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homology.tex
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\input{preamble}
% OK, start here.
%
\begin{document}
\title{Homological Algebra}
\maketitle
\phantomsection
\label{section-phantom}
\tableofcontents
\section{Introduction}
\label{section-introduction}
\noindent
Basic homological algebra will be explained in this document.
We add as needed in the other parts, since there is clearly
an infinite amount of this stuff around.
A reference is \cite{Maclane}.
\section{Basic notions}
\label{section-topology-basic}
\noindent
The following notions are considered basic and will not be defined,
and or proved. This does not mean they are all necessarily easy or
well known.
\begin{enumerate}
\item Nothing yet.
\end{enumerate}
\section{Abelian categories}
\label{section-abelian-categories}
\noindent
An abelian category will be a category satisfying
just enough axioms so the snake lemma holds.
\begin{definition}
\label{definition-preadditive}
A category $\mathcal{A}$ is called {\it preadditive} if each
morphism set $\Mor_\mathcal{A}(x, y)$ is endowed
with the structure of an abelian group such that the
compositions
$$
\Mor(x, y) \times \Mor(y, z)
\longrightarrow
\Mor(x, z)
$$
are bilinear. A functor $F : \mathcal{A} \to \mathcal{B}$ of
preadditive categories is called {\it additive} if and only
if $F : \Mor(x, y) \to \Mor(F(x), F(y))$
is a homomorphism of abelian groups for all
$x, y \in \Ob(\mathcal{A})$.
\end{definition}
\noindent
In particular for every $x, y$ there exists at least
one morphism $x \to y$, namely the zero map.
\begin{lemma}
\label{lemma-preadditive-zero}
Let $\mathcal{A}$ be a preadditive category.
Let $x$ be an object of $\mathcal{A}$.
The following are equivalent
\begin{enumerate}
\item $x$ is an initial object,
\item $x$ is a final object, and
\item $\text{id}_x = 0$ in $\Mor_\mathcal{A}(x, x)$.
\end{enumerate}
Furthermore, if such an object $0$ exists, then a morphism
$\alpha : x \to y$ factors through $0$ if and only if $\alpha = 0$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-zero-object}
In a preadditive category $\mathcal{A}$ we call
{\it zero object}, and we denote it $0$
any final and initial object as in Lemma \ref{lemma-preadditive-zero} above.
\end{definition}
\begin{lemma}
\label{lemma-preadditive-direct-sum}
Let $\mathcal{A}$ be a preadditive category.
Let $x, y \in \Ob(\mathcal{A})$.
If the product $x \times y$ exists, then so does
the coproduct $x \coprod y$.
If the coproduct $x \coprod y$ exists, then so does
the product $x \times y$. In this case
also $x \coprod y \cong x \times y$.
\end{lemma}
\begin{proof}
Suppose that $z = x \times y$ with projections
$p : z \to x$ and $q : z \to y$. Denote $i : x \to z$
the morphism corresponding to $(1, 0)$. Denote $j : y \to z$
the morphism corresponding to $(0, 1)$. Thus we have the
commutative diagram
$$
\xymatrix{
x \ar[rr]^1 \ar[rd]^i & & x \\
& z \ar[ru]^p \ar[rd]^q & \\
y \ar[rr]^1 \ar[ru]^j & & y
}
$$
where the diagonal compositions are zero. It follows that
$i \circ p + j \circ q : z \to z$ is the identity since
it is a morphism which upon composing with $p$ gives $p$
and upon composing with $q$ gives $q$.
Suppose given morphisms $a : x \to w$ and $b : y \to w$.
Then we can form the map $a \circ p + b \circ q : z \to w$.
In this way we get a bijection $\Mor(z, w)
= \Mor(x, w) \times \Mor(y, w)$ which
show that $z = x \coprod y$.
\medskip\noindent
We leave it to the reader to construct the morphisms
$p, q$ given a coproduct $x \coprod y$ instead of a
product.
\end{proof}
\begin{definition}
\label{definition-direct-sum}
Given a pair of objects $x, y$
in a preadditive category $\mathcal{A}$ we call
{\it direct sum}, and we denote it $x \oplus y$ the
product $x \times y$ endowed with the morphisms
$i, j, p, q$ as in Lemma \ref{lemma-preadditive-direct-sum} above.
\end{definition}
\begin{remark}
\label{remark-direct-sum}
Note that the proof of Lemma \ref{lemma-preadditive-direct-sum}
shows that given $p$ and $q$ the morphisms $i$, $j$ are uniquely
determined by the rules $p \circ i = \text{id}_x$,
$q \circ j = \text{id}_y$, $p \circ j = 0$, $q \circ i = 0$.
Moreover, we automatically have
$i \circ p + j \circ q = \text{id}_{x \oplus y}$.
Similarly, given $i$, $j$ the morphisms $p$ and $q$ are uniquely determined.
Finally, given objects $x, y, z$ and morphisms
$i : x \to z$, $j : y \to z$, $p : z \to x$ and
$q : z \to y$ such that $p \circ i = \text{id}_x$,
$q \circ j = \text{id}_y$, $p \circ j = 0$, $q \circ i = 0$
and $i \circ p + j \circ q = \text{id}_z$, then $z$
is the direct sum of $x$ and $y$ with the four morphisms
equal to $i, j, p, q$.
\end{remark}
\begin{lemma}
\label{lemma-additive-additive}
Let $\mathcal{A}$, $\mathcal{B}$ be preadditive categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor.
Then $F$ transforms direct sums to direct sums and zero to zero.
\end{lemma}
\begin{proof}
Suppose $F$ is additive. A direct sum $z$
of $x$ and $y$ is characterized by having morphisms
$i : x \to z$, $j : y \to z$, $p : z \to x$ and
$q : z \to y$ such that $p \circ i = \text{id}_x$,
$q \circ j = \text{id}_y$, $p \circ j = 0$, $q \circ i = 0$
and $i \circ p + j \circ q = \text{id}_z$, according
to Remark \ref{remark-direct-sum}. Clearly $F(x), F(y), F(z)$
and the morphisms $F(i), F(j), F(p), F(q)$ satisfy exactly the
same relations (by additivity) and we see that $F(z)$ is
a direct sum of $F(x)$ and $F(y)$.
\end{proof}
\begin{definition}
\label{definition-additive-category}
A category $\mathcal{A}$ is called {\it additive}
if it is preadditive and finite products exist, in other
words it has a zero object and direct sums.
\end{definition}
\noindent
Namely the empty product is a finite product and
if it exists, then it is a final object.
\begin{definition}
\label{definition-kernel}
Let $\mathcal{A}$ be a preadditive category.
Let $f : x \to y$ be a morphism.
\begin{enumerate}
\item A {\it kernel} of $f$ is a morphism
$i : z \to x$ such that (a) $f \circ i = 0$ and (b)
for any $i' : z' \to x$ such that $f \circ i' = 0$ there
exists a unique morphism $g : z' \to z$ such that
$i' = i \circ g$.
\item If the kernel of $f$ exists, then we denote
this $\text{Ker}(f) \to x$.
\item A {\it cokernel} of $f$ is a morphism
$p : y \to z$ such that (a) $p \circ f = 0$ and (b)
for any $p' : y \to z'$ such that $p' \circ f = 0$ there
exists a unique morphism $g : z \to z'$ such that
$p' = g \circ p$.
\item If a cokernel of $f$ exists we denote this
$y \to \text{Coker}(f)$.
\item If a kernel of $f$ exists, then a {\it coimage
of $f$} is a cokernel for the morphism $\text{Ker}(f) \to x$.
\item If a kernel and coimage exist then we denote this
$x \to \text{Coim}(f)$.
\item If a cokernel of $f$ exists, then the {\it image of
$f$} is a kernel of the morphism $y \to \text{Coker}(f)$.
\item If a cokernel and image of $f$ exist then we denote
this $\text{Im}(f) \to y$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-coim-im-map}
Let $f : x \to y$ be a morphism in a preadditive category
such that the kernel, cokernel, image and coimage all exist.
Then $f$ can be factored uniquely as
$x \to \text{Coim}(f) \to \text{Im}(f) \to y$.
\end{lemma}
\begin{proof}
There is a canonical morphism $\text{Coim}(f) \to y$
because $\text{Ker}(f) \to x \to y$ is zero.
The composition $\text{Coim}(f) \to y \to \text{Coker}(f)$
is zero, because it is the unique morphism which gives
rise to the morphism $x \to y \to \text{Coker}(f)$ which
is zero. Hence $\text{Coim}(f) \to y$ factors uniquely through
$\text{Im}(f) \to y$, which gives us the desired map.
\end{proof}
\begin{example}
\label{example-not-abelian}
Let $k$ be a field.
Consider the category
of filtered vector spaces over $k$.
(See Definition \ref{definition-filtered}.)
Consider the filtered vector spaces $(V, F)$ and $(W, F)$ with
$V = W = k$ and
$$
F^iV
=
\left\{
\begin{matrix}
V & \text{if} & i < 0 \\
0 & \text{if} & i \geq 0
\end{matrix}
\right.
\text{ and }
F^iW
=
\left\{
\begin{matrix}
W & \text{if} & i \leq 0 \\
0 & \text{if} & i > 0
\end{matrix}
\right.
$$
The map $f : V \to W$ corresponding to $\text{id}_k$ on the underlying
vector spaces has trivial kernel and cokernel but is not
an isomorphism. Note also that $\text{Coim}(f) = V$ and $\text{Im}(f) = W$.
This means that the category of filtered vector spaces over $k$
is not abelian.
\end{example}
\begin{definition}
\label{definition-abelian-category}
A category $\mathcal{A}$ is {\it abelian} if
it is additive, if all kernels and cokernels exist,
and if the natural map $\text{Coim}(f) \to \text{Im}(f)$
is an isomorphism for all morphisms $f$ of
$\mathcal{A}$.
\end{definition}
\begin{lemma}
\label{lemma-abelian-opposite}
Let $\mathcal{A}$ be a preadditive category.
The additions on sets of morphisms make
$\mathcal{A}^{opp}$ into a preadditive category.
Furthermore, $\mathcal{A}$ is additive if and only if $\mathcal{A}^{opp}$
is additive, and
$\mathcal{A}$ is abelian if and only if $\mathcal{A}^{opp}$ is abelian.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-injective-surjective}
Let $f : x \to y$ be a morphism in an abelian category.
\begin{enumerate}
\item We say $f$ is {\it injective} if $\text{Ker}(f) = 0$.
\item We say $f$ is {\it surjective} if $\text{Coker}(f) = 0$.
\end{enumerate}
If $x \to y$ is injective, then we say that $x$ is a {\it subobject}
of $y$ and we use the notation $x \subset y$. If $x \to y$ is
surjective, then we say that $y$ is a {\it quotient} of $x$.
\end{definition}
\begin{lemma}
\label{lemma-characterize-injective}
Let $f : x \to y$ be a morphism in an abelian category. Then
\begin{enumerate}
\item $f$ is injective if and only if $f$ is a monomorphism, and
\item $f$ is surjective if and only if $f$ is an epimorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
In an abelian category, if $x \subset y$ is a subobject,
then we denote
$$
x/y = \text{Coker}(x \to y).
$$
\begin{lemma}
\label{lemma-colimit-abelian-category}
Let $\mathcal{A}$ be an abelian category.
All finite limits and finite colimits exist in $\mathcal{A}$.
\end{lemma}
\begin{proof}
To show that finite limits exist it suffices to show
that finite products and equalizers exist, see
Categories, Lemma \ref{categories-lemma-finite-limits-exist}.
Finite products exist
by definition and the equalizer of $a, b : x \to y$ is
the kernel of $a - b$. The argument for finite colimits
is similar but dual to this.
\end{proof}
\begin{example}
\label{example-fibre-product-pushouts}
Let $\mathcal{A}$ be an abelian category.
Pushouts and fibre products in $\mathcal{A}$ have the following
simple descriptions:
\begin{enumerate}
\item If $a : x \to y$, $b : z \to y$ are morphisms in $\mathcal{A}$, then
we have the fibre product:
$x \times_y z = \text{Ker}((a, -b) : x \oplus z \to y)$.
\item If $a : y \to x$, $b : y \to z$ are morphisms in $\mathcal{A}$, then
we have the pushout:
$x \amalg_y z = \text{Coker}((a, -b) : y \to x \oplus z)$.
\end{enumerate}
\end{example}
\begin{definition}
\label{definition-exact}
Let $\mathcal{A}$ be an additive category.
We say a sequence of morphisms
$$
\ldots \to x \to y \to z \to \ldots
$$
in $\mathcal{A}$
is a {\it complex} if the composition of any two (drawn)
arrows is zero. If $\mathcal{A}$ is abelian then
we say a sequence as above is {\it exact at $y$} if
$\text{Im}(x \to y) = \text{Ker}(y \to z)$. We say it is {\it exact}
if it is exact at every object. A {\it short exact sequence}
is an exact complex of the form
$$
0 \to A \to B \to C \to 0.
$$
\end{definition}
\noindent
In the following lemma we assume the reader knows what it means
for a sequence of abelian groups to be exact.
\begin{lemma}
\label{lemma-check-exactness}
Let $\mathcal{A}$ be an abelian category.
Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a complex of $\mathcal{A}$.
\begin{enumerate}
\item $M_1 \to M_2 \to M_3 \to 0$ is exact if and only if
$$
0 \to \Hom_\mathcal{A}(M_3, N) \to
\Hom_\mathcal{A}(M_2, N) \to \Hom_\mathcal{A}(M_1, N)
$$
is an exact sequence of abelian groups for all objects $N$ of
$\mathcal{A}$, and
\item $0 \to M_1 \to M_2 \to M_3$ is exact if and only if
$$
0 \to \Hom_\mathcal{A}(N, M_1) \to \Hom_\mathcal{A}(N, M_2) \to
\Hom_\mathcal{A}(N, M_1)
$$
is an exact sequence of abelian groups for all objects $N$ of $\mathcal{A}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Hint: See
Algebra, Lemma \ref{algebra-lemma-hom-exact}.
\end{proof}
\begin{definition}
\label{definition-ses-split}
Let $\mathcal{A}$ be an abelian category.
Let $i : A \to B$ and $q : B \to C$ be morphisms
of $\mathcal{A}$ such that
$0 \to A \to B \to C \to 0$ is a short
exact sequence. We say the short exact
sequence is {\it split} if there exist
morphisms $j : C \to B$ and $p : B \to A$ such
that $(B, i, j, p, q)$ is the direct sum of $A$ and $C$.
\end{definition}
\begin{lemma}
\label{lemma-ses-split}
Let $\mathcal{A}$ be an abelian category.
Let $0 \to A \to B \to C \to 0$
be a short exact sequence.
\begin{enumerate}
\item Given a morphism $s : C \to B$ left inverse to
$B \to C$, there exists a unique $\pi : B \to A$
such that $(s, \pi)$ splits the short exact sequence
as in Definition \ref{definition-ses-split}.
\item Given a morphism $\pi : B \to A$ right inverse to
$A \to B$, there exists a unique $s : C \to B$
such that $(s, \pi)$ splits the short exact sequence
as in Definition \ref{definition-ses-split}.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-characterize-cartesian}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
w\ar[r]^f\ar[d]_g
& y\ar[d]^h\\
x\ar[r]^k
& z
}
$$
be a commutative diagram.
\begin{enumerate}
\item The diagram is cartesian if and only if
$$
0\to w\xrightarrow{(g,-f)}x\oplus y\xrightarrow{(k,-h)}z
$$
is exact.
\item The diagram is cocartesian if and only if
$$
w\xrightarrow{(g,-f)}x\oplus y\xrightarrow{(k,-h)}z\to 0
$$
is exact.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $u=(g,-f):w\to x\oplus y$ and $v=(k,-h):x\oplus y\to z$.
Let $p:x\oplus y\to x$ and $q:x\oplus y\to y$ be the canonical
projections. Let $i:\text{Ker}(v)\to x\oplus v$ be the canonical
injection. By Example \ref{example-fibre-product-pushouts}, the diagram is
cartesian if and only if there exists an isomorphism
$r:\text{Ker}(v)\to w$ with $-f\circ r=q\circ i$ and
$g\circ r=p\circ i$. The sequence
$0\to w\overset{u}\to x\oplus y\overset{v}\to z$ is exact if and
only if there exists an isomorphism $r:\text{Ker}(v)\to w$ with
$u\circ r=i$. But given $r:\text{Ker}(v)\to w$, we have
$-f\circ r=q\circ i$ and $g\circ r=p\circ i$ if and
only if $q\circ u\circ r=-f\circ r=q\circ i$ and
$p\circ u\circ r=g\circ r=p\circ i$, hence if and only if $u\circ r=i$.
This proves (1), and then (2) follows by duality.
\end{proof}
\begin{lemma}
\label{lemma-cartesian-kernel}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
w\ar[r]^f\ar[d]_g
& y\ar[d]^h\\
x\ar[r]^k
& z
}
$$
be a commutative diagram.
\begin{enumerate}
\item If the diagram is cartesian, then the morphism
$\text{Ker}(f)\to\text{Ker}(k)$ induced by $g$ is an isomorphism.
\item If the diagram is cocartesian, then the morphism
$\text{Coker}(f)\to\text{Coker}(k)$ induced by $h$ is an isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose the diagram is cartesian. Let
$e:\text{Ker}(f)\to\text{Ker}(k)$ be induced by $g$. Let
$i:\text{Ker}(f)\to w$ and $j:\text{Ker}(k)\to x$ be the canonical
injections. There exists $t:\text{Ker}(k)\to w$ with $f\circ t=0$
and $g\circ t=j$. Hence, there exists $u:\text{Ker}(k)\to\text{Ker}(f)$
with $i\circ u=t$. It follows
$g\circ i\circ u\circ e=g\circ t\circ e=j\circ e=g\circ i$ and
$f\circ i\circ u\circ e=0=f\circ i$, hence $i\circ u\circ e=i$. Since
$i$ is a monomorphism this implies $u\circ e=\text{id}_{\text{Ker}(f)}$.
Furthermore, we have $j\circ e\circ u=g\circ i\circ u=g\circ t=j$.
Since $j$ is a monomorphism this implies $e\circ u=\text{id}_{\text{Ker}(k)}$.
This proves (1). Now, (2) follows by duality.
\end{proof}
\begin{lemma}
\label{lemma-cartesian-cocartesian}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
w\ar[r]^f\ar[d]_g
& y\ar[d]^h\\
x\ar[r]^k
& z
}
$$
be a commutative diagram.
\begin{enumerate}
\item If the diagram is cartesian and $k$ is an epimorphism,
then the diagram is cocartesian and $f$ is an epimorphism.
\item If the diagram is cocartesian and $g$ is a monomorphism,
then the diagram is cartesian and $h$ is a monomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose the diagram is cartesian and $k$ is an epimorphism.
Let $u=(g,-f):w\to x\oplus y$ and let $v=(k,-h):x\oplus y\to z$.
As $k$ is an epimorphism, $v$ is an epimorphism, too. Therefore
and by Lemma \ref{lemma-characterize-cartesian}, the sequence
$0\to w\overset{u}\to x\oplus y\overset{v}\to z\to 0$ is exact. Thus, the
diagram is cocartesian by Lemma \ref{lemma-characterize-cartesian}. Finally,
$f$ is an epimorphism by Lemma \ref{lemma-cartesian-kernel} and
Lemma \ref{lemma-characterize-injective}. This proves (1), and (2)
follows by duality.
\end{proof}
\begin{lemma}
\label{lemma-epimorphism-universal-abelian-category}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item If $x \to y$ is surjective, then for every $z \to y$ the
projection $x \times_y z \to z$ is surjective.
\item If $x \to y$ is injective, then for every $x \to z$ the
morphism $z \to z \amalg_x y$ is injective.
\end{enumerate}
\end{lemma}
\begin{proof}
Immediately from Lemma \ref{lemma-characterize-injective} and
Lemma \ref{lemma-cartesian-cocartesian}.
\end{proof}
\begin{lemma}
\label{lemma-check-exactness-fibre-product}
Let $\mathcal{A}$ be an abelian category. Let $f:x\to y$ and $g:y\to z$
be morphisms with $g\circ f=0$. Then, the following statements are equivalent:
\begin{enumerate}
\item The sequence $x\overset{f}\to y\overset{g}\to z$ is exact.
\item For every $h:w\to y$ with $g\circ h=0$ there exist an object $v$,
an epimorphism $k:v\to w$ and a morphism $l:v\to x$ with $h\circ k=f\circ l$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $i:\text{Ker}(g)\to y$ be the canonical injection. Let
$p:x\to\text{Coim}(f)$ be the canonical projection. Let
$j:\text{Im}(f)\to\text{Ker}(g)$ be the canonical injection.
\medskip\noindent
Suppose (1) holds. Let $h:w\to y$ with $g\circ h=0$. There exists
$c:w\to\text{Ker}(g)$ with $i\circ c=h$.
Let $v=x\times_{\text{Ker}(g)}w$ with canonical projections
$k:v\to w$ and $l:v\to x$, so that $c\circ k=p\circ l$.
Then, $h\circ k=i\circ c\circ k=i\circ j\circ p\circ l=f\circ l$.
As $j\circ p$ is an epimorphism by hypothesis, $k$ is an
epimorphism by Lemma \ref{lemma-cartesian-cocartesian}. This implies (2).
\medskip\noindent
Suppose (2) holds. Then, $g\circ i=0$. So, there are an object
$w$, an epimorphism $k:w\to\text{Ker}(g)$ and a morphism
$l:w\to x$ with $f\circ l=i\circ k$. It follows
$i\circ j\circ p\circ l=f\circ l=i\circ k$. Since $i$ is a
monomorphism we see that $j\circ p\circ l=k$ is an epimorphism.
So, $j$ is an epimorphisms and thus an isomorphism. This implies (1).
\end{proof}
\begin{lemma}
\label{lemma-exact-kernel-sequence}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
x \ar[r]^f \ar[d]^\alpha &
y \ar[r]^g \ar[d]^\beta &
z \ar[d]^\gamma\\
u \ar[r]^k & v \ar[r]^l & w
}
$$
be a commutative diagram.
\begin{enumerate}
\item If the first row is exact and $k$ is a monomorphism, then the induced
sequence $\text{Ker}(\alpha) \to \text{Ker}(\beta) \to \text{Ker}(\gamma)$
is exact.
\item If the second row is exact and $g$ is an epimorphism, then the induced
sequence
$\text{Coker}(\alpha) \to \text{Coker}(\beta) \to \text{Coker}(\gamma)$
is exact.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose the first row is exact and $k$ is a monomorphism. Let
$a:\text{Ker}(\alpha)\to\text{Ker}(\beta)$ and
$b:\text{Ker}(\beta)\to\text{Ker}(\gamma)$ be the induced morphisms.
Let $h:\text{Ker}(\alpha)\to x$, $i:\text{Ker}(\beta)\to y$ and
$j:\text{Ker}(\gamma)\to z$ be the canonical injections. As $j$ is
a monomorphism we have $b\circ a=0$. Let $c:s\to\text{Ker}(\beta)$
with $b\circ c=0$. Then, $g\circ i\circ c=j\circ b\circ c=0$. By
Lemma \ref{lemma-check-exactness-fibre-product} there are an object $t$, an
epimorphism $d:t\to s$ and a morphism $e:t\to x$ with
$i\circ c\circ d=f\circ e$. Then,
$k\circ \alpha\circ e=\beta\circ f\circ e=\beta\circ i\circ c\circ d=0$.
As $k$ is a monomorphism we get $\alpha\circ e=0$. So, there exists
$m:t\to\text{Ker}(\alpha)$ with $h\circ m=e$. It follows
$i\circ a\circ m=f\circ h\circ m=f\circ e=i\circ c\circ d$.
As $i$ is a monomorphism we get $a\circ m=c\circ d$. Thus,
Lemma \ref{lemma-check-exactness-fibre-product} implies (1), and then
(2) follows by duality.
\end{proof}
\begin{lemma}
\label{lemma-snake}
Let $\mathcal{A}$ be an abelian category.
Let
$$
\xymatrix{
& x \ar[r]^f \ar[d]^\alpha &
y \ar[r]^g \ar[d]^\beta &
z \ar[r] \ar[d]^\gamma &
0 \\
0 \ar[r] & u \ar[r]^k & v \ar[r]^l & w
}
$$
be a commutative diagram with exact rows.
\begin{enumerate}
\item There exists a unique morphism
$\delta : \text{Ker}(\gamma) \rightarrow \text{Coker}(\alpha)$
such that the diagram
$$
\xymatrix{
y \ar[d]_\beta &
y \times_z \text{Ker}(\gamma) \ar[l]_{\pi'} \ar[r]^{\pi} &
\text{Ker}(\gamma) \ar[d]^\delta \\
v \ar[r]^{\iota'} & \text{Coker}(\alpha) \amalg_u v &
\text{Coker}(\alpha) \ar[l]_\iota
}
$$
commutes, where $\pi$ and $\pi'$ are the canonical projections
and $\iota$ and $\iota'$ are the canonical coprojections.
\item The induced sequence
$$
\text{Ker}(\alpha)\overset{f'} \to \text{Ker}(\beta)
\overset{g'}\to \text{Ker}(\gamma)\overset{\delta}\to
\text{Coker}(\alpha) \overset{k'}\to \text{Coker}(\beta)
\overset{l'}\to \text{Coker}(\gamma)
$$
is exact. If $f$ is injective then so is $f'$, and if $l$ is
surjective then so is $l'$.
\end{enumerate}
\end{lemma}
\begin{proof}
As $\pi$ is an epimorphism and $\iota$ is a monomorphism by
Lemma \ref{lemma-cartesian-cocartesian}, uniqueness of $\delta$ is clear.
Let $p=y\times_z\text{Ker}(\gamma)$ and $q=\text{Coker}(\alpha)\amalg_uv$.
Let $h:\text{Ker}(\beta)\to y$, $i:\text{Ker}(\gamma)\to z$ and
$j:\text{Ker}(\pi)\rightarrow p$ be the canonical injections.
Let $p:u\to\text{Coker}(\alpha)$ be the canonical projection.
Keeping in mind Lemma \ref{lemma-cartesian-cocartesian} we get a commutative
diagram with exact rows
$$
\xymatrix{
0 \ar[r] &
\text{Ker}(\pi) \ar[r]^j &
p \ar[r]^{\pi} \ar[d]_{\pi'} &
\text{Ker}(\gamma) \ar[d]_i \ar[r] & 0 \\
& x \ar[r]^f \ar[d]_\alpha & y \ar[r]^g \ar[d]_\beta &
z \ar[d]_\gamma \ar[r] & 0 \\
0 \ar[r] & u \ar[r]^k \ar[d]_p &
v \ar[r]^l \ar[d]_{\iota'} & w & \\
0 \ar[r] & \text{Coker}(\alpha) \ar[r]^\iota & q & &
}
$$
As $l\circ\beta\circ\pi'=\gamma\circ i\circ\pi=0$ and as the third
row of the diagram above is exact, there is an $a:p\to u$
with $k\circ a=\beta\circ\pi'$. As the upper right quadrangle of the
diagram above is cartesian, Lemma \ref{lemma-cartesian-kernel} yields an
epimorphism $b:x\to\text{Ker}(\pi)$ with $\pi'\circ j\circ b=f$.
It follows
$k\circ a\circ j\circ b=\beta\circ\pi'\circ j\circ b=\beta\circ f=k\circ\alpha$.
As $k$ is a monomorphism this implies $a\circ j\circ b=\alpha$. It follows
$p\circ a\circ j\circ b=p\circ \alpha=0$. As $b$ is an epimorphism this
implies $p\circ a\circ j=0$. Therefore, as the top row of the diagram
above is exact, there exists
$\delta:\text{Ker}(\gamma)\to\text{Coker}(\alpha)$ with
$\delta\circ\pi=p\circ a$. It follows
$\iota\circ\delta\circ\pi=\iota\circ p\circ a=\iota'\circ k\circ a=
\iota'\circ\beta\circ\pi'$
as desired.
\medskip\noindent
As the upper right quadrangle in the diagram above is cartesian there
is a $c:\text{Ker}(\beta)\to p$ with $\pi'\circ c=h$ and $\pi\circ c=g'$.
It follows
$\iota\circ\delta\circ g'=\iota\circ\delta\circ\pi\circ c=
\iota'\circ\beta\circ \pi'\circ c=\iota'\circ\beta\circ h=0$.
As $\iota$ is a monomorphism this implies $\delta\circ g'=0$.
\medskip\noindent
Next, let $d:r\to\text{Ker}(\gamma)$ with $\delta\circ d=0$. Applying
Lemma \ref{lemma-check-exactness-fibre-product} to the exact sequence
$p\overset{\pi}\to\text{Ker}(\gamma)\to 0$ and $d$ yields an object $s$,
an epimorphism $m:s\to r$ and a morphism $n:s\to p$ with
$\pi\circ n=d\circ m$. As $p\circ a\circ n=\delta\circ d\circ m=0$,
applying Lemma \ref{lemma-check-exactness-fibre-product} to the exact sequence
$x\overset{\alpha}\to u\overset{p}\to\text{Coker}(\alpha)$ and
$a\circ n$ yields an object $t$, an epimorphism $\varepsilon:t\to s$ and
a morphism $\zeta:t\to x$ with $a\circ n\circ \varepsilon=\alpha\circ\zeta$.
It holds
$\beta\circ\pi'\circ n\circ\varepsilon=k\circ\alpha\circ\zeta=
\beta\circ f\circ\zeta$.
Let $\eta=\pi'\circ n\circ \varepsilon-f\circ \zeta:t\to y$. Then,
$\beta\circ \eta=0$. It follows that there is a
$\vartheta:t\to\text{Ker}(\beta)$ with $\eta=h\circ \vartheta$. It holds
$i\circ g'\circ\vartheta=g\circ h\circ\vartheta=
g\circ\pi'\circ n\circ\varepsilon-g\circ f\circ\zeta=
i\circ\pi\circ n\circ\varepsilon=i\circ d\circ m\circ\varepsilon$.
As $i$ is a monomorphism we get $g'\circ\vartheta=d\circ m\circ\varepsilon$.
Thus, as $m\circ \varepsilon$ is an epimorphism,
Lemma \ref{lemma-check-exactness-fibre-product} implies that
$\text{Ker}(\beta)\overset{g'}\to\text{Ker}(\gamma)\overset{\delta}\to
\text{Coker}(\alpha)$
is exact. Then, the claim follows by Lemma \ref{lemma-exact-kernel-sequence}
and duality.
\end{proof}
\begin{lemma}
\label{lemma-snake-natural}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
& & & x\ar[ld]\ar[rr]\ar[dd]^(.4)\alpha
& & y\ar[ld]\ar[rr]\ar[dd]^(.4)\beta
& & z\ar[ld]\ar[rr]\ar[dd]^(.4)\gamma
& & 0\\
& & x'\ar[rr]\ar[dd]^(.4){\alpha'}
& & y'\ar[rr]\ar[dd]^(.4){\beta'}
& & z'\ar[rr]\ar[dd]^(.4){\gamma'}
& & 0
& \\
& 0\ar[rr]
& & u\ar[ld]\ar[rr]
& & v\ar[ld]\ar[rr]
& & w\ar[ld]
& & \\
0\ar[rr]
& & u'\ar[rr]
& & v'\ar[rr]
& & w'
& & &
}
$$
be a commutative diagram with exact rows. Then, the induced diagram
$$
\xymatrix@C=15pt{
\text{Ker}(\alpha) \ar[r] \ar[d] &
\text{Ker}(\beta) \ar[r] \ar[d] &
\text{Ker}(\gamma) \ar[r]^(.45){\delta} \ar[d] &
\text{Coker}(\alpha) \ar[r] \ar[d] &
\text{Coker}(\beta) \ar[r] \ar[d] &
\text{Coker}(\gamma) \ar[d] \\
\text{Ker}(\alpha') \ar[r] &
\text{Ker}(\beta') \ar[r] &
\text{Ker}(\gamma') \ar[r]^(.45){\delta'} &
\text{Coker}(\alpha') \ar[r] &
\text{Coker}(\beta') \ar[r] &
\text{Coker}(\gamma')
}
$$
commutes.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-four-lemma}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
w \ar[r] \ar[d]^\alpha & x \ar[r] \ar[d]^\beta & y \ar[r] \ar[d]^\gamma &
z \ar[d]^\delta \\
w' \ar[r] & x' \ar[r] & y' \ar[r] & z'
}
$$
be a commutative diagram with exact rows.
\begin{enumerate}
\item If $\alpha, \gamma$ are surjective and $\delta$ is injective, then
$\beta$ is surjective.
\item If $\beta, \delta$ are injective and $\alpha$ is surjective, then
$\gamma$ is injective.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $\alpha, \gamma$ are surjective and $\delta$ is injective.
We may replace $w'$ by $\text{Im}(w' \to x')$, i.e., we may assume
that $w' \to x'$ is injective.
We may replace $z$ by $\text{Im}(y \to z)$, i.e., we may assume that
$y \to z$ is surjective. Then we may apply
Lemma \ref{lemma-snake}
to
$$
\xymatrix{
& \text{Ker}(y \to z) \ar[r] \ar[d] & y \ar[r] \ar[d] & z \ar[r] \ar[d] & 0 \\
0 \ar[r] & \text{Ker}(y' \to z') \ar[r] & y' \ar[r] & z'
}
$$
to conclude that $\text{Ker}(y \to z) \to \text{Ker}(y' \to z')$ is
surjective. Finally, we apply
Lemma \ref{lemma-snake}
to
$$
\xymatrix{
& w \ar[r] \ar[d] & x \ar[r] \ar[d] & \text{Ker}(y \to z) \ar[r] \ar[d] & 0 \\
0 \ar[r] & w' \ar[r] & x' \ar[r] & \text{Ker}(y' \to z')
}
$$
to conclude that $x \to x'$ is surjective. This proves (1). The proof
of (2) is dual to this.
\end{proof}
\begin{lemma}
\label{lemma-five-lemma}
Let $\mathcal{A}$ be an abelian category. Let
$$
\xymatrix{
v \ar[r] \ar[d]^\alpha &
w \ar[r] \ar[d]^\beta &
x \ar[r] \ar[d]^\gamma &
y \ar[r] \ar[d]^\delta &
z \ar[d]^\epsilon \\
v' \ar[r] & w' \ar[r] & x' \ar[r] & y' \ar[r] & z'
}
$$
be a commutative diagram with exact rows. If $\beta, \delta$
are isomorphisms, $\epsilon$ is injective, and $\alpha$ is surjective
then $\gamma$ is an isomorphism.
\end{lemma}
\begin{proof}
Immediate consequence of
Lemma \ref{lemma-four-lemma}.
\end{proof}
\section{Extensions}
\label{section-extensions}
\begin{definition}
\label{definition-extension}
Let $\mathcal{A}$ be an abelian category.
Let $A, C \in \Ob(\mathcal{A})$.
An {\it extension $E$ of $B$ by $A$} is a short
exact sequence
$$
0 \to A \to E \to B \to 0.
$$
\end{definition}
\noindent
By abuse of language we often omit mention of the
morphisms $A \to E$ and $E \to B$, although they are
definitively part of the structure of an extension.
\begin{definition}
\label{definition-ext-group}
Let $\mathcal{A}$ be an abelian category.
Let $A, C \in \Ob(\mathcal{A})$.
The set of isomorphism classes of extensions
of $B$ by $A$ is denoted
$$
\text{Ext}_\mathcal{A}(B, A).
$$
This is called the {\it $\text{Ext}$-group}.
\end{definition}
\noindent
This definition works, because by our conventions
$\mathcal{A}$ is a set, and hence
$\text{Ext}_\mathcal{A}(B, A)$ is a set.
In any of the cases of ``big'' abelian categories
listed in Categories, Remark \ref{categories-remark-big-categories}.
one can check by hand that $\text{Ext}_\mathcal{A}(B, A)$
is a set as well. Also, we will see later that this is
always the case when $\mathcal{A}$ has either enough projectives
or enough injectives. Insert future reference here.
\medskip\noindent
Actually we can turn $\text{Ext}_\mathcal{A}(-, -)$ into a
functor
$$
\mathcal{A}^{opp} \times \mathcal{A} \longrightarrow \textit{Sets}, \quad
(A, B) \longmapsto \text{Ext}_\mathcal{A}(A, B)
$$
as follows:
\begin{enumerate}
\item Given a morphism $B' \to B$ and an extension
$E$ of $B$ by $A$ we define $E' = E \times_B B'$
so that we have the following commutative diagram
of short exact sequences
$$
\xymatrix{
0 \ar[r] & A \ar[r] \ar[d] & E' \ar[r] \ar[d] & B' \ar[r] \ar[d] & 0 \\
0 \ar[r] & A \ar[r] & E \ar[r] & B \ar[r] & 0
}
$$
The extension $E'$ is called the {\it pullback of $E$ via
$B' \to B$}.
\item Given a morphism $A \to A'$ and an extension
$E$ of $B$ by $A$ we define $E' = A' \coprod_A E$
so that we have the following commutative diagram
of short exact sequences
$$
\xymatrix{
0 \ar[r] & A \ar[r] \ar[d] & E \ar[r] \ar[d] & B \ar[r] \ar[d] & 0 \\
0 \ar[r] & A' \ar[r] & E' \ar[r] & B \ar[r] & 0
}
$$
The extension $E'$ is called the {\it pushout of $E$ via
$A \to A'$}.
\end{enumerate}
To see that this defines a functor as indicated above
there are several things to verify. First of all
functoriality in the variable $B$ requires that
$(E \times_B B') \times_{B'} B'' = E \times_B B''$
which is a general property of fibre products.
Dually one deals with functoriality in the
variable $A$. Finally, given $A \to A'$ and
$B' \to B$ we have to show that
$$
A' \coprod\nolimits_A (E \times_B B')
\cong
(A' \coprod\nolimits_A E)\times_B B'
$$
as extensions of $B'$ by $A'$. Recall that $A' \coprod_A E$
is a quotient of $A' \oplus E$.
Thus the right hand side is a quotient of
$A' \oplus E \times_B B'$, and it is straightforward to see that
the kernel is exactly what you need in order to
get the left hand side.
\medskip\noindent
Note that if $E_1$ and $E_2$ are extensions of
$B$ by $A$, then $E_1\oplus E_2$ is an extension
of $B \oplus B$ by $A\oplus A$. We pull back by
the diagonal map $B \to B \oplus B$ and we push
out by the sum map $A \oplus A \to A$ to get
an extension $E_1 + E_2$ of $B$ by $A$.
$$
\xymatrix{
0 \ar[r] &
A \oplus A \ar[r] \ar[d]^{\sum} &
E_1 \oplus E_2 \ar[r] \ar[d] &
B \oplus B \ar[r] \ar[d] &
0 \\
0 \ar[r] &
A \ar[r] &
E' \ar[r] &
B \oplus B \ar[r] &