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 \input{preamble} % OK, start here. % \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 $\mathcal{C}_{X/Y} = 0$