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defos.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Deformation Theory}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
The goal of this chapter is to give a (relatively) gentle introduction to
deformation theory of modules, morphisms, etc. In this chapter we deal with
those results that can be proven using the naive cotangent complex. In
the chapter on the cotangent complex we will extend these results a little
bit. The advanced reader may wish to consult the treatise by Illusie on this
subject, see \cite{cotangent}.
\section{Deformations of rings and the naive cotangent complex}
\label{section-deformations}
\noindent
In this section we use the naive cotangent complex to do a little bit
of deformation theory. We start with a surjective ring map $A' \to A$
whose kernel is an ideal $I$ of square zero. Moreover we assume
given a ring map $A \to B$, a $B$-module $N$, and an $A$-module map
$c : I \to N$. In this section we ask ourselves whether we can find
the question mark fitting into the following diagram
\begin{equation}
\label{equation-to-solve}
\vcenter{
\xymatrix{
0 \ar[r] & N \ar[r] & {?} \ar[r] & B \ar[r] & 0 \\
0 \ar[r] & I \ar[u]^c \ar[r] & A' \ar[u] \ar[r] & A \ar[u] \ar[r] & 0
}
}
\end{equation}
and moreover how unique the solution is (if it exists). More precisely,
we look for a surjection of $A'$-algebras $B' \to B$ whose kernel is
identified with $N$ such that $A' \to B'$ induces the given map $c$.
We will say $B'$ is a {\it solution} to (\ref{equation-to-solve}).
\begin{lemma}
\label{lemma-huge-diagram}
Given a commutative diagram
$$
\xymatrix{
& 0 \ar[r] & N_2 \ar[r] & B'_2 \ar[r] & B_2 \ar[r] & 0 \\
& 0 \ar[r]|\hole & I_2 \ar[u]_{c_2} \ar[r] &
A'_2 \ar[u] \ar[r]|\hole & A_2 \ar[u] \ar[r] & 0 \\
0 \ar[r] & N_1 \ar[ruu] \ar[r] & B'_1 \ar[r] & B_1 \ar[ruu] \ar[r] & 0 \\
0 \ar[r] & I_1 \ar[ruu]|\hole \ar[u]^{c_1} \ar[r] &
A'_1 \ar[ruu]|\hole \ar[u] \ar[r] & A_1 \ar[ruu]|\hole \ar[u] \ar[r] & 0
}
$$
with front and back solutions to (\ref{equation-to-solve}) we have
\begin{enumerate}
\item There exist a canonical element in
$\text{Ext}^1_{B_1}(\NL_{B_1/A_1}, N_2)$
whose vanishing is a necessary and sufficient condition for the existence
of a ring map $B'_1 \to B'_2$ fitting into the diagram.
\item If there exists a map $B'_1 \to B'_2$ fitting into the diagram
the set of all such maps is a principal homogeneous space under
$\Hom_{B_1}(\Omega_{B_1/A_1}, N_2)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $E = B_1$ viewed as a set.
Consider the surjection $A_1[E] \to B_1$ with kernel $J$ used
to define the naive cotangent complex by the formula
$$
\NL_{B_1/A_1} = (J/J^2 \to \Omega_{A_1[E]/A_1} \otimes_{A_1[E]} B_1)
$$
in
Algebra, Section \ref{algebra-section-netherlander}.
Since $\Omega_{A_1[E]/A_1} \otimes B_1$ is a free
$B_1$-module we have
$$
\text{Ext}^1_{B_1}(\NL_{B_1/A_1}, N_2) =
\frac{\Hom_{B_1}(J/J^2, N_2)}
{\Hom_{B_1}(\Omega_{A_1[E]/A_1} \otimes B_1, N_2)}
$$
We will construct an obstruction in the module on the right.
Let $J' = \text{Ker}(A'_1[E] \to B_1)$. Note that there is a surjection
$J' \to J$ whose kernel is $I_1A_1[E]$.
For every $e \in E$ denote $x_e \in A_1[E]$ the corresponding variable.
Choose a lift $y_e \in B'_1$ of the image of $x_e$ in $B_1$ and
a lift $z_e \in B'_2$ of the image of $x_e$ in $B_2$.
These choices determine $A'_1$-algebra maps
$$
A'_1[E] \to B'_1 \quad\text{and}\quad A'_1[E] \to B'_2
$$
The first of these gives a map $J' \to N_1$, $f' \mapsto f'(y_e)$
and the second gives a map $J' \to N_2$, $f' \mapsto f'(z_e)$.
A calculation shows that these maps annihilate $(J')^2$.
Because the left square of the diagram (involving $c_1$ and $c_2$)
commutes we see that these maps agree on $I_1A_1[E]$ as maps into $N_2$.
Observe that $B'_1$ is the pushout of $J' \to A'_1[B_1]$ and $J' \to N_1$.
Thus, if the maps $J' \to N_1 \to N_2$ and $J' \to N_2$ agree, then we
obtain a map $B'_1 \to B'_2$ fitting into the diagram.
Thus we let the obstruction be the class of the map
$$
J/J^2 \to N_2,\quad f \mapsto f'(z_e) - \nu(f'(y_e))
$$
where $\nu : N_1 \to N_2$ is the given map and where $f' \in J'$
is a lift of $f$. This is well defined by our remarks above.
Note that we have the freedom
to modify our choices of $z_e$ into $z_e + \delta_{2, e}$
and $y_e$ into $y_e + \delta_{1, e}$ for some $\delta_{i, e} \in N_i$.
This will modify the map above into
$$
f \mapsto f'(z_e + \delta_{2, e}) - \nu(f'(y_e + \delta_{1, e})) =
f'(z_e) - \nu(f'(z_e)) +
\sum (\delta_{2, e} - \nu(\delta_{1, e}))\frac{\partial f}{\partial x_e}
$$
This means exactly that we are modifying the map $J/J^2 \to N_2$
by the composition $J/J^2 \to \Omega_{A_1[E]/A_1} \otimes B_1 \to N_2$
where the second map sends $\text{d}x_e$ to
$\delta_{2, e} - \nu(\delta_{1, e})$. Thus our obstruction is well defined
and is zero if and only if a lift exists.
\medskip\noindent
Part (2) comes from the observation that given two maps
$\varphi, \psi : B'_1 \to B'_2$ fitting into the diagram, then
$\varphi - \psi$ factors through a map $D : B_1 \to N_2$ which
is an $A_1$-derivation:
\begin{align*}
D(fg) & = \varphi(f'g') - \psi(f'g') \\
& =
\varphi(f')\varphi(g') - \psi(f')\psi(g') \\
& =
(\varphi(f') - \psi(f'))\varphi(g') + \psi(f')(\varphi(g') - \psi(g')) \\
& =
gD(f) + fD(g)
\end{align*}
Thus $D$ corresponds to a unique $B_1$-linear map
$\Omega_{B_1/A_1} \to N_2$. Conversely, given such a linear map
we get a derivation $D$ and given a ring map $\psi : B'_1 \to B'_2$
fitting into the diagram
the map $\psi + D$ is another ring map fitting into the diagram.
\end{proof}
\noindent
The naive cotangent complex isn't good enough to contain all information
regarding obstructions to finding solutions to (\ref{equation-to-solve}).
However, if the ring map is a local complete intersection, then the
obstruction vanishes. This is a kind of lifting result; observe that
for syntomic ring maps we have proved a rather strong lifting result in
Smoothing Ring Maps, Proposition \ref{smoothing-proposition-lift-smooth}.
\begin{lemma}
\label{lemma-existence-lci}
If $A \to B$ is a local complete intersection ring map, then
there exists a solution to (\ref{equation-to-solve}).
\end{lemma}
\begin{proof}
Write $B = A[x_1, \ldots, x_n]/J$. Let $J' \subset A'[x_1, \ldots, x_n]$
be the inverse image of $J$. Denote $I[x_1, \ldots, x_n]$ the
kernel of $A'[x_1, \ldots, x_n] \to A[x_1, \ldots, x_n]$.
By More on Algebra, Lemma
\ref{more-algebra-lemma-conormal-sequence-H1-regular-ideal} we have
$I[x_1, \ldots, x_n] \cap (J')^2 = J'I[x_1, \ldots, x_n] =
JI[x_1, \ldots, x_n]$. Hence we obtain a short exact sequence
$$
0 \to I \otimes_A B \to J'/(J')^2 \to J/J^2 \to 0
$$
Since $J/J^2$ is projective (More on Algebra, Lemma
\ref{more-algebra-lemma-quasi-regular-ideal-finite-projective})
we can choose a splitting of this sequence
$$
J'/(J')^2 = I \otimes_A B \oplus J/J^2
$$
Let $(J')^2 \subset J'' \subset J'$ be the elements which map to the
second summand in the decomposition above. Then
$$
0 \to I \otimes_A B \to A'[x_1, \ldots, x_n]/J'' \to B \to 0
$$
is a solution to (\ref{equation-to-solve}) with $N = I \otimes_A B$.
The general case is obtained by doing a pushout along the given
map $I \otimes_A B \to N$.
\end{proof}
\begin{lemma}
\label{lemma-choices}
If there exists a solution to (\ref{equation-to-solve}), then the set of
isomorphism classes of solutions is principal homogeneous under
$\text{Ext}^1_B(\NL_{B/A}, N)$.
\end{lemma}
\begin{proof}
We observe right away that given two solutions $B'_1$ and $B'_2$
to (\ref{equation-to-solve}) we obtain by Lemma \ref{lemma-huge-diagram} an
obstruction element $o(B'_1, B'_2) \in \text{Ext}^1_B(\NL_{B/A}, N)$
to the existence of a map $B'_1 \to B'_2$. Clearly, this element
is the obstruction to the existence of an isomorphism, hence separates
the isomorphism classes. To finish the proof it therefore suffices to
show that given a solution $B'$ and an element
$\xi \in \text{Ext}^1_B(\NL_{B/A}, N)$
we can find a second solution $B'_\xi$ such that
$o(B', B'_\xi) = \xi$.
\medskip\noindent
Let $E = B$ viewed as a set. Consider the surjection $A[E] \to B$ with kernel
$J$ used to define the naive cotangent complex by the formula
$$
\NL_{B/A} = (J/J^2 \to \Omega_{A[E]/A} \otimes_{A[E]} B)
$$
in Algebra, Section \ref{algebra-section-netherlander}.
Since $\Omega_{A[E]/A} \otimes B$ is a free $B$-module we have
$$
\text{Ext}^1_B(\NL_{B/A}, N) =
\frac{\Hom_B(J/J^2, N)}
{\Hom_B(\Omega_{A[E]/A} \otimes B, N)}
$$
Thus we may represent $\xi$ as the class of a morphism $\delta : J/J^2 \to N$.
\medskip\noindent
For every $e \in E$ denote $x_e \in A[E]$ the corresponding variable.
Choose a lift $y_e \in B'$ of the image of $x_e$ in $B$.
These choices determine an $A'$-algebra map $\varphi : A'[E] \to B'$.
Let $J' = \text{Ker}(A'[E] \to B)$. Observe that $\varphi$ induces a map
$\varphi|_{J'} : J' \to N$ and that $B'$ is the pushout, as in the following
diagram
$$
\xymatrix{
0 \ar[r] & N \ar[r] & B' \ar[r] & B \ar[r] & 0 \\
0 \ar[r] & J' \ar[u]^{\varphi|_{J'}} \ar[r] & A'[E] \ar[u] \ar[r] &
B \ar[u]_{=} \ar[r] & 0
}
$$
Let $\psi : J' \to N$ be the sum of the map $\varphi|_{J'}$ and the
composition
$$
J' \to J'/(J')^2 \to J/J^2 \xrightarrow{\delta} N.
$$
Then the pushout along $\psi$ is an other ring extension $B'_\xi$
fitting into a diagram as above. A calculation shows that
$o(B', B'_\xi) = \xi$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-extensions-of-rings}
Let $A$ be a ring and let $I$ be an $A$-module.
\begin{enumerate}
\item The set of extensions of rings $0 \to I \to A' \to A \to 0$
where $I$ is an ideal of square zero is canonically bijective to
$\text{Ext}^1_A(\NL_{A/\mathbf{Z}}, I)$.
\item Given a ring map $A \to B$, a $B$-module $N$, an $A$-module
map $c : I \to N$, and given extensions of rings with square zero kernels:
\begin{enumerate}
\item[(a)] $0 \to I \to A' \to A \to 0$ corresponding to
$\alpha \in \text{Ext}^1_A(\NL_{A/\mathbf{Z}}, I)$, and
\item[(b)] $0 \to N \to B' \to B \to 0$ corresponding to
$\beta \in \text{Ext}^1_B(\NL_{B/\mathbf{Z}}, N)$
\end{enumerate}
then there is a map $A' \to B'$ fitting into a diagram
(\ref{equation-to-solve}) if and only if $\beta$ and $\alpha$
map to the same element of
$\text{Ext}^1_A(\NL_{A/\mathbf{Z}}, N)$.
\end{enumerate}
\end{lemma}
\begin{proof}
To prove this we apply the previous results where we work over
$0 \to 0 \to \mathbf{Z} \to \mathbf{Z} \to 0$, in order words,
we work over the extension of $\mathbf{Z}$ by $0$.
Part (1) follows from Lemma \ref{lemma-choices}
and the fact that there exists a solution, namely $I \oplus A$.
Part (2) follows from Lemma \ref{lemma-huge-diagram}
and a compatibility between the constructions in the proofs
of Lemmas \ref{lemma-choices} and \ref{lemma-huge-diagram}
whose statement and proof we omit.
\end{proof}
\section{Thickenings of ringed spaces}
\label{section-thickenings-spaces}
\noindent
In the following few sections we will use the following notions:
\begin{enumerate}
\item A sheaf of ideals $\mathcal{I} \subset \mathcal{O}_{X'}$ on
a ringed space $(X', \mathcal{O}_{X'})$ is {\it locally nilpotent}
if any local section of $\mathcal{I}$ is locally nilpotent.
Compare with Algebra, Item \ref{algebra-item-ideal-locally-nilpotent}.
\item A {\it thickening} of ringed spaces is a morphism
$i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$ of ringed spaces
such that
\begin{enumerate}
\item $i$ induces a homeomorphism $X \to X'$,
\item the map $i^\sharp : \mathcal{O}_{X'} \to i_*\mathcal{O}_X$
is surjective, and
\item the kernel of $i^\sharp$ is a locally nilpotent sheaf of ideals.
\end{enumerate}
\item A {\it first order thickening} of ringed spaces is a thickening
$i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$ of ringed spaces
such that $\text{Ker}(i^\sharp)$ has square zero.
\item It is clear how to define {\it morphisms of thickenings},
{\it morphisms of thickenings over a base ringed space}, etc.
\end{enumerate}
If $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$ is a thickening
of ringed spaces then we identify the underlying topological spaces
and think of $\mathcal{O}_X$, $\mathcal{O}_{X'}$, and
$\mathcal{I} = \text{Ker}(i^\sharp)$ as sheaves on $X = X'$. We obtain
a short exact sequence
$$
0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0
$$
of $\mathcal{O}_{X'}$-modules. By
Modules, Lemma \ref{modules-lemma-i-star-equivalence}
the category of $\mathcal{O}_X$-modules is equivalent to the category
of $\mathcal{O}_{X'}$-modules annihilated by $\mathcal{I}$. In particular,
if $i$ is a first order thickening, then
$\mathcal{I}$ is a $\mathcal{O}_X$-module.
\begin{situation}
\label{situation-morphism-thickenings}
A morphism of thickenings $(f, f')$ is given by a commutative diagram
\begin{equation}
\label{equation-morphism-thickenings}
\vcenter{
\xymatrix{
(X, \mathcal{O}_X) \ar[r]_i \ar[d]_f & (X', \mathcal{O}_{X'}) \ar[d]^{f'} \\
(S, \mathcal{O}_S) \ar[r]^t & (S', \mathcal{O}_{S'})
}
}
\end{equation}
of ringed spaces whose horizontal arrows are thickenings. In this
situation we set
$\mathcal{I} = \text{Ker}(i^\sharp) \subset \mathcal{O}_{X'}$ and
$\mathcal{J} = \text{Ker}(t^\sharp) \subset \mathcal{O}_{S'}$.
As $f = f'$ on underlying topological spaces we will identify
the (topological) pullback functors $f^{-1}$ and $(f')^{-1}$.
Observe that $(f')^\sharp : f^{-1}\mathcal{O}_{S'} \to \mathcal{O}_{X'}$
induces in particular a map $f^{-1}\mathcal{J} \to \mathcal{I}$
and therefore a map of $\mathcal{O}_{X'}$-modules
$$
(f')^*\mathcal{J} \longrightarrow \mathcal{I}
$$
If $i$ and $t$ are first order thickenings, then
$(f')^*\mathcal{J} = f^*\mathcal{J}$ and the map above becomes a
map $f^*\mathcal{J} \to \mathcal{I}$.
\end{situation}
\begin{definition}
\label{definition-strict-morphism-thickenings}
In Situation \ref{situation-morphism-thickenings} we say that $(f, f')$ is a
{\it strict morphism of thickenings}
if the map $(f')^*\mathcal{J} \longrightarrow \mathcal{I}$ is surjective.
\end{definition}
\noindent
The following lemma in particular shows that a morphism
$(f, f') : (X \subset X') \to (S \subset S')$ of
thickenings of schemes is strict if and only if $X = S \times_{S'} X'$.
\begin{lemma}
\label{lemma-strict-morphism-thickenings}
In Situation \ref{situation-morphism-thickenings} the morphism $(f, f')$
is a strict morphism of thickenings if and only if
(\ref{equation-morphism-thickenings}) is cartesian in the category
of ringed spaces.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\section{Modules on first order thickenings of ringed spaces}
\label{section-modules-thickenings}
\noindent
In this section we discuss some preliminaries to the deformation theory
of modules. Let $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$
be a first order thickening of ringed spaces. We will freely use the notation
introduced in Section \ref{section-thickenings-spaces}, in particular we will
identify the underlying topological spaces.
In this section we consider short exact sequences
\begin{equation}
\label{equation-extension}
0 \to \mathcal{K} \to \mathcal{F}' \to \mathcal{F} \to 0
\end{equation}
of $\mathcal{O}_{X'}$-modules, where $\mathcal{F}$, $\mathcal{K}$ are
$\mathcal{O}_X$-modules and $\mathcal{F}'$ is an $\mathcal{O}_{X'}$-module.
In this situation we have a canonical $\mathcal{O}_X$-module map
$$
c_{\mathcal{F}'} :
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F}
\longrightarrow
\mathcal{K}
$$
where $\mathcal{I} = \text{Ker}(i^\sharp)$.
Namely, given local sections $f$ of $\mathcal{I}$ and $s$
of $\mathcal{F}$ we set $c_{\mathcal{F}'}(f \otimes s) = fs'$
where $s'$ is a local section of $\mathcal{F}'$ lifting $s$.
\begin{lemma}
\label{lemma-inf-map}
Let $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$
be a first order thickening of ringed spaces. Assume given
extensions
$$
0 \to \mathcal{K} \to \mathcal{F}' \to \mathcal{F} \to 0
\quad\text{and}\quad
0 \to \mathcal{L} \to \mathcal{G}' \to \mathcal{G} \to 0
$$
as in (\ref{equation-extension})
and maps $\varphi : \mathcal{F} \to \mathcal{G}$ and
$\psi : \mathcal{K} \to \mathcal{L}$.
\begin{enumerate}
\item If there exists an $\mathcal{O}_{X'}$-module
map $\varphi' : \mathcal{F}' \to \mathcal{G}'$ compatible with $\varphi$
and $\psi$, then the diagram
$$
\xymatrix{
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]_-{c_{\mathcal{F}'}} \ar[d]_{1 \otimes \varphi} &
\mathcal{K} \ar[d]^\psi \\
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{G}
\ar[r]^-{c_{\mathcal{G}'}} &
\mathcal{L}
}
$$
is commutative.
\item The set of $\mathcal{O}_{X'}$-module
maps $\varphi' : \mathcal{F}' \to \mathcal{G}'$ compatible with $\varphi$
and $\psi$ is, if nonempty, a principal homogeneous space under
$\Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{L})$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) is immediate from the description of the maps.
For (2), if $\varphi'$ and $\varphi''$ are two maps
$\mathcal{F}' \to \mathcal{G}'$ compatible with $\varphi$
and $\psi$, then $\varphi' - \varphi''$ factors as
$$
\mathcal{F}' \to \mathcal{F} \to \mathcal{L} \to \mathcal{G}'
$$
The map in the middle comes from a unique element of
$\Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{L})$ by
Modules, Lemma \ref{modules-lemma-i-star-equivalence}.
Conversely, given an element $\alpha$ of this group we can add the
composition (as displayed above with $\alpha$ in the middle)
to $\varphi'$. Some details omitted.
\end{proof}
\begin{lemma}
\label{lemma-inf-obs-map}
Let $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$
be a first order thickening of ringed spaces. Assume given
extensions
$$
0 \to \mathcal{K} \to \mathcal{F}' \to \mathcal{F} \to 0
\quad\text{and}\quad
0 \to \mathcal{L} \to \mathcal{G}' \to \mathcal{G} \to 0
$$
as in (\ref{equation-extension})
and maps $\varphi : \mathcal{F} \to \mathcal{G}$ and
$\psi : \mathcal{K} \to \mathcal{L}$. Assume the diagram
$$
\xymatrix{
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]_-{c_{\mathcal{F}'}} \ar[d]_{1 \otimes \varphi} &
\mathcal{K} \ar[d]^\psi \\
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{G}
\ar[r]^-{c_{\mathcal{G}'}} &
\mathcal{L}
}
$$
is commutative. Then there exists an element
$$
o(\varphi, \psi) \in
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{L})
$$
whose vanishing is a necessary and sufficient condition for the existence
of a map $\varphi' : \mathcal{F}' \to \mathcal{G}'$ compatible with
$\varphi$ and $\psi$.
\end{lemma}
\begin{proof}
We can construct explicitly an extension
$$
0 \to \mathcal{L} \to \mathcal{H} \to \mathcal{F} \to 0
$$
by taking $\mathcal{H}$ to be the cohomology of the complex
$$
\mathcal{K}
\xrightarrow{1, - \psi}
\mathcal{F}' \oplus \mathcal{G}' \xrightarrow{\varphi, 1}
\mathcal{G}
$$
in the middle (with obvious notation). A calculation with local sections
using the assumption that the diagram of the lemma commutes
shows that $\mathcal{H}$ is annihilated by $\mathcal{I}$. Hence
$\mathcal{H}$ defines a class in
$$
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{L})
\subset
\text{Ext}^1_{\mathcal{O}_{X'}}(\mathcal{F}, \mathcal{L})
$$
Finally, the class of $\mathcal{H}$ is the difference of the pushout
of the extension $\mathcal{F}'$ via $\psi$ and the pullback
of the extension $\mathcal{G}'$ via $\varphi$ (calculations omitted).
Thus the vanishing of the class of $\mathcal{H}$ is equivalent to the
existence of a commutative diagram
$$
\xymatrix{
0 \ar[r] &
\mathcal{K} \ar[r] \ar[d]_{\psi} &
\mathcal{F}' \ar[r] \ar[d]_{\varphi'} &
\mathcal{F} \ar[r] \ar[d]_\varphi & 0\\
0 \ar[r] &
\mathcal{L} \ar[r] &
\mathcal{G}' \ar[r] &
\mathcal{G} \ar[r] & 0
}
$$
as desired.
\end{proof}
\begin{lemma}
\label{lemma-inf-ext}
Let $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$ be a first order
thickening of ringed spaces.
Assume given $\mathcal{O}_X$-modules $\mathcal{F}$, $\mathcal{K}$
and an $\mathcal{O}_X$-linear map
$c : \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{K}$.
If there exists a sequence (\ref{equation-extension}) with
$c_{\mathcal{F}'} = c$ then the set of isomorphism classes of these
extensions is principal homogeneous under
$\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K})$.
\end{lemma}
\begin{proof}
Assume given extensions
$$
0 \to \mathcal{K} \to \mathcal{F}'_1 \to \mathcal{F} \to 0
\quad\text{and}\quad
0 \to \mathcal{K} \to \mathcal{F}'_2 \to \mathcal{F} \to 0
$$
with $c_{\mathcal{F}'_1} = c_{\mathcal{F}'_2} = c$. Then the difference
(in the extension group, see
Homology, Section \ref{homology-section-extensions})
is an extension
$$
0 \to \mathcal{K} \to \mathcal{E} \to \mathcal{F} \to 0
$$
where $\mathcal{E}$ is annihilated by $\mathcal{I}$ (local computation
omitted). Hence the sequence is an extension of $\mathcal{O}_X$-modules,
see Modules, Lemma \ref{modules-lemma-i-star-equivalence}.
Conversely, given such an extension $\mathcal{E}$ we can add the extension
$\mathcal{E}$ to the $\mathcal{O}_{X'}$-extension $\mathcal{F}'$ without
affecting the map $c_{\mathcal{F}'}$. Some details omitted.
\end{proof}
\begin{lemma}
\label{lemma-inf-obs-ext}
Let $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$
be a first order thickening of ringed spaces. Assume given
$\mathcal{O}_X$-modules $\mathcal{F}$, $\mathcal{K}$
and an $\mathcal{O}_X$-linear map
$c : \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{K}$.
Then there exists an element
$$
o(\mathcal{F}, \mathcal{K}, c) \in
\text{Ext}^2_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K})
$$
whose vanishing is a necessary and sufficient condition for the existence
of a sequence (\ref{equation-extension}) with $c_{\mathcal{F}'} = c$.
\end{lemma}
\begin{proof}
We first show that if $\mathcal{K}$ is an injective $\mathcal{O}_X$-module,
then there does exist a sequence (\ref{equation-extension}) with
$c_{\mathcal{F}'} = c$. To do this, choose a flat
$\mathcal{O}_{X'}$-module $\mathcal{H}'$ and a surjection
$\mathcal{H}' \to \mathcal{F}$
(Modules, Lemma \ref{modules-lemma-module-quotient-flat}).
Let $\mathcal{J} \subset \mathcal{H}'$ be the kernel. Since $\mathcal{H}'$
is flat we have
$$
\mathcal{I} \otimes_{\mathcal{O}_{X'}} \mathcal{H}' =
\mathcal{I}\mathcal{H}'
\subset \mathcal{J} \subset \mathcal{H}'
$$
Observe that the map
$$
\mathcal{I}\mathcal{H}' =
\mathcal{I} \otimes_{\mathcal{O}_{X'}} \mathcal{H}'
\longrightarrow
\mathcal{I} \otimes_{\mathcal{O}_{X'}} \mathcal{F} =
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F}
$$
annihilates $\mathcal{I}\mathcal{J}$. Namely, if $f$ is a local section
of $\mathcal{I}$ and $s$ is a local section of $\mathcal{H}$, then
$fs$ is mapped to $f \otimes \overline{s}$ where $\overline{s}$ is
the image of $s$ in $\mathcal{F}$. Thus we obtain
$$
\xymatrix{
\mathcal{I}\mathcal{H}'/\mathcal{I}\mathcal{J}
\ar@{^{(}->}[r] \ar[d] &
\mathcal{J}/\mathcal{I}\mathcal{J} \ar@{..>}[d]_\gamma \\
\mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \ar[r]^-c &
\mathcal{K}
}
$$
a diagram of $\mathcal{O}_X$-modules. If $\mathcal{K}$ is injective
as an $\mathcal{O}_X$-module, then we obtain the dotted arrow.
Denote $\gamma' : \mathcal{J} \to \mathcal{K}$ the composition
of $\gamma$ with $\mathcal{J} \to \mathcal{J}/\mathcal{I}\mathcal{J}$.
A local calculation shows the pushout
$$
\xymatrix{
0 \ar[r] &
\mathcal{J} \ar[r] \ar[d]_{\gamma'} &
\mathcal{H}' \ar[r] \ar[d] &
\mathcal{F} \ar[r] \ar@{=}[d] &
0 \\
0 \ar[r] &
\mathcal{K} \ar[r] &
\mathcal{F}' \ar[r] &
\mathcal{F} \ar[r] &
0
}
$$
is a solution to the problem posed by the lemma.
\medskip\noindent
General case. Choose an embedding $\mathcal{K} \subset \mathcal{K}'$
with $\mathcal{K}'$ an injective $\mathcal{O}_X$-module. Let $\mathcal{Q}$
be the quotient, so that we have an exact sequence
$$
0 \to \mathcal{K} \to \mathcal{K}' \to \mathcal{Q} \to 0
$$
Denote
$c' : \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{K}'$
be the composition. By the paragraph above there exists a sequence
$$
0 \to \mathcal{K}' \to \mathcal{E}' \to \mathcal{F} \to 0
$$
as in (\ref{equation-extension}) with $c_{\mathcal{E}'} = c'$.
Note that $c'$ composed with the map $\mathcal{K}' \to \mathcal{Q}$
is zero, hence the pushout of $\mathcal{E}'$ by
$\mathcal{K}' \to \mathcal{Q}$ is an extension
$$
0 \to \mathcal{Q} \to \mathcal{D}' \to \mathcal{F} \to 0
$$
as in (\ref{equation-extension}) with $c_{\mathcal{D}'} = 0$.
This means exactly that $\mathcal{D}'$ is annihilated by
$\mathcal{I}$, in other words, the $\mathcal{D}'$ is an extension
of $\mathcal{O}_X$-modules, i.e., defines an element
$$
o(\mathcal{F}, \mathcal{K}, c) \in
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{Q}) =
\text{Ext}^2_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K})
$$
(the equality holds by the long exact cohomology sequence associated
to the exact sequence above and the vanishing of higher ext groups
into the injective module $\mathcal{K}'$). If
$o(\mathcal{F}, \mathcal{K}, c) = 0$, then we can choose a splitting
$s : \mathcal{F} \to \mathcal{D}'$ and we can set
$$
\mathcal{F}' = \text{Ker}(\mathcal{E}' \to \mathcal{D}'/s(\mathcal{F}))
$$
so that we obtain the following diagram
$$
\xymatrix{
0 \ar[r] &
\mathcal{K} \ar[r] \ar[d] &
\mathcal{F}' \ar[r] \ar[d] &
\mathcal{F} \ar[r] \ar@{=}[d] &
0 \\
0 \ar[r] &
\mathcal{K}' \ar[r] &
\mathcal{E}' \ar[r] &
\mathcal{F} \ar[r] & 0
}
$$
with exact rows which shows that $c_{\mathcal{F}'} = c$. Conversely, if
$\mathcal{F}'$ exists, then the pushout of $\mathcal{F}'$ by the map
$\mathcal{K} \to \mathcal{K}'$ is isomorphic to $\mathcal{E}'$ by
Lemma \ref{lemma-inf-ext} and the vanishing of higher ext groups
into the injective module $\mathcal{K}'$. This gives a diagram
as above, which implies that $\mathcal{D}'$ is split as an extension, i.e.,
the class $o(\mathcal{F}, \mathcal{K}, c)$ is zero.
\end{proof}
\begin{remark}
\label{remark-trivial-thickening}
Let $(X, \mathcal{O}_X)$ be a ringed space. A first order thickening
$i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$ is said
to be {\it trivial} if there exists a morphism of ringed spaces
$\pi : (X', \mathcal{O}_{X'}) \to (X, \mathcal{O}_X)$ which is a
left inverse to $i$. The choice of such a morphism
$\pi$ is called a {\it trivialization} of the first order thickening.
Given $\pi$ we obtain a splitting
\begin{equation}
\label{equation-splitting}
\mathcal{O}_{X'} = \mathcal{O}_X \oplus \mathcal{I}
\end{equation}
as sheaves of algebras on $X$ by using $\pi^\sharp$ to split the surjection
$\mathcal{O}_{X'} \to \mathcal{O}_X$. Conversely, such a splitting determines
a morphism $\pi$. The category of trivialized first order thickenings of
$(X, \mathcal{O}_X)$ is equivalent to the category of
$\mathcal{O}_X$-modules.
\end{remark}
\begin{remark}
\label{remark-trivial-extension}
Let $i : (X, \mathcal{O}_X) \to (X', \mathcal{O}_{X'})$
be a trivial first order thickening of ringed spaces
and let $\pi : (X', \mathcal{O}_{X'}) \to (X, \mathcal{O}_X)$
be a trivialization. Then given any triple
$(\mathcal{F}, \mathcal{K}, c)$ consisting of a pair of
$\mathcal{O}_X$-modules and a map
$c : \mathcal{I} \otimes_{\mathcal{O}_X} \mathcal{F} \to \mathcal{K}$
we may set
$$
\mathcal{F}'_{c, triv} = \mathcal{F} \oplus \mathcal{K}
$$
and use the splitting (\ref{equation-splitting}) associated to $\pi$
and the map $c$ to define the $\mathcal{O}_{X'}$-module structure
and obtain an extension (\ref{equation-extension}). We will call
$\mathcal{F}'_{c, triv}$ the {\it trivial extension} of $\mathcal{F}$
by $\mathcal{K}$ corresponding
to $c$ and the trivialization $\pi$. Given any extension
$\mathcal{F}'$ as in (\ref{equation-extension}) we can use
$\pi^\sharp : \mathcal{O}_X \to \mathcal{O}_{X'}$ to think of $\mathcal{F}'$
as an $\mathcal{O}_X$-module extension, hence a class $\xi_{\mathcal{F}'}$
in $\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K})$.
Lemma \ref{lemma-inf-ext} assures that
$\mathcal{F}' \mapsto \xi_{\mathcal{F}'}$
induces a bijection
$$
\left\{
\begin{matrix}
\text{isomorphism classes of extensions}\\
\mathcal{F}'\text{ as in (\ref{equation-extension}) with }c = c_{\mathcal{F}'}
\end{matrix}
\right\}
\longrightarrow
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K})
$$
Moreover, the trivial extension $\mathcal{F}'_{c, triv}$ maps to the zero class.
\end{remark}
\begin{remark}
\label{remark-extension-functorial}
Let $(X, \mathcal{O}_X)$ be a ringed space. Let
$(X, \mathcal{O}_X) \to (X'_i, \mathcal{O}_{X'_i})$, $i = 1, 2$
be first order thickenings with ideal sheaves $\mathcal{I}_i$.
Let $h : (X'_1, \mathcal{O}_{X'_1}) \to (X'_2, \mathcal{O}_{X'_2})$
be a morphism of first order thickenings of $(X, \mathcal{O}_X)$.
Picture
$$
\xymatrix{
& (X, \mathcal{O}_X) \ar[ld] \ar[rd] & \\
(X'_1, \mathcal{O}_{X'_1}) \ar[rr]^h & &
(X'_2, \mathcal{O}_{X'_2})
}
$$
Observe that $h^\sharp : \mathcal{O}_{X'_2} \to \mathcal{O}_{X'_1}$
in particular induces an $\mathcal{O}_X$-module map
$\mathcal{I}_2 \to \mathcal{I}_1$.
Let $\mathcal{F}$ be an
$\mathcal{O}_X$-module. Let $(\mathcal{K}_i, c_i)$, $i = 1, 2$ be a pair
consisting of an $\mathcal{O}_X$-module $\mathcal{K}_i$ and a map
$c_i : \mathcal{I}_i \otimes_{\mathcal{O}_X} \mathcal{F} \to
\mathcal{K}_i$. Assume furthermore given a map
of $\mathcal{O}_X$-modules $\mathcal{K}_2 \to \mathcal{K}_1$
such that
$$
\xymatrix{
\mathcal{I}_2 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]_-{c_2} \ar[d] &
\mathcal{K}_2 \ar[d] \\
\mathcal{I}_1 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]^-{c_1} &
\mathcal{K}_1
}
$$
is commutative. Then there is a canonical functoriality
$$
\left\{
\begin{matrix}
\mathcal{F}'_2\text{ as in (\ref{equation-extension}) with }\\
c_2 = c_{\mathcal{F}'_2}\text{ and }\mathcal{K} = \mathcal{K}_2
\end{matrix}
\right\}
\longrightarrow
\left\{
\begin{matrix}
\mathcal{F}'_1\text{ as in (\ref{equation-extension}) with }\\
c_1 = c_{\mathcal{F}'_1}\text{ and }\mathcal{K} = \mathcal{K}_1
\end{matrix}
\right\}
$$
Namely, thinking of all sheaves $\mathcal{O}_X$, $\mathcal{O}_{X'_i}$,
$\mathcal{F}$, $\mathcal{K}_i$, etc as sheaves on $X$, we set
given $\mathcal{F}'_2$ the sheaf $\mathcal{F}'_1$ equal to the
pushout, i.e., fitting into the following diagram of extensions
$$
\xymatrix{
0 \ar[r] &
\mathcal{K}_2 \ar[r] \ar[d] &
\mathcal{F}'_2 \ar[r] \ar[d] &
\mathcal{F} \ar@{=}[d] \ar[r] & 0 \\
0 \ar[r] &
\mathcal{K}_1 \ar[r] &
\mathcal{F}'_1 \ar[r] &
\mathcal{F} \ar[r] & 0
}
$$
We omit the construction of the $\mathcal{O}_{X'_1}$-module structure
on the pushout (this uses the commutativity of the diagram
involving $c_1$ and $c_2$).
\end{remark}
\begin{remark}
\label{remark-trivial-extension-functorial}
Let $(X, \mathcal{O}_X)$, $(X, \mathcal{O}_X) \to (X'_i, \mathcal{O}_{X'_i})$,
$\mathcal{I}_i$, and
$h : (X'_1, \mathcal{O}_{X'_1}) \to (X'_2, \mathcal{O}_{X'_2})$
be as in Remark \ref{remark-extension-functorial}. Assume that we are
given given trivializations $\pi_i : X'_i \to X$ such that
$\pi_1 = h \circ \pi_2$. In other words, assume $h$ is a morphism
of trivialized first order thickening of $(X, \mathcal{O}_X)$. Let
$(\mathcal{K}_i, c_i)$, $i = 1, 2$ be a pair consisting of an
$\mathcal{O}_X$-module $\mathcal{K}_i$ and a map
$c_i : \mathcal{I}_i \otimes_{\mathcal{O}_X} \mathcal{F} \to
\mathcal{K}_i$. Assume furthermore given a map
of $\mathcal{O}_X$-modules $\mathcal{K}_2 \to \mathcal{K}_1$
such that
$$
\xymatrix{
\mathcal{I}_2 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]_-{c_2} \ar[d] &
\mathcal{K}_2 \ar[d] \\
\mathcal{I}_1 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]^-{c_1} &
\mathcal{K}_1
}
$$
is commutative. In this situation the construction of
Remark \ref{remark-trivial-extension} induces
a commutative diagram
$$
\xymatrix{
\{\mathcal{F}'_2\text{ as in (\ref{equation-extension}) with }
c_2 = c_{\mathcal{F}'_2}\text{ and }\mathcal{K} = \mathcal{K}_2\}
\ar[d] \ar[rr] & &
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}_2) \ar[d] \\
\{\mathcal{F}'_1\text{ as in (\ref{equation-extension}) with }
c_1 = c_{\mathcal{F}'_1}\text{ and }\mathcal{K} = \mathcal{K}_1\}
\ar[rr] & &
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}_1)
}
$$
where the vertical map on the right is given by functoriality of $\text{Ext}$
and the map $\mathcal{K}_2 \to \mathcal{K}_1$ and the vertical map on the left
is the one from Remark \ref{remark-extension-functorial}.
\end{remark}
\begin{remark}
\label{remark-short-exact-sequence-thickenings}
Let $(X, \mathcal{O}_X)$ be a ringed space. We define a sequence of morphisms
of first order thickenings
$$
(X'_1, \mathcal{O}_{X'_1}) \to
(X'_2, \mathcal{O}_{X'_2}) \to
(X'_3, \mathcal{O}_{X'_3})
$$
of $(X, \mathcal{O}_X)$ to be a {\it complex}
if the corresponding maps between
the ideal sheaves $\mathcal{I}_i$
give a complex of $\mathcal{O}_X$-modules
$\mathcal{I}_3 \to \mathcal{I}_2 \to \mathcal{I}_1$
(i.e., the composition is zero). In this case the composition
$(X'_1, \mathcal{O}_{X'_1}) \to (X_3', \mathcal{O}_{X'_3})$ factors through
$(X, \mathcal{O}_X) \to (X'_3, \mathcal{O}_{X'_3})$, i.e.,
the first order thickening $(X'_1, \mathcal{O}_{X'_1})$ of
$(X, \mathcal{O}_X)$ is trivial and comes with
a canonical trivialization
$\pi : (X'_1, \mathcal{O}_{X'_1}) \to (X, \mathcal{O}_X)$.
\medskip\noindent
We say a sequence of morphisms of first order thickenings
$$
(X'_1, \mathcal{O}_{X'_1}) \to
(X'_2, \mathcal{O}_{X'_2}) \to
(X'_3, \mathcal{O}_{X'_3})
$$
of $(X, \mathcal{O}_X)$ is {\it a short exact sequence} if the
corresponding maps between ideal sheaves is a short exact sequence
$$
0 \to \mathcal{I}_3 \to \mathcal{I}_2 \to \mathcal{I}_1 \to 0
$$
of $\mathcal{O}_X$-modules.
\end{remark}
\begin{remark}
\label{remark-complex-thickenings-and-ses-modules}
Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{F}$ be an
$\mathcal{O}_X$-module. Let
$$
(X'_1, \mathcal{O}_{X'_1}) \to
(X'_2, \mathcal{O}_{X'_2}) \to
(X'_3, \mathcal{O}_{X'_3})
$$
be a complex first order thickenings of $(X, \mathcal{O}_X)$, see
Remark \ref{remark-short-exact-sequence-thickenings}.
Let $(\mathcal{K}_i, c_i)$, $i = 1, 2, 3$ be pairs consisting of
an $\mathcal{O}_X$-module $\mathcal{K}_i$ and a map
$c_i : \mathcal{I}_i \otimes_{\mathcal{O}_X} \mathcal{F} \to
\mathcal{K}_i$. Assume given a short exact sequence
of $\mathcal{O}_X$-modules
$$
0 \to \mathcal{K}_3 \to \mathcal{K}_2 \to \mathcal{K}_1 \to 0
$$
such that
$$
\vcenter{
\xymatrix{
\mathcal{I}_2 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]_-{c_2} \ar[d] &
\mathcal{K}_2 \ar[d] \\
\mathcal{I}_1 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]^-{c_1} &
\mathcal{K}_1
}
}
\quad\text{and}\quad
\vcenter{
\xymatrix{
\mathcal{I}_3 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]_-{c_3} \ar[d] &
\mathcal{K}_3 \ar[d] \\
\mathcal{I}_2 \otimes_{\mathcal{O}_X} \mathcal{F}
\ar[r]^-{c_2} &
\mathcal{K}_2
}
}
$$
are commutative. Finally, assume given an extension
$$
0 \to \mathcal{K}_2 \to \mathcal{F}'_2 \to \mathcal{F} \to 0
$$
as in (\ref{equation-extension}) with $\mathcal{K} = \mathcal{K}_2$
of $\mathcal{O}_{X'_2}$-modules with $c_{\mathcal{F}'_2} = c_2$.
In this situation we can apply the functoriality of
Remark \ref{remark-extension-functorial} to obtain an extension
$\mathcal{F}'_1$ on $X'_1$ (we'll describe $\mathcal{F}'_1$
in this special case below). By
Remark \ref{remark-trivial-extension}
using the canonical splitting
$\pi : (X'_1, \mathcal{O}_{X'_1}) \to (X, \mathcal{O}_X)$ of
Remark \ref{remark-short-exact-sequence-thickenings}
we obtain
$\xi_{\mathcal{F}'_1} \in
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}_1)$.
Finally, we have the obstruction
$$
o(\mathcal{F}, \mathcal{K}_3, c_3) \in
\text{Ext}^2_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}_3)
$$
see Lemma \ref{lemma-inf-obs-ext}.
In this situation we {\bf claim} that the canonical map
$$
\partial :
\text{Ext}^1_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}_1)
\longrightarrow
\text{Ext}^2_{\mathcal{O}_X}(\mathcal{F}, \mathcal{K}_3)
$$
coming from the short exact sequence
$0 \to \mathcal{K}_3 \to \mathcal{K}_2 \to \mathcal{K}_1 \to 0$