to Y'$ and we win.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening}
Let $S$ be a scheme.
Let $h : Z \to X$ be a formally unramified morphism of algebraic
spaces over $S$.
There exists a universal first order thickening $Z \subset Z'$ of
$Z$ over $X$.
\end{lemma}
\begin{proof}
Choose any commutative diagram
$$
\xymatrix{
V \ar[d] \ar[r] & U \ar[d] \\
Z \ar[r] & X
}
$$
where $V$ and $U$ are schemes and the vertical arrows are \'etale.
Note that $V \to U$ is a formally unramified morphism of schemes, see
Lemma \ref{lemma-formally-unramified}.
Combining
Lemma \ref{lemma-check-universal-first-order-thickening}
and
More on Morphisms, Lemma \ref{more-morphisms-lemma-universal-thickening}
we see that a universal first order thickening $V \subset V'$
of $V$ over $U$ exists. By
Lemma \ref{lemma-universal-thickening-over-formally-etale} part (1)
$V'$ is a universal first order thickening of $V$ over $X$.
\medskip\noindent
Fix a scheme $U$ and a surjective \'etale morphism $U \to X$.
The argument above shows that for any $V \to Z$ \'etale with $V$
a scheme such that $V \to Z \to X$ factors through $U$ a
universal first order thickening $V \subset V'$ of $V$ over $X$
exists (but does not depend on the chosen factorization of $V \to X$
through $U$). Now we may choose $V$ such that $V \to Z$ is surjective
\'etale (see
Spaces, Lemma \ref{spaces-lemma-lift-morphism-presentations}).
Then $R = V \times_Z V$ a scheme \'etale over $Z$ such that
$R \to X$ factors through $U$ also.
Hence we obtain universal first order thickenings
$V \subset V'$ and $R \subset R'$ over $X$.
As $V \subset V'$ is a universal first order thickening,
the two projections $s, t : R \to V$ lift to morphisms
$s', t': R' \to V'$. By
Lemma \ref{lemma-etale-morphism-of-universal-thickenings}
as $R'$ is the universal first order thickening of $R$ over $X$
these morphisms are \'etale.
Then $(t', s') : R' \to V'$ is an \'etale equivalence relation
and we can set $Z' = V'/R'$. Since $V' \to Z'$ is surjective \'etale
and $v'$ is the universal first order thickening of $V$ over $X$
we conclude from
Lemma \ref{lemma-universal-thickening-over-formally-etale} part (2)
that $Z'$ is a universal first order thickening of $Z$ over $X$.
\end{proof}
\begin{definition}
\label{definition-universal-thickening}
Let $S$ be a scheme.
Let $h : Z \to X$ be a formally unramified morphism of
algebraic spaces over $S$.
\begin{enumerate}
\item The {\it universal first order thickening} of $Z$ over $X$
is the thickening $Z \subset Z'$ constructed in
Lemma \ref{lemma-universal-thickening}.
\item The {\it conormal sheaf of $Z$ over $X$} is the conormal sheaf
of $Z$ in its universal first order thickening $Z'$ over $X$.
\end{enumerate}
We often denote the conormal sheaf $\mathcal{C}_{Z/X}$ in this situation.
\end{definition}
\noindent
Thus we see that there is a short exact sequence of sheaves
$$
0 \to \mathcal{C}_{Z/X} \to \mathcal{O}_{Z'} \to \mathcal{O}_Z \to 0
$$
on $Z_\etale$ and $\mathcal{C}_{Z/X}$ is a quasi-coherent
$\mathcal{O}_Z$-module. The following lemma proves that there is no
conflict between this definition and the definition in case $Z \to X$
is an immersion.
\begin{lemma}
\label{lemma-immersion-universal-thickening}
Let $S$ be a scheme.
Let $i : Z \to X$ be an immersion of algebraic spaces over $S$. Then
\begin{enumerate}
\item $i$ is formally unramified,
\item the universal first order thickening of $Z$ over $X$ is the first order
infinitesimal neighbourhood of $Z$ in $X$ of
Definition \ref{definition-first-order-infinitesimal-neighbourhood},
\item the conormal sheaf of $i$ in the sense of
Definition \ref{definition-conormal-sheaf}
agrees with the conormal sheaf of $i$ in the sense of
Definition \ref{definition-universal-thickening}.
\end{enumerate}
\end{lemma}
\begin{proof}
An immersion of algebraic spaces is by definition a representable morphism.
Hence by
Morphisms, Lemmas \ref{morphisms-lemma-open-immersion-unramified} and
\ref{morphisms-lemma-closed-immersion-unramified}
an immersion is unramified (via the abstract principle of
Spaces, Lemma
\ref{spaces-lemma-representable-transformations-property-implication}).
Hence it is formally unramified by
Lemma \ref{lemma-unramified-formally-unramified}.
The other assertions follow by combining
Lemmas \ref{lemma-first-order-infinitesimal-neighbourhood} and
\ref{lemma-infinitesimal-neighbourhood-conormal}
and the definitions.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-unramified}
Let $S$ be a scheme.
Let $Z \to X$ be a formally unramified morphism of algebraic spaces over $S$.
Then the universal first order thickening $Z'$ is formally
unramified over $X$.
\end{lemma}
\begin{proof}
Let $T \subset T'$ be a first order thickening of affine schemes over $X$.
Let
$$
\xymatrix{
Z' \ar[d] & T \ar[l]^c \ar[d] \\
X & T' \ar[l] \ar[lu]^{a, b}
}
$$
be a commutative diagram. Set $T_0 = c^{-1}(Z) \subset T$ and
$T'_a = a^{-1}(Z)$ (scheme theoretically).
Since $Z'$ is a first order thickening of $Z$, we see that $T'$
is a first order thickening of $T'_a$. Moreover, since $c = a|_T$ we see that
$T_0 = T \cap T'_a$ (scheme theoretically). As $T'$ is a first order
thickening of $T$ it follows that $T'_a$
is a first order thickening of $T_0$. Now $a|_{T'_a}$ and $b|_{T'_a}$
are morphisms of $T'_a$ into $Z'$ over $X$ which agree on $T_0$ as
morphisms into $Z$. Hence by the universal property of $Z'$ we conclude that
$a|_{T'_a} = b|_{T'_a}$. Thus $a$ and $b$ are morphism from
the first order thickening $T'$ of $T'_a$ whose restrictions to
$T'_a$ agree as morphisms into $Z$. Thus using the universal property of
$Z'$ once more we conclude that $a = b$. In other words, the defining
property of a formally unramified morphism holds for $Z' \to X$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-functorial}
Let $S$ be a scheme
Consider a commutative diagram of algebraic spaces over $S$
$$
\xymatrix{
Z \ar[r]_h \ar[d]_f & X \ar[d]^g \\
W \ar[r]^{h'} & Y
}
$$
with $h$ and $h'$ formally unramified. Let $Z \subset Z'$ be the universal
first order thickening of $Z$ over $X$. Let $W \subset W'$ be the universal
first order thickening of $W$ over $Y$. There exists a canonical morphism
$(f, f') : (Z, Z') \to (W, W')$ of thickenings over $Y$ which fits into
the following commutative diagram
$$
\xymatrix{
& & & Z' \ar[ld] \ar[d]^{f'} \\
Z \ar[rr] \ar[d]_f \ar[rrru] & & X \ar[d] & W' \ar[ld] \\
W \ar[rrru]|!{[rr];[rruu]}\hole \ar[rr] & & Y
}
$$
In particular the morphism $(f, f')$ of thickenings induces a morphism
of conormal sheaves $f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$.
\end{lemma}
\begin{proof}
The first assertion is clear from the universal property of $W'$.
The induced map on conormal sheaves is the map of
Lemma \ref{lemma-conormal-functorial}
applied to $(Z \subset Z') \to (W \subset W')$.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-fibre-product}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_h \ar[d]_f & X \ar[d]^g \\
W \ar[r]^{h'} & Y
}
$$
be a fibre product diagram of algebraic spaces over $S$ with
$h'$ formally unramified. Then $h$ is formally unramified and if
$W \subset W'$ is the universal first order thickening of $W$ over $Y$,
then $Z = X \times_Y W \subset X \times_Y W'$ is the universal
first order thickening of $Z$ over $X$. In particular the canonical map
$f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ of
Lemma \ref{lemma-universal-thickening-functorial}
is surjective.
\end{lemma}
\begin{proof}
The morphism $h$ is formally unramified by
Lemma \ref{lemma-base-change-formally-unramified}.
It is clear that $X \times_Y W'$ is a first order thickening.
It is straightforward to check that it has the universal property
because $W'$ has the universal property (by mapping properties of
fibre products). See
Lemma \ref{lemma-conormal-functorial-flat}
for why this implies that the map of conormal sheaves is surjective.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-fibre-product-flat}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_h \ar[d]_f & X \ar[d]^g \\
W \ar[r]^{h'} & Y
}
$$
be a fibre product diagram of algebraic spaces over $S$ with
$h'$ formally unramified and $g$ flat. In this case the corresponding
map $Z' \to W'$ of universal first order thickenings is flat, and
$f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ is an isomorphism.
\end{lemma}
\begin{proof}
Flatness is preserved under base change, see
Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-flat}.
Hence the first statement follows from the description of $W'$ in
Lemma \ref{lemma-universal-thickening-fibre-product}.
It is clear that $X \times_Y W'$ is a first order thickening.
It is straightforward to check that it has the universal property
because $W'$ has the universal property (by mapping properties of
fibre products). See
Lemma \ref{lemma-conormal-functorial-flat}
for why this implies that the map of conormal sheaves is an isomorphism.
\end{proof}
\begin{lemma}
\label{lemma-universal-thickening-localize}
Taking the universal first order thickenings commutes with \'etale
localization. More precisely, let $h : Z \to X$ be a formally unramified
morphism of algebraic spaces over a base scheme $S$.
Let
$$
\xymatrix{
V \ar[d] \ar[r] & U \ar[d] \\
Z \ar[r] & X
}
$$
be a commutative diagram with \'etale vertical arrows.
Let $Z'$ be the universal first order thickening of $Z$ over $X$.
Then $V \to U$ is formally unramified and the universal first
order thickening $V'$ of $V$ over $U$ is \'etale over $Z'$.
In particular, $\mathcal{C}_{Z/X}|_V = \mathcal{C}_{V/U}$.
\end{lemma}
\begin{proof}
The first statement is
Lemma \ref{lemma-formally-unramified}.
The compatibility of universal first order thickenings is
a consequence of
Lemmas \ref{lemma-universal-thickening-over-formally-etale} and
\ref{lemma-etale-morphism-of-universal-thickenings}.
\end{proof}
\begin{lemma}
\label{lemma-differentials-universally-unramified}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $h : Z \to X$ be a formally unramified morphism of algebraic spaces
over $B$. Let $Z \subset Z'$ be the universal first order thickening of $Z$
over $X$ with structure morphism $h' : Z' \to X$. The canonical map
$$
\text{d}h' : (h')^*\Omega_{X/B} \to \Omega_{Z'/B}
$$
induces an isomorphism
$h^*\Omega_{X/B} \to \Omega_{Z'/B} \otimes \mathcal{O}_Z$.
\end{lemma}
\begin{proof}
The map $c_{h'}$ is the map defined in
Lemma \ref{lemma-functoriality-differentials}.
If $i : Z \to Z'$ is the given closed immersion, then
$i^*c_{h'}$ is a map
$h^*\Omega_{X/S} \to \Omega_{Z'/S} \otimes \mathcal{O}_Z$.
Checking that it is an isomorphism reduces to the case of schemes
by \'etale localization, see
Lemma \ref{lemma-universal-thickening-localize}
and
Lemma \ref{lemma-localize-differentials}.
In this case the result is
More on Morphisms,
Lemma \ref{more-morphisms-lemma-differentials-universally-unramified}.
\end{proof}
\begin{lemma}
\label{lemma-universally-unramified-differentials-sequence}
Let $S$ be a scheme. Let $B$ be an algebraic space over $S$.
Let $h : Z \to X$ be a formally unramified morphism of algebraic
spaces over $B$.
There is a canonical exact sequence
$$
\mathcal{C}_{Z/X} \to h^*\Omega_{X/B} \to \Omega_{Z/B} \to 0.
$$
The first arrow is induced by $\text{d}_{Z'/B}$ where
$Z'$ is the universal first order neighbourhood of $Z$ over $X$.
\end{lemma}
\begin{proof}
We know that there is a canonical exact sequence
$$
\mathcal{C}_{Z/Z'} \to
\Omega_{Z'/S} \otimes \mathcal{O}_Z \to
\Omega_{Z/S} \to 0.
$$
see
Lemma \ref{lemma-differentials-relative-immersion}.
Hence the result follows on applying
Lemma \ref{lemma-differentials-universally-unramified}.
\end{proof}
\begin{lemma}
\label{lemma-two-unramified-morphisms}
Let $S$ be a scheme. Let
$$
\xymatrix{
Z \ar[r]_i \ar[rd]_j & X \ar[d] \\
& Y
}
$$
be a commutative diagram of algebraic spaces over $S$
where $i$ and $j$ are formally unramified. Then there is a
canonical exact sequence
$$
\mathcal{C}_{Z/Y} \to
\mathcal{C}_{Z/X} \to
i^*\Omega_{X/Y} \to 0
$$
where the first arrow comes from
Lemma \ref{lemma-universal-thickening-functorial}
and the second from
Lemma \ref{lemma-universally-unramified-differentials-sequence}.
\end{lemma}
\begin{proof}
Since the maps have been defined, checking the sequence is exact
reduces to the case of schemes by \'etale localization, see
Lemma \ref{lemma-universal-thickening-localize}
and
Lemma \ref{lemma-localize-differentials}.
In this case the result is
More on Morphisms,
Lemma \ref{more-morphisms-lemma-two-unramified-morphisms}.
\end{proof}
\begin{lemma}
\label{lemma-transitivity-conormal-unramified}
Let $S$ be a scheme.
Let $Z \to Y \to X$ be formally unramified morphisms of
algebraic spaces over $S$.
\begin{enumerate}
\item If $Z \subset Z'$ is the universal first order thickening of $Z$
over $X$ and $Y \subset Y'$ is the universal first order thickening of $Y$
over $X$, then there is a morphism $Z' \to Y'$ and $Y \times_{Y'} Z'$ is
the universal first order thickening of $Z$ over $Y$.
\item There is a canonical exact sequence
$$
i^*\mathcal{C}_{Y/X} \to
\mathcal{C}_{Z/X} \to
\mathcal{C}_{Z/Y} \to 0
$$
where the maps come from
Lemma \ref{lemma-universal-thickening-functorial}
and $i : Z \to Y$ is the first morphism.
\end{enumerate}
\end{lemma}
\begin{proof}
The map $h : Z' \to Y'$ in (1) comes from
Lemma \ref{lemma-universal-thickening-functorial}.
The assertion that $Y \times_{Y'} Z'$ is the universal first order
thickening of $Z$ over $Y$ is clear from the universal properties
of $Z'$ and $Y'$. By
Lemma \ref{lemma-transitivity-conormal}
we have an exact sequence
$$
(i')^*\mathcal{C}_{Y \times_{Y'} Z'/Z'} \to
\mathcal{C}_{Z/Z'} \to
\mathcal{C}_{Z/Y \times_{Y'} Z'} \to 0
$$
where $i' : Z \to Y \times_{Y'} Z'$ is the given morphism. By
Lemma \ref{lemma-conormal-functorial-flat}
there exists a surjection
$h^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{Y \times_{Y'} Z'/Z'}$.
Combined with the equalities
$\mathcal{C}_{Y/Y'} = \mathcal{C}_{Y/X}$,
$\mathcal{C}_{Z/Z'} = \mathcal{C}_{Z/X}$, and
$\mathcal{C}_{Z/Y \times_{Y'} Z'} = \mathcal{C}_{Z/Y}$
this proves the lemma.
\end{proof}
\section{Formally \'etale morphisms}
\label{section-formally-etale}
\noindent
In this section we work out what it means that a morphism of algebraic spaces
is formally \'etale.
\begin{definition}
\label{definition-formally-etale}
Let $S$ be a scheme. A morphism $f : X \to Y$ of algebraic spaces over $S$
is said to be {\it formally \'etale} if it is formally \'etale as a
transformation of functors as in
Definition \ref{definition-formally-smooth-etale-unramified}.
\end{definition}
\noindent
We will not restate the results proved in the more general setting of
formally \'etale transformations of functors in
Section \ref{section-formally-smooth-etale-unramified}.
\begin{lemma}
\label{lemma-formally-etale}
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 formally \'etale,
\item for every diagram
$$
\xymatrix{
U \ar[d] \ar[r]_\psi & V \ar[d] \\
X \ar[r]^f & Y
}
$$
where $U$ and $V$ are schemes and the vertical arrows are \'etale
the morphism of schemes $\psi$ is formally \'etale (as in
More on Morphisms,
Definition \ref{more-morphisms-definition-formally-etale}), and
\item for one such diagram with surjective vertical arrows the morphism
$\psi$ is formally \'etale.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $f$ is formally \'etale. By
Lemma \ref{lemma-representable-property-formally-property}
the morphisms $U \to X$ and $V \to Y$ are formally \'etale. Thus by
Lemma \ref{lemma-composition-formally-smooth-etale-unramified}
the composition $U \to Y$ is formally \'etale. Then it follows from
Lemma \ref{lemma-formally-permanence}
that $U \to V$ is formally \'etale. Thus (1) implies (2). And (2)
implies (3) trivially
\medskip\noindent
Assume given a diagram as in (3). By
Lemma \ref{lemma-representable-property-formally-property}
the morphism $V \to Y$ is formally \'etale. Thus by
Lemma \ref{lemma-composition-formally-smooth-etale-unramified}
the composition $U \to Y$ is formally \'etale. Then it follows from
Lemma \ref{lemma-etale-on-top}
that $X \to Y$ is formally \'etale, i.e., (1) holds.
\end{proof}
\begin{lemma}
\label{lemma-formally-etale-not-affine}
Let $S$ be a scheme.
Let $f : X \to Y$ be a formally \'etale morphism of algebraic spaces over $S$.
Then given any solid commutative diagram
$$
\xymatrix{
X \ar[d]_f & T \ar[d]^i \ar[l]_a \\
Y & T' \ar[l] \ar@{-->}[lu]
}
$$
where $T \subset T'$ is a first order thickening of algebraic spaces
over $Y$ there exists exactly one dotted arrow making the diagram commute.
In other words, in
Definition \ref{definition-formally-etale}
the condition that $T$ be affine may be dropped.
\end{lemma}
\begin{proof}
Let $U' \to T'$ be a surjective \'etale morphism where $U' = \coprod U'_i$
is a disjoint union of affine schemes. Let
$U_i = T \times_{T'} U'_i$. Then we get morphisms
$a'_i : U'_i \to X$ such that $a'_i|_{U_i}$ equals the composition
$U_i \to T \to X$. By uniqueness (see
Lemma \ref{lemma-formally-unramified-not-affine})
we see that $a'_i$ and $a'_j$ agree on the fibre product
$U'_i \times_{T'} U'_j$. Hence $\coprod a'_i : U' \to X$
descends to give a unique morphism $a' : T' \to X$.
\end{proof}
\begin{lemma}
\label{lemma-composition-formally-etale}
A composition of formally \'etale morphisms is formally \'etale.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-base-change-formally-etale}
A base change of a formally \'etale morphism is formally \'etale.
\end{lemma}
\begin{proof}
This is formal.
\end{proof}
\begin{lemma}
\label{lemma-characterize-formally-etale}
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 formally \'etale,
\item $f$ is formally unramified and the universal first order thickening
of $X$ over $Y$ is equal to $X$,
\item $f$ is formally unramified and $\mathcal{C}_{X/Y} = 0$