Permalink
Find file
Fetching contributors…
Cannot retrieve contributors at this time
6522 lines (5875 sloc) 210 KB
\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{Preadditive and additive categories}
\label{section-additive-categories}
\noindent
Here is the definition of a preadditive category.
\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 $x \amalg y$.
If the coproduct $x \amalg y$ exists, then so does
the product $x \times y$. In this case
also $x \amalg y \cong x \times y$.
\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 $z = x \amalg y$.
\medskip\noindent
We leave it to the reader to construct the morphisms
$p, q$ given a coproduct $x \amalg y$ 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 $\Ker(f) \to x$.
\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
$y \to \Coker(f)$.
\item If a kernel of $f$ exists, then a {\it coimage
of $f$} is a cokernel for the morphism $\Ker(f) \to x$.
\item If a kernel and coimage exist then we denote this
$x \to \Coim(f)$.
\item If a cokernel of $f$ exists, then the {\it image of
$f$} is a kernel of the morphism $y \to \Coker(f)$.
\item If a cokernel and image of $f$ exist then we denote
this $\Im(f) \to y$.
\end{enumerate}
\end{definition}
\noindent
We first relate the direct sum to kernels as follows.
\begin{lemma}
\label{lemma-additive-cat-biproduct-kernel}
Let $\mathcal{C}$ be a preadditive category.
Let $x \oplus y$ with morphisms $i, j, p, q$ as in
Lemma \ref{lemma-preadditive-direct-sum}
be a direct sum in $\mathcal{C}$. Then $i : x \to x \oplus y$
is a kernel of $q : x \oplus y \rightarrow y$. Dually, $p$ is
a cokernel for $j$.
\end{lemma}
\begin{proof}
Let $f : z \to x \oplus y$ be a morphism such that $q \circ f = 0$.
We have to show that there exists a unique morphism $g : z \to x$
such that $f = i \circ g$. Since $i \circ p + j \circ q$ is the identity on
$x \oplus y$ we see that
$$
f = (i \circ p + j \circ q) \circ f = i \circ p \circ f
$$
and hence $g = p \circ f$ works. Uniqueness holds because $p \circ i$
is the identity on $x$. The proof of the second statement is dual.
\end{proof}
\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
$x \to \Coim(f) \to \Im(f) \to y$.
\end{lemma}
\begin{proof}
There is a canonical morphism $\Coim(f) \to y$
because $\Ker(f) \to x \to y$ is zero.
The composition $\Coim(f) \to y \to \Coker(f)$
is zero, because it is the unique morphism which gives
rise to the morphism $x \to y \to \Coker(f)$ which
is zero. Hence $\Coim(f) \to y$ factors uniquely through
$\Im(f) \to y$, 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 $\Coim(f) = V$ and $\Im(f) = W$.
This means that the category of filtered vector spaces over $k$
is not abelian.
\end{example}
\section{Karoubian categories}
\label{section-karoubian}
\noindent
Skip this section on a first reading.
\begin{definition}
\label{definition-karoubian}
Let $\mathcal{C}$ be a preadditive category. We say $\mathcal{C}$
is {\it Karoubian} if every idempotent endomorphism of an object
of $\mathcal{C}$ has a kernel.
\end{definition}
\noindent
The dual notion would be that every idempotent endomorphism of an
object has a cokernel. However, in view of the (dual of the)
following lemma that would be an equivalent notion.
\begin{lemma}
\label{lemma-karoubian}
Let $\mathcal{C}$ be a preadditive category. The following
are equivalent
\begin{enumerate}
\item $\mathcal{C}$ is Karoubian,
\item every idempotent endomorphism of an object of $\mathcal{C}$ has a
cokernel, and
\item given an idempotent endomorphism $p : z \to z$ of $\mathcal{C}$
there exists a direct sum decomposition $z = x \oplus y$ such
that $p$ corresponds to the projection onto $y$.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume (1) and let $p : z \to z$ be as in (3).
Let $x = \Ker(p)$ and $y = \Ker(1 - p)$. There are maps
$x \to z$ and $y \to z$. Since $(1 - p)p = 0$ we see that $p : z \to z$
factors through $y$, hence we obtain a morphism $z \to y$. Similarly
we obtain a morphism $z \to x$. We omit the verification that these
four morphisms induce an isomorphism $x = y \oplus z$ as in
Remark \ref{remark-direct-sum}.
Thus (1) $\Rightarrow$ (3). The implication (2) $\Rightarrow$ (3)
is dual. Finally, condition (3) implies (1) and (2) by
Lemma \ref{lemma-additive-cat-biproduct-kernel}.
\end{proof}
\begin{lemma}
\label{lemma-projectors-have-images}
Let $\mathcal{D}$ be a preadditive category.
\begin{enumerate}
\item If $\mathcal{D}$ has countable products and kernels of maps which
have a right inverse, then $\mathcal{D}$ is Karoubian.
\item If $\mathcal{D}$ has countable coproducts and cokernels of
maps which have a left inverse, then $\mathcal{D}$ is Karoubian.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $X$ be an object of $\mathcal{D}$ and let $e : X \to X$ be an idempotent.
The functor
$$
W \longmapsto \Ker(
\Mor_\mathcal{D}(W, X)
\xrightarrow{e}
\Mor_\mathcal{D}(W, X)
)
$$
if representable if and only if $e$ has a kernel. Note that for any
abelian group $A$ and idempotent endomorphism $e : A \to A$ we have
$$
\Ker(e : A \to A)
= \Ker(\Phi :
\prod\nolimits_{n \in \mathbf{N}} A
\to
\prod\nolimits_{n \in \mathbf{N}} A
)
$$
where
$$
\Phi(a_1, a_2, a_3, \ldots) = (ea_1 + (1 - e)a_2, ea_2 + (1 - e)a_3, \ldots)
$$
Moreover, $\Phi$ has the right inverse
$$
\Psi(a_1, a_2, a_3, \ldots) =
(a_1, (1 - e)a_1 + ea_2, (1 - e)a_2 + ea_3, \ldots).
$$
Hence (1) holds. The proof of (2) is dual (using the dual definition
of a Karoubian category, namely condition (2) of
Lemma \ref{lemma-karoubian}).
\end{proof}
\section{Abelian categories}
\label{section-abelian-categories}
\noindent
An abelian category is a category satisfying just enough axioms so the
snake lemma holds. An axiom (that is sometimes forgotten)
is that the canonical map $\Coim(f) \to \Im(f)$
of Lemma \ref{lemma-coim-im-map} is always an isomorphism.
Example \ref{example-not-abelian} shows that it is necessary.
\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 $\Coim(f) \to \Im(f)$
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 $\Ker(f) = 0$.
\item We say $f$ is {\it surjective} if $\Coker(f) = 0$.
\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
$$
y/x = \Coker(x \to y).
$$
\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:
$x \times_y z = \Ker((a, -b) : x \oplus z \to y)$.
\item If $a : y \to x$, $b : y \to z$ are morphisms in $\mathcal{A}$, then
we have the pushout:
$x \amalg_y z = \Coker((a, -b) : y \to x \oplus z)$.
\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
$\Im(x \to y) = \Ker(y \to z)$. 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 $i : \Ker(v) \to x \oplus y$ be the canonical
injection. By Example \ref{example-fibre-product-pushouts}, the diagram is
cartesian if and only if there exists an isomorphism
$r : \Ker(v) \to w$ 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 $r : \Ker(v) \to w$ with
$u \circ r = i$. But given $r : \Ker(v) \to w$, 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
$\Ker(f)\to\Ker(k)$ induced by $g$ is an isomorphism.
\item If the diagram is cocartesian, then the morphism
$\Coker(f)\to\Coker(k)$ induced by $h$ is an isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose the diagram is cartesian. Let
$e:\Ker(f)\to\Ker(k)$ be induced by $g$. Let
$i:\Ker(f)\to w$ and $j:\Ker(k)\to x$ be the canonical
injections. There exists $t:\Ker(k)\to w$ with $f\circ t=0$
and $g\circ t=j$. Hence, there exists $u:\Ker(k)\to\Ker(f)$
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 $u\circ e=\text{id}_{\Ker(f)}$.
Furthermore, we have $j\circ e\circ u=g\circ i\circ u=g\circ t=j$.
Since $j$ is a monomorphism this implies $e\circ u=\text{id}_{\Ker(k)}$.
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 $i:\Ker(g)\to y$ be the canonical injection. Let
$p:x\to\Coim(f)$ be the canonical projection. Let
$j:\Im(f)\to\Ker(g)$ be the canonical injection.
\medskip\noindent
Suppose (1) holds. Let $h:w\to y$ with $g\circ h=0$. There exists
$c:w\to\Ker(g)$ with $i\circ c=h$.
Let $v=x\times_{\Ker(g)}w$ with canonical projections
$k:v\to w$ and $l:v\to x$, so that $c\circ k=j\circ p\circ l$.
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 $k:w\to\Ker(g)$ 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 $\Ker(\alpha) \to \Ker(\beta) \to \Ker(\gamma)$
is exact.
\item If the second row is exact and $g$ is an epimorphism, then the induced
sequence
$\Coker(\alpha) \to \Coker(\beta) \to \Coker(\gamma)$
is exact.
\end{enumerate}
\end{lemma}
\begin{proof}
Suppose the first row is exact and $k$ is a monomorphism. Let
$a:\Ker(\alpha)\to\Ker(\beta)$ and
$b:\Ker(\beta)\to\Ker(\gamma)$ be the induced morphisms.
Let $h:\Ker(\alpha)\to x$, $i:\Ker(\beta)\to y$ and
$j:\Ker(\gamma)\to z$ be the canonical injections. As $j$ is
a monomorphism we have $b\circ a=0$. Let $c:s\to\Ker(\beta)$
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
$m:t\to\Ker(\alpha)$ 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
$\delta : \Ker(\gamma) \rightarrow \Coker(\alpha)$
such that the diagram
$$
\xymatrix{
y \ar[d]_\beta &
y \times_z \Ker(\gamma) \ar[l]_{\pi'} \ar[r]^{\pi} &
\Ker(\gamma) \ar[d]^\delta \\
v \ar[r]^{\iota'} & \Coker(\alpha) \amalg_u v &
\Coker(\alpha) \ar[l]_\iota
}
$$
commutes, where $\pi$ and $\pi'$ are the canonical projections
and $\iota$ and $\iota'$ are the canonical coprojections.
\item The induced sequence
$$
\Ker(\alpha)\overset{f'} \to \Ker(\beta)
\overset{g'}\to \Ker(\gamma)\overset{\delta}\to
\Coker(\alpha) \overset{k'}\to \Coker(\beta)
\overset{l'}\to \Coker(\gamma)
$$
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 $p=y\times_z\Ker(\gamma)$ and $q=\Coker(\alpha)\amalg_uv$.
Let $h:\Ker(\beta)\to y$, $i:\Ker(\gamma)\to z$ and
$j:\Ker(\pi)\rightarrow p$ be the canonical injections.
Let $p:u\to\Coker(\alpha)$ be the canonical projection.
Keeping in mind Lemma \ref{lemma-cartesian-cocartesian} we get a commutative
diagram with exact rows
$$
\xymatrix{
0 \ar[r] &
\Ker(\pi) \ar[r]^j &
p \ar[r]^{\pi} \ar[d]_{\pi'} &
\Ker(\gamma) \ar[d]_i \ar[r] & 0 \\
& x \ar[r]^f \ar[d]_\alpha & y \ar[r]^g \ar[d]_\beta &
z \ar[d]_\gamma \ar[r] & 0 \\
0 \ar[r] & u \ar[r]^k \ar[d]_p &
v \ar[r]^l \ar[d]_{\iota'} & w & \\
0 \ar[r] & \Coker(\alpha) \ar[r]^\iota & q & &
}
$$
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 $b : x \to \Ker(\pi)$ 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
$\delta : \Ker(\gamma) \to \Coker(\alpha)$ 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 $c : \Ker(\beta) \to p$ 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 $d : r \to \Ker(\gamma)$ with $\delta \circ d = 0$. Applying
Lemma \ref{lemma-check-exactness-fibre-product} to the exact sequence
$p \overset{\pi}\to \Ker(\gamma) \to 0$ 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
$x \overset{\alpha}\to u \overset{p}\to \Coker(\alpha)$ 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
$\vartheta : t \to \Ker(\beta)$ 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
$\Ker(\beta) \overset{g'}\to \Ker(\gamma) \overset{\delta}\to \Coker(\alpha)$
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
$$
\xymatrix@C=15pt{
\Ker(\alpha) \ar[r] \ar[d] &
\Ker(\beta) \ar[r] \ar[d] &
\Ker(\gamma) \ar[r]^(.45){\delta} \ar[d] &
\Coker(\alpha) \ar[r] \ar[d] &
\Coker(\beta) \ar[r] \ar[d] &
\Coker(\gamma) \ar[d] \\
\Ker(\alpha') \ar[r] &
\Ker(\beta') \ar[r] &
\Ker(\gamma') \ar[r]^(.45){\delta'} &
\Coker(\alpha') \ar[r] &
\Coker(\beta') \ar[r] &
\Coker(\gamma')
}
$$
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 $\Im(w' \to x')$, i.e., we may assume
that $w' \to x'$ is injective.
We may replace $z$ by $\Im(y \to z)$, i.e., we may assume that
$y \to z$ is surjective. Then we may apply
Lemma \ref{lemma-snake}
to
$$
\xymatrix{
& \Ker(y \to z) \ar[r] \ar[d] & y \ar[r] \ar[d] & z \ar[r] \ar[d] & 0 \\
0 \ar[r] & \Ker(y' \to z') \ar[r] & y' \ar[r] & z'
}
$$
to conclude that $\Ker(y \to z) \to \Ker(y' \to z')$ is
surjective. Finally, we apply
Lemma \ref{lemma-snake}
to
$$
\xymatrix{
& w \ar[r] \ar[d] & x \ar[r] \ar[d] & \Ker(y \to z) \ar[r] \ar[d] & 0 \\
0 \ar[r] & w' \ar[r] & x' \ar[r] & \Ker(y' \to z')
}
$$
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}
\begin{reference}
\cite[Lemma 4.5 page 16]{Eilenberg-Steenrod}
\end{reference}
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 $A, B \in \Ob(\mathcal{A})$.
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 $A, B \in \Ob(\mathcal{A})$.
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 $E' = A' \amalg_A E$
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
$$
A' \amalg_A (E \times_B B')
\cong
(A' \amalg_A E)\times_B B'
$$
as extensions of $B'$ by $A'$. Recall that $A' \amalg_A E$
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 $A$.
$$
\xymatrix{
0 \ar[r] &
A \oplus A \ar[r] \ar[d]^{\sum} &
E_1 \oplus E_2 \ar[r] \ar[d] &
B \oplus B \ar[r] \ar[d] &
0 \\
0 \ar[r] &
A \ar[r] &
E' \ar[r] &
B \oplus B \ar[r] &
0\\
0 \ar[r] &
A \ar[r] \ar[u] &
E_1 + E_2 \ar[r] \ar[u] &
B \ar[r] \ar[u]^{\Delta} &
0
}
$$
The extension $E_1 + E_2$ is called the {\it Baer sum} of the
given extensions.
\begin{lemma}
\label{lemma-baer-sum}
The construction $(E_1, E_2) \mapsto E_1 + E_2$
above defines a commutative group
law on $\text{Ext}_\mathcal{A}(B, A)$ which is
functorial in both variables.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-six-term-sequence-ext}
Let $\mathcal{A}$ be an abelian category.
Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence
in $\mathcal{A}$.
\begin{enumerate}
\item There is a canonical six term exact sequence of abelian groups
$$
\xymatrix{
0 \ar[r] &
\Hom_\mathcal{A}(M_3, N) \ar[r] &
\Hom_\mathcal{A}(M_2, N) \ar[r] &
\Hom_\mathcal{A}(M_1, N) \ar[lld] \\
& \text{Ext}_\mathcal{A}(M_3, N) \ar[r] &
\text{Ext}_\mathcal{A}(M_2, N) \ar[r] &
\text{Ext}_\mathcal{A}(M_1, N)
}
$$
for all objects $N$ of $\mathcal{A}$, and
\item there is a canonical six term exact sequence of abelian groups
$$
\xymatrix{
0 \ar[r] &
\Hom_\mathcal{A}(N, M_1) \ar[r] &
\Hom_\mathcal{A}(N, M_2) \ar[r] &
\Hom_\mathcal{A}(N, M_3) \ar[lld] \\
& \text{Ext}_\mathcal{A}(N, M_1) \ar[r] &
\text{Ext}_\mathcal{A}(N, M_2) \ar[r] &
\text{Ext}_\mathcal{A}(N, M_3)
}
$$
for all objects $N$ of $\mathcal{A}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted. Hint: The boundary maps are defined using either the pushout
or pullback of the given short exact sequence.
\end{proof}
\section{Additive functors}
\label{section-functors}
\noindent
Recall that we defined, in
Categories, Definition \ref{categories-definition-exact}
the notion of a ``right exact'', ``left exact'' and
``exact'' functor in the setting of a functor between
categories that have finite (co)limits. Thus this
applies in particular to functors between abelian
categories.
\begin{lemma}
\label{lemma-exact-functor}
Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
\begin{enumerate}
\item If $F$ is either left or right exact, then it is additive.
\item If $F$ is additive then it is left exact if and only if
for every short exact sequence
$0 \to A \to B \to C \to 0$
the sequence $0 \to F(A) \to F(B) \to F(C)$
is exact.
\item If $F$ is additive then it is right exact if and only if
for every short exact sequence
$0 \to A \to B \to C \to 0$
the sequence $F(A) \to F(B) \to F(C) \to 0$
is exact.
\item If $F$ is additive then it is exact if and only if
for every short exact sequence
$0 \to A \to B \to C \to 0$
the sequence $0 \to F(A) \to F(B) \to F(C) \to 0$
is exact.
\end{enumerate}
\end{lemma}
\begin{proof}
Let us first note that if $F$ commutes with the empty limit or
the empty colimit, then $F(0) = 0$. In particular $F$ applied
to the zero morphism is zero. We will use this below without mention.
\medskip\noindent
Suppose that $F$ is left exact, i.e., commutes with finite limits.
Then $F(A \times A) = F(A) \times F(A)$ with
projections $F(p)$ and $F(q)$. Hence
$F(A \oplus A) = F(A) \oplus F(A)$ with all
four morphisms $F(i), F(j), F(p), F(q)$ equal to their
counterparts in $\mathcal{B}$ as they satisfy the same
relations, see Remark \ref{remark-direct-sum}.
Then $f = F(p + q)$ is a morphism $f : F(A) \oplus F(A) \to F(A)$
such that $f \circ F(i) = F(p \circ i + q \circ i) = F(\text{id}_A)
= \text{id}_{F(A)}$. And similarly $f \circ F(j) = \text{id}_A$.
We conclude that $F(p + q) = F(p) + F(q)$. For
any pair of morphisms $a, b : B \to A$ the map
$g = F(i \circ a + j \circ b) : F(B) \to F(A) \oplus F(A)$
is a morphism such that $F(p) \circ g =
F(p \circ (i \circ a + j \circ b)) = F(a)$ and similarly
$F(q) \circ g = F(b)$. Hence $g = F(i) \circ F(a) + F(j) \circ F(b)$.
The sum of $a$ and $b$ is the composition
$$
\xymatrix{
B \ar[rr]^-{i \circ a + j \circ b} & &
A \oplus A \ar[r]^-{p + q} & A.
}
$$
Applying $F$ we get
$$
\xymatrix{
F(B) \ar[rrr]^-{F(i) \circ F(a) + F(j) \circ F(b)} & & &
F(A) \oplus F(A) \ar[rr]^-{F(p) + F(q)} & &
A.
}
$$
where we used the expressions for $f$ and $g$ obtained above.
Hence $F$ is additive.\footnote{I'm sure there is an infinitely
slicker proof of this.}
\medskip\noindent
Denote $f : B \to C$ a map from $B$ to $C$.
Exactness of $0 \to A \to B \to C$ just means that
$A = \Ker(f)$. Clearly the kernel of $f$ is
the equalizer of the two maps $f$ and $0$ from $B$ to $C$.
Hence if $F$ commutes with limits, then $F(\Ker(f))
= \Ker(F(f))$ which exactly means that
$0 \to F(A) \to F(B) \to F(C)$ is exact.
\medskip\noindent
Conversely, suppose that $F$ is additive and
transforms any short exact sequence $0 \to A \to B \to C \to 0$ into
an exact sequence $0 \to F(A) \to F(B) \to F(C)$.
Because it is additive it commutes with direct sums
and hence finite products in $\mathcal{A}$. To show
it commutes with finite limits it therefore
suffices to show that it commutes with
equalizers. But equalizers in an abelian category
are the same as the kernel of the difference map,
hence it suffices to show that $F$ commutes with
taking kernels. Let $f : A \to B$ be a morphism.
Factor $f$ as $A \to I \to B$ with $f' : A \to I$ surjective
and $i : I \to B$ injective. (This is possible by the
definition of an abelian category.) Then it is
clear that $\Ker(f) = \Ker(f')$. Also
$0 \to \Ker(f') \to A \to I \to 0$
and
$0 \to I \to B \to B/I \to 0$
are short exact. By the condition imposed on $F$
we see that
$0 \to F(\Ker(f')) \to F(A) \to F(I)$
and
$0 \to F(I) \to F(B) \to F(B/I)$
are exact. Hence it is also the case that
$F(\Ker(f'))$ is the kernel of the map
$F(A) \to F(B)$, and we win.
\medskip\noindent
The proof of (3) is similar to the proof of (2).
Statement (4) is a combination of (2) and (3).
\end{proof}
\begin{lemma}
\label{lemma-exact-functor-ext}
Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor.
For every pair of objects $A, B$ of $\mathcal{A}$ the
functor $F$ induces an abelian group homomorphism
$$
\text{Ext}_\mathcal{A}(B, A)
\longrightarrow
\text{Ext}_\mathcal{B}(F(B), F(A))
$$
which maps the extension $E$ to $F(E)$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
The following lemma is used in the proof that the category of abelian
sheaves on a site is abelian, where the functor $b$ is sheafification.
\begin{lemma}
\label{lemma-adjoint-get-abelian}
Let $a : \mathcal{A} \to \mathcal{B}$ and $b : \mathcal{B} \to \mathcal{A}$
be functors. Assume that
\begin{enumerate}
\item $\mathcal{A}$, $\mathcal{B}$ are additive categories,
$a$, $b$ are additive functors, and $a$ is right adjoint to $b$,
\item $\mathcal{B}$ is abelian and $b$ is left exact, and
\item $ba \cong \text{id}_\mathcal{A}$.
\end{enumerate}
Then $\mathcal{A}$ is abelian.
\end{lemma}
\begin{proof}
As $\mathcal{B}$ is abelian we see that all finite limits and colimits
exist in $\mathcal{B}$ by Lemma \ref{lemma-colimit-abelian-category}.
Since $b$ is a left adjoint we see that $b$ is also right exact
and hence exact, see
Categories, Lemma \ref{categories-lemma-exact-adjoint}.
Let $\varphi : B_1 \to B_2$ be a morphism of $\mathcal{B}$.
In particular, if $K = \Ker(B_1 \to B_2)$, then $K$ is
the equalizer of $0$ and $\varphi$ and hence
$bK$ is the equalizer of $0$ and $b\varphi$, hence
$bK$ is the kernel of $b\varphi$. Similarly, if
$Q = \Coker(B_1 \to B_2)$, then $Q$ is
the coequalizer of $0$ and $\varphi$ and hence
$bQ$ is the coequalizer of $0$ and $b\varphi$, hence
$bQ$ is the cokernel of $b\varphi$. Thus we see that every morphism
of the form $b\varphi$ in $\mathcal{A}$ has a kernel and a cokernel.
However, since $ba \cong \text{id}$ we see that every morphism of
$\mathcal{A}$ is of this form, and we conclude that kernels and
cokernels exist in $\mathcal{A}$. In fact, the argument shows that
if $\psi : A_1 \to A_2$ is a morphism then
$$
\Ker(\psi) = b\Ker(a\psi),
\quad\text{and}\quad
\Coker(\psi) = b\Coker(a\psi).
$$
Now we still have to show that $\Coim(\psi)= \Im(\psi)$.
We do this as follows.
First note that since $\mathcal{A}$ has kernels and cokernels it
has all finite limits and colimits (see proof of
Lemma \ref{lemma-colimit-abelian-category}).
Hence we see by Categories, Lemma \ref{categories-lemma-exact-adjoint}
that $a$ is left exact and
hence transforms kernels (=equalizers) into kernels.
\begin{align*}
\Coim(\psi)
& =
\Coker(\Ker(\psi) \to A_1)
& \text{by definition} \\
& =
b\Coker(a(\Ker(\psi) \to A_1))
& \text{by formula above} \\
& =
b\Coker(\Ker(a\psi) \to aA_1))
& a\text{ preserves kernels} \\
& =
b\Coim(a\psi)
& \text{by definition} \\
& =
b\Im(a\psi)
& \mathcal{B}\text{ is abelian} \\
& =
b\Ker(aA_2 \to \Coker(a\psi))
& \text{by definition} \\
& =
\Ker(baA_2 \to b\Coker(a\psi))
& b\text{ preserves kernels} \\
& =
\Ker(A_2 \to b\Coker(a\psi))
& ba = \text{id}_\mathcal{A} \\
& =
\Ker(A_2 \to \Coker(\psi))
& \text{by formula above} \\
& =
\Im(\psi)
& \text{by definition}
\end{align*}
Thus the lemma holds.
\end{proof}
\section{Localization}
\label{section-localization}
\noindent
In this section we note how Gabriel-Zisman localization interacts with
the additive structure on a category.
\begin{lemma}
\label{lemma-localization-preadditive}
Let $\mathcal{C}$ be a preadditive category.
Let $S$ be a left or right multiplicative system.
There exists a canonical preadditive structure on
$S^{-1}\mathcal{C}$ such that the localization functor
$Q : \mathcal{C} \to S^{-1}\mathcal{C}$ is additive.
\end{lemma}
\begin{proof}
We will prove this in the case $S$ is a left multiplicative system.
The case where $S$ is a right multiplicative system is dual.
Suppose that $X, Y$ are objects of $\mathcal{C}$ and that
$\alpha, \beta : X \to Y$ are morphisms in $S^{-1}\mathcal{C}$. According to
Categories, Lemma \ref{categories-lemma-morphisms-left-localization}
we may represent these by pairs $s^{-1}f, s^{-1}g$ with common denominator
$s$. In this case we define $\alpha + \beta$ to be the equivalence class of
$s^{-1}(f + g)$. In the rest of the proof we show that this is well defined
and that composition is bilinear. Once this is done it is clear that
$Q$ is an additive functor.
\medskip\noindent
Let us show construction above is well defined.
An abstract way of saying this is that filtered colimits of
abelian groups agree with filtered colimits of sets and to use
Categories,
Equation (\ref{categories-equation-left-localization-morphisms-colimit}).
We can work this out in a bit more detail as follows.
Say $s : Y \to Y_1$ and $f, g : X \to Y_1$. Suppose we have a second
representation of $\alpha, \beta$ as $(s')^{-1}f', (s')^{-1}g'$ with
$s' : Y \to Y_2$ and $f', g' : X \to Y_2$. By
Categories, Remark \ref{categories-remark-left-localization-morphisms-colimit}
we can find a morphism $s_3 : Y \to Y_3$ and morphisms
$a_1 : Y_1 \to Y_3$, $a_2 : Y_2 \to Y_3$ such that
$a_1 \circ s = s_3 = a_2 \circ s'$ and also
$a_1 \circ f = a_2 \circ f'$ and $a_1 \circ g = a_2 \circ g'$.
Hence we see that $s^{-1}(f + g)$ is equivalent to
\begin{align*}
s_3^{-1}(a_1 \circ (f + g)) & =
s_3^{-1}(a_1 \circ f + a_1 \circ g) \\
& = s_3^{-1}(a_2 \circ f' + a_2 \circ g') \\
& = s_3^{-1}(a_2 \circ (f' + g'))
\end{align*}
which is equivalent to $(s')^{-1}(f' + g')$.
\medskip\noindent
Fix $s : Y \to Y'$ and $f, g : X \to Y'$ with
$\alpha = s^{-1}f$ and $\beta = s^{-1}g$ as morphisms $X \to Y$
in $S^{-1}\mathcal{C}$.
To show that composition is bilinear first consider the case of a
morphism $\gamma : Y \to Z$ in $S^{-1}\mathcal{C}$. Say $\gamma = t^{-1}h$
for some $h : Y \to Z'$ and $t : Z \to Z'$ in $S$. Using LMS2 we
choose morphisms $a : Y' \to Z''$ and $t' : Z' \to Z''$ in $S$ such
that $a \circ s = t' \circ h$. Picture
$$
\xymatrix{
& & Z \ar[d]^t \\
& Y \ar[r]^h \ar[d]^s & Z' \ar[d]^{t'} \\
X \ar[r]^{f, g} & Y' \ar[r]^a & Z''
}
$$
Then
$\gamma \circ \alpha = (t' \circ t)^{-1}(a \circ f)$ and
$\gamma \circ \beta = (t' \circ t)^{-1}(a \circ g)$.
Hence we see that $\gamma \circ (\alpha + \beta)$ is represented
by $(t' \circ t)^{-1}(a \circ (f + g)) =
(t' \circ t)^{-1}(a \circ f + a \circ g)$ which represents
$\gamma \circ \alpha + \gamma \circ \beta$.
\medskip\noindent
Finally, assume that $\delta : W \to X$ is another morphism of
$S^{-1}\mathcal{C}$. Say $\delta = r^{-1}i$ for some
$i : W \to X'$ and $r : X \to X'$ in $S$. We claim that we can find
a morphism $s : Y' \to Y''$ in $S$ and morphisms $a'', b'' : X' \to Y''$
such that the following diagram commutes
$$
\xymatrix{
& & & Y \ar[d]^s \\
& X \ar[rr]^{f, g, f + g} \ar[d]^s & & Y' \ar[d]^{s'} \\
W \ar[r]^i & X' \ar[rr]^{a'', b'', a'' + b''} & & Y''
}
$$
Namely, using LMS2 we can first choose
$s_1 : Y' \to Y_1$, $s_2 : Y' \to Y_2$ in $S$ and
$a : X' \to Y_1$, $b : X' \to Y_2$ such that
$a \circ s = s_1 \circ f$ and $b \circ s = s_2 \circ f$.
Then using that the category $Y'/S$ is filtered (see
Categories, Remark \ref{categories-remark-left-localization-morphisms-colimit}),
we can
find a $s' : Y' \to Y''$ and morphisms $a' : Y_1 \to Y''$, $b' : Y_2 \to Y''$
such that $s' = a' \circ s_1$ and $s' = b' \circ s_2$. Setting
$a'' = a' \circ a$ and $b'' = b' \circ b$ works.
At this point we see that the compositions
$\alpha \circ \delta$ and $\beta \circ \delta$ are represented by
$(s' \circ s)^{-1}a''$ and $(s' \circ s)^{-1}b''$.
Hence $\alpha \circ \delta + \beta \circ \delta$ is represented
by $(s' \circ s)^{-1}(a'' + b'')$ which by the diagram again
is a representative of $(\alpha + \beta) \circ \delta$.
\end{proof}
\begin{lemma}
\label{lemma-localization-additive}
Let $\mathcal{C}$ be an additive category.
Let $S$ be a left or right multiplicative system.
Then $S^{-1}\mathcal{C}$ is an additive category and the localization functor
$Q : \mathcal{C} \to S^{-1}\mathcal{C}$ is additive.
\end{lemma}
\begin{proof}
By
Lemma \ref{lemma-localization-preadditive}
we see that $S^{-1}\mathcal{C}$ is preadditive and that $Q$ is additive.
Recall that the functor $Q$ commutes with finite colimits
(resp.\ finite limits), see
Categories, Lemmas \ref{categories-lemma-left-localization-limits} and
\ref{categories-lemma-right-localization-limits}.
We conclude that $S^{-1}\mathcal{C}$ has a zero object and
direct sums, see
Lemmas \ref{lemma-preadditive-zero} and
\ref{lemma-preadditive-direct-sum}.
\end{proof}
\noindent
The following lemma describes the kernel (see
Definition \ref{definition-kernel-category})
of the localization functor in case we invert a multiplicative system.
\begin{lemma}
\label{lemma-kernel-localization}
Let $\mathcal{C}$ be an additive category. Let $S$ be a multiplicative
system. Let $X$ be an object
of $\mathcal{C}$. The following are equivalent
\begin{enumerate}
\item $Q(X) = 0$ in $S^{-1}\mathcal{C}$,
\item there exists $Y \in \Ob(\mathcal{C})$ such that
$0 : X \to Y$ is an element of $S$, and
\item there exists $Z \in \Ob(\mathcal{C})$ such that
$0 : Z \to X$ is an element of $S$.
\end{enumerate}
\end{lemma}
\begin{proof}
If (2) holds we see that $0 = Q(0) : Q(X) \to Q(Y)$ is an isomorphism.
In the additive category $S^{-1}\mathcal{C}$ this implies that $Q(X) = 0$.
Hence (2) $\Rightarrow$ (1). Similarly, (3) $\Rightarrow$ (1).
Suppose that $Q(X) = 0$. This implies that the morphism
$f : 0 \to X$ is transformed into an isomorphism in $S^{-1}\mathcal{C}$.
Hence by
Categories, Lemma \ref{categories-lemma-what-gets-inverted}
there exists a morphism $g : Z \to 0$ such that $fg \in S$. This proves
(1) $\Rightarrow$ (3). Similarly, (1) $\Rightarrow$ (2).
\end{proof}
\begin{lemma}
\label{lemma-localization-abelian}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item If $S$ is a left multiplicative system, then
the category $S^{-1}\mathcal{A}$ has cokernels and the functor
$Q : \mathcal{A} \to S^{-1}\mathcal{A}$ commutes with them.
\item If $S$ is a right multiplicative system, then
the category $S^{-1}\mathcal{A}$ has kernels and the functor
$Q : \mathcal{A} \to S^{-1}\mathcal{A}$ commutes with them.
\item If $S$ is a multiplicative system, then the category
$S^{-1}\mathcal{A}$ is abelian and the functor
$Q : \mathcal{A} \to S^{-1}\mathcal{A}$ is exact.
\end{enumerate}
\end{lemma}
\begin{proof}
Assume $S$ is a left multiplicative system. Let $a : X \to Y$ be a morphism
of $S^{-1}\mathcal{A}$. Then $a = s^{-1}f$ for some $s : Y \to Y'$
in $S$ and $f : X \to Y'$. Since $Q(s)$ is an isomorphism we see that
the existence of $\Coker(a : X \to Y)$ is equivalent to the existence
of $\Coker(Q(f) : X \to Y')$. Since $\Coker(Q(f))$ is the
coequalizer of $0$ and $Q(f)$ we see that $\Coker(Q(f))$ is
represented by $Q(\Coker(f))$ by
Categories, Lemma \ref{categories-lemma-left-localization-limits}.
This proves (1).
\medskip\noindent
Part (2) is dual to part (1).
\medskip\noindent
If $S$ is a multiplicative system, then $S$ is both a left and a right
multiplicative system. Thus we see that $S^{-1}\mathcal{A}$ has
kernels and cokernels and $Q$ commutes with kernels and cokernels.
To finish the proof of (3) we have to show that $\Coim = \Im$ in
$S^{-1}\mathcal{A}$. Again using that any arrow in $S^{-1}\mathcal{A}$
is isomorphic to an arrow $Q(f)$ we see that the result follows
from the result for $\mathcal{A}$.
\end{proof}
\section{Serre subcategories}
\label{section-serre-subcategories}
\noindent
In \cite[Chapter I, Section 1]{Serre_homotopie_classes}
a notion of a ``class'' of abelian groups is defined.
This notion has been extended to abelian categories by many authors
(in slightly different ways). We will use the following variant
which is virtually identical to Serre's original definition.
\begin{definition}
\label{definition-serre-subcategory}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item A {\it Serre subcategory} of $\mathcal{A}$ is a
nonempty full subcategory $\mathcal{C}$ of $\mathcal{A}$
such that given an exact sequence
$$
A \to B \to C
$$
with $A, C \in \Ob(\mathcal{C})$, then also
$B \in \Ob(\mathcal{C})$.
\item A {\it weak Serre subcategory} of $\mathcal{A}$ is a nonempty
full subcategory $\mathcal{C}$ of $\mathcal{A}$ such that given an
exact sequence
$$
A_0 \to A_1 \to A_2 \to A_3 \to A_4
$$
with $A_0, A_1, A_3, A_4$ in $\mathcal{C}$, then also $A_2$ in $\mathcal{C}$.
\end{enumerate}
\end{definition}
\noindent
In some references the second notion is called a ``thick'' subcategory
and in other references the first notion is called a ``thick'' subcategory.
However, it seems that the notion of a Serre subcategory is universally
accepted to be the one defined above. Note that in both cases the category
$\mathcal{C}$ is abelian and that the inclusion functor
$\mathcal{C} \to \mathcal{A}$ is a fully faithful exact functor.
Let's characterize these types of subcategories in more detail.
\begin{lemma}
\label{lemma-characterize-serre-subcategory}
Let $\mathcal{A}$ be an abelian category.
Let $\mathcal{C}$ be a subcategory of $\mathcal{A}$.
Then $\mathcal{C}$ is a Serre subcategory if and only if
the following conditions are satisfied:
\begin{enumerate}
\item $0 \in \Ob(\mathcal{C})$,
\item $\mathcal{C}$ is a strictly full subcategory of $\mathcal{A}$,
\item any subobject or quotient of an object of $\mathcal{C}$ is an object
of $\mathcal{C}$,
\item if $A \in \Ob(\mathcal{A})$ is an extension of objects of $\mathcal{C}$
then also $A \in \Ob(\mathcal{C})$.
\end{enumerate}
Moreover, a Serre subcategory is an abelian category and
the inclusion functor is exact.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-characterize-weak-serre-subcategory}
Let $\mathcal{A}$ be an abelian category.
Let $\mathcal{C}$ be a subcategory of $\mathcal{A}$.
Then $\mathcal{C}$ is a weak Serre subcategory if and only if
the following conditions are satisfied:
\begin{enumerate}
\item $0 \in \Ob(\mathcal{C})$,
\item $\mathcal{C}$ is a strictly full subcategory of $\mathcal{A}$,
\item kernels and cokernels in $\mathcal{A}$ of morphisms
between objects of $\mathcal{C}$ are in $\mathcal{C}$,
\item if $A \in \Ob(\mathcal{A})$ is an extension of objects of $\mathcal{C}$
then also $A \in \Ob(\mathcal{C})$.
\end{enumerate}
Moreover, a weak Serre subcategory is an abelian category and
the inclusion functor is exact.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-kernel-exact-functor}
Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor.
Then the full subcategory of objects $C$ of $\mathcal{A}$
such that $F(C) = 0$ forms a Serre subcategory of $\mathcal{A}$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-kernel-category}
Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor.
Then the full subcategory of objects $C$ of $\mathcal{A}$
such that $F(C) = 0$ is called the {\it kernel of the functor $F$},
and is sometimes denoted $\Ker(F)$.
\end{definition}
\noindent
Any Serre subcategory of an abelian category is the kernel of
an exact functor. In
Examples, Section \ref{examples-section-serre-quotient-modulo-torsion-modules}
we discuss this for Serre's original example of torsion groups.
\begin{lemma}
\label{lemma-serre-subcategory-is-kernel}
Let $\mathcal{A}$ be an abelian category.
Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory.
There exists an abelian category $\mathcal{A}/\mathcal{C}$
and an exact functor
$$
F : \mathcal{A} \longrightarrow \mathcal{A}/\mathcal{C}
$$
which is essentially surjective and whose kernel is $\mathcal{C}$.
The category $\mathcal{A}/\mathcal{C}$ and the functor $F$ are
characterized by the following universal property: For any exact
functor $G : \mathcal{A} \to \mathcal{B}$ such that
$\mathcal{C} \subset \Ker(G)$ there exists a factorization
$G = H \circ F$ for a unique exact functor
$H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$.
\end{lemma}
\begin{proof}
Consider the set of arrows of $\mathcal{A}$ defined by
the following formula
$$
S = \{f \in \text{Arrows}(\mathcal{A}) \mid
\Ker(f), \Coker(f) \in \Ob(\mathcal{C}) \}.
$$
We claim that $S$ is a multiplicative system. To prove this we have
to check MS1, MS2, MS3, see
Categories, Definition \ref{categories-definition-multiplicative-system}.
\medskip\noindent
It is clear that identities are elements of $S$. Suppose that
$f : A \to B$ and $g : B \to C$ are elements of $S$.
There are exact sequences
$$
\begin{matrix}
0 \to \Ker(f) \to \Ker(gf) \to \Ker(g) \\
\Coker(f) \to \Coker(gf) \to \Coker(g) \to 0
\end{matrix}
$$
Hence it follows that $gf \in S$. This proves MS1. (In fact, a similar
argument will show that $S$ is a saturated multiplicative system, see
Categories, Definition
\ref{categories-definition-saturated-multiplicative-system}.)
\medskip\noindent
Consider a solid diagram
$$
\xymatrix{
A \ar[d]_t \ar[r]_g & B \ar@{..>}[d]^s \\
C \ar@{..>}[r]^f & C \amalg_A B
}
$$
with $t \in S$. Set
$W = C \amalg_A B = \Coker((t, -g) : A \to C \oplus B)$.
Then $\Ker(t) \to \Ker(s)$ is surjective and
$\Coker(t) \to \Coker(s)$ is an isomorphism. Hence
$s$ is an element of $S$. This proves LMS2 and the proof of RMS2 is dual.
\medskip\noindent
Finally, consider morphisms $f, g : B \to C$ and a morphism $s : A \to B$
in $S$ such that $f \circ s = g \circ s$. This means that
$(f - g) \circ s = 0$. In turn this means that
$I = \Im(f - g) \subset C$ is a quotient of $\Coker(s)$
hence an object of $\mathcal{C}$. Thus $t : C \to C' = C/I$ is an
element of $S$ such that $t \circ (f - g) = 0$, i.e., such that
$t \circ f = t \circ g$. This proves LMS3 and the proof of
RMS3 is dual.
\medskip\noindent
Having proved that $S$ is a multiplicative system we set
$\mathcal{A}/\mathcal{C} = S^{-1}\mathcal{A}$, and we set
$F$ equal to the localization functor $Q$. By
Lemma \ref{lemma-localization-abelian}
the category $\mathcal{A}/\mathcal{C}$ is abelian and $F$ is exact.
If $X$ is in the kernel of $F = Q$, then by
Lemma \ref{lemma-kernel-localization}
we see that $0 : X \to Z$ is an element of $S$ and hence
$X$ is an object of $\mathcal{C}$, i.e., the kernel of
$F$ is $\mathcal{C}$.
Finally, if $G$ is as in the statement of the lemma, then $G$ turns
every element of $S$ into an isomorphism. Hence we obtain the
functor $H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ from
the universal property of localization, see
Categories, Lemma \ref{categories-lemma-properties-left-localization}.
\end{proof}
\begin{lemma}
\label{lemma-quotient-by-kernel-exact-functor}
Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor.
Then $\mathcal{C} = \Ker(F)$ if and only if the induced functor
$\overline{F} : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ is
faithful.
\end{lemma}
\begin{proof}
The ``only if'' direction is true because the kernel of $\overline{F}$ is zero
by construction. Namely, if $f : X \to Y$ is a morphism in
$\mathcal{A}/\mathcal{C}$ such that $\overline{F}(f) = 0$, then
$\overline{F}(\Im(f)) = \Im(\overline{F}(f)) = 0$, hence $\Im(f) = 0$ by the
assumption on the kernel of $F$. Thus $f = 0$.
\medskip\noindent
For the ``if'' direction, let $X$ be an object of $\mathcal{A}$ such that $F(X)
= 0$. Then $\overline{F}(\text{id}_X) = \text{id}_{\overline{F}(X)} = 0$, thus
$\text{id}_X = 0$ in $\mathcal{A}/\mathcal{C}$ by faithfulness of
$\overline{F}$. Hence $X = 0$ in $\mathcal{A}/\mathcal{C}$, that is $X \in
\Ob(\mathcal{C})$.
\end{proof}
\section{K-groups}
\label{section-K-groups}
\begin{definition}
\label{definition-K-zero}
Let $\mathcal{A}$ be an abelian category.
We denote $K_0(\mathcal{A})$ the
{\it zeroth $K$-group of $\mathcal{A}$}.
It is the abelian group constructed as follows.
Take the free abelian group
on the objects on $\mathcal{A}$
and for every short exact sequence
$0 \to A \to B \to C \to 0$
impose the relation $[B] - [A] - [C] = 0$.
\end{definition}
\noindent
Another way to say this is that there is a presentation
$$
\bigoplus_{A \to B \to C\text{ ses}}
\mathbf{Z}[A \to B \to C]
\longrightarrow
\bigoplus_{A \in \Ob(\mathcal{A})}
\mathbf{Z}[A]
\longrightarrow
K_0(\mathcal{A})
\longrightarrow
0
$$
with $[A \to B \to C] \mapsto [B] - [A] - [C]$ of $K_0(\mathcal{A})$.
The short exact sequence $0 \to 0 \to 0 \to 0 \to 0$
leads to the relation $[0] = 0$ in $K_0(\mathcal{A})$.
There are no set-theoretical issues as all of our categories
are ``small'' if not mentioned otherwise.
Some examples of $K$-groups for categories of modules
over rings where computed in
Algebra, Section \ref{algebra-section-K-groups}.
\begin{lemma}
\label{lemma-exact-functor-K-groups}
Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor between
abelian categories. Then $F$ induces a homomorphism of $K$-groups
$K_0(F) : K_0(\mathcal{A}) \to K_0(\mathcal{B})$ by simply setting
$K_0(F)([A]) = [F(A)]$.
\end{lemma}
\begin{proof}
Proves itself.
\end{proof}
\noindent
Suppose we are given an object $M$ of an abelian category $\mathcal{A}$
and a complex of the form
\begin{equation}
\label{equation-cyclic-complex}
\xymatrix{
\ldots \ar[r] &
M \ar[r]^\varphi &
M \ar[r]^\psi &
M \ar[r]^\varphi &
M \ar[r] & \ldots
}
\end{equation}
In this situation we define
$$
H^0(M, \varphi, \psi) = \Ker(\psi)/\Im(\varphi)
, \quad\text{and}\quad
H^1(M, \varphi, \psi) = \Ker(\varphi)/\Im(\psi).
$$
\begin{lemma}
\label{lemma-serre-subcategory-K-groups}
Let $\mathcal{A}$ be an abelian category.
Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory and
set $\mathcal{B} = \mathcal{A}/\mathcal{C}$.
\begin{enumerate}
\item The exact functors $\mathcal{C} \to \mathcal{A}$ and
$\mathcal{A} \to \mathcal{B}$ induce an exact sequence
$$
K_0(\mathcal{C}) \to
K_0(\mathcal{A}) \to
K_0(\mathcal{B}) \to
0
$$
of $K$-groups, and
\item the kernel of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$ is equal
to the collection of elements of the form
$$
[H^0(M, \varphi, \psi)] - [H^1(M, \varphi, \psi)]
$$
where $(M, \varphi, \psi)$ is a complex as in (\ref{equation-cyclic-complex})
with the property that it becomes exact in $\mathcal{B}$; in other words
that $H^0(M, \varphi, \psi)$ and $H^1(M, \varphi, \psi)$ are
objects of $\mathcal{C}$.
\end{enumerate}
\end{lemma}
\begin{proof}
We omit the proof of (1). The proof of (2) is in a sense completely
combinatorial. First we remark that any class of the type
$[H^0(M, \varphi, \psi)] - [H^1(M, \varphi, \psi)]$ is zero
in $K_0(\mathcal{A})$ by the following calculation
\begin{align*}
0 & = [M] - [M] \\
& = [\Ker(\varphi)] + [\Im(\varphi)]
- [\Ker(\psi)] - [\Im(\psi)] \\
& =
[\Ker(\varphi)/\Im(\psi)] -
[\Ker(\psi)/\Im(\varphi)] \\
& = [H^1(M, \varphi, \psi)] - [H^0(M, \varphi, \psi)]
\end{align*}
as desired. Hence it suffices to show that any element in the kernel
of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$ is of this form.
\medskip\noindent
Any element $x$ in $K_0(\mathcal{C})$ can be represented as the
difference $x = [P] - [Q]$ of two objects of $\mathcal{C}$ (fun exercise).
Suppose that this element maps to zero in $K_0(\mathcal{A})$.
This means that there exist
\begin{enumerate}
\item a finite set $I = I^{+} \amalg I^{-}$,
\item for each $i \in I$ a short exact sequence
$$
0 \to A_i \to B_i \to C_i \to 0
$$
in the abelian category $\mathcal{A}$
\end{enumerate}
such that
$$
[P] - [Q] =
\sum\nolimits_{i \in I^{+}} ([B_i] - [A_i] - [C_i])
-
\sum\nolimits_{i \in I^{-}} ([B_i] - [A_i] - [C_i])
$$
in the free abelian group on the objects of $\mathcal{A}$.
We can rewrite this as
$$
[P]
+ \sum\nolimits_{i \in I^{+}} ([A_i] + [C_i])
+ \sum\nolimits_{i \in I^{-}} [B_i]
=
[Q]
+ \sum\nolimits_{i \in I^{-}} ([A_i] + [C_i])
+ \sum\nolimits_{i \in I^{+}} [B_i].
$$
Since the right and left hand side should contain the same objects
of $\mathcal{A}$ counted with multiplicity, this means there should be
a bijection $\tau$ between the terms which occur above. Set
$$
T^{+} =
\{p\}\ \amalg\ \{a, c\} \times I^{+}\ \amalg\ \{b\} \times I^{-}
$$
and
$$
T^{-} =
\{q\}\ \amalg\ \{a, c\} \times I^{-}\ \amalg\ \{b\} \times I^{+}.
$$
Set $T = T^{+} \amalg T^{-} = \{p, q\} \amalg \{a, b, c\} \times I$.
For $t \in T$ define
$$
O(t)
=
\left\{
\begin{matrix}
P & \text{if} & t = p \\
Q & \text{if} & t = q \\
A_i & \text{if} & t = (a, i) \\
B_i & \text{if} & t = (b, i) \\
C_i & \text{if} & t = (c, i)
\end{matrix}
\right.
$$
Hence we can view $\tau : T^{+} \to T^{-}$ as a bijection
such that $O(t) = O(\tau(t))$ for all $t \in T^{+}$.
Let $t^{-}_0 = \tau(p)$ and let $t^{+}_0 \in T^{+}$ be the
unique element such that $\tau(t^{+}_0) = q$.
Consider the object
$$
M^{+} = \bigoplus\nolimits_{t \in T^{+}} O(t)
$$
By using $\tau$ we see that it is equal to the object
$$
M^{-} = \bigoplus\nolimits_{t \in T^{-}} O(t)
$$
Consider the map
$$
\varphi : M^{+} \longrightarrow M^{-}
$$
which on the summand $O(t) = A_i$ corresponding to $t = (a, i)$, $i \in I^{+}$
uses the map $A_i \to B_i$ into the summand $O((b, i)) = B_i$ of $M^{-}$
and on the summand $O(t) = B_i$ corresponding to $(b, i)$, $i \in I^{-}$
uses the map $B_i \to C_i$ into the summand $O((c, i)) = C_i$ of $M^{-}$.
The map is zero on the summands corresponding to $p$
and $(c, i)$, $i \in I^{+}$.
Similarly, consider the map
$$
\psi : M^{-} \longrightarrow M^{+}
$$
which on the summand $O(t) = A_i$ corresponding to $t = (a, i)$, $i \in I^{-}$
uses the map $A_i \to B_i$ into the summand $O((b, i)) = B_i$ of $M^{+}$
and on the summand $O(t) = B_i$ corresponding to $(b, i)$, $i \in I^{+}$
uses the map $B_i \to C_i$ into the summand $O((c, i)) = C_i$ of $M^{+}$.
The map is zero on the summands corresponding to $q$ and
$(c, i)$, $i \in I^{-}$.
\medskip\noindent
Note that the kernel of $\varphi$ is equal to the direct sum of the
summand $P$ and the summands $O((c, i)) = C_i$, $i \in I^{+}$ and
the subobjects $A_i$ inside the summands $O((b, i)) = B_i$, $i \in I^{-}$.
The image of $\psi$ is equal to the direct sum of the
summands $O((c, i)) = C_i$, $i \in I^{+}$ and
the subobjects $A_i$ inside the summands $O((b, i)) = B_i$, $i \in I^{-}$.
In other words we see that
$$
P \cong \Ker(\varphi)/\Im(\psi).
$$
In exactly the same way we see that
$$
Q \cong \Ker(\psi)/\Im(\varphi).
$$
Since as we remarked above the existence of the bijection
$\tau$ shows that $M^{+} = M^{-}$ we see that the lemma follows.
\end{proof}
\section{Cohomological delta-functors}
\label{section-cohomological-delta-functor}
\begin{definition}
\label{definition-cohomological-delta-functor}
Let $\mathcal{A}, \mathcal{B}$ be abelian categories.
A {\it cohomological $\delta$-functor} or simply a
{\it $\delta$-functor} from $\mathcal{A}$
to $\mathcal{B}$ is given by the following data:
\begin{enumerate}
\item a collection $F^n : \mathcal{A} \to \mathcal{B}$, $n \geq 0$ of additive
functors, and
\item for every short exact sequence $0 \to A \to B \to C \to 0$
of $\mathcal{A}$
a collection $\delta_{A \to B \to C} : F^n(C) \to F^{n + 1}(A)$, $n \geq 0$
of morphisms of $\mathcal{B}$.
\end{enumerate}
These data are assumed to satisfy the following axioms
\begin{enumerate}
\item for every short exact sequence as above the sequence
$$
\xymatrix{
0 \ar[r] &
F^0(A) \ar[r] &
F^0(B) \ar[r] &
F^0(C) \ar[lld]^{\delta_{A \to B \to C}} \\
&
F^1(A) \ar[r] &
F^1(B) \ar[r] &
F^1(C) \ar[lld]^{\delta_{A \to B \to C}} \\
&
F^2(A) \ar[r] &
F^2(B) \ar[r] &
\ldots
}
$$
is exact, and
\item for every morphism $(A \to B \to C) \to (A' \to B' \to C')$
of short exact sequences of $\mathcal{A}$ the diagrams
$$
\xymatrix{
F^n(C) \ar[d] \ar[rr]_{\delta_{A \to B \to C}} & & F^{n + 1}(A) \ar[d] \\
F^n(C') \ar[rr]^{\delta_{A' \to B' \to C'}} & & F^{n + 1}(A')
}
$$
are commutative.
\end{enumerate}
\end{definition}
\noindent
Note that this in particular implies that $F^0$ is left exact.
\begin{definition}
\label{definition-morphism-delta-functors}
Let $\mathcal{A}, \mathcal{B}$ be abelian categories.
Let $(F^n, \delta_F)$ and $(G^n, \delta_G)$ be $\delta$-functors
from $\mathcal{A}$ to $\mathcal{B}$. A {\it morphism of $\delta$-functors
from $F$ to $G$} is a collection of
transformation of functors $t^n : F^n \to G^n$, $n \geq 0$ such
that for every short exact sequence $0 \to A \to B \to C \to 0$
of $\mathcal{A}$ the diagrams
$$
\xymatrix{
F^n(C) \ar[d]_{t^n} \ar[rr]_{\delta_{F, A \to B \to C}} &
& F^{n + 1}(A) \ar[d]^{t^{n + 1}} \\
G^n(C) \ar[rr]^{\delta_{G, A \to B \to C}} & & G^{n + 1}(A)
}
$$
are commutative.
\end{definition}
\begin{definition}
\label{definition-universal-delta-functor}
Let $\mathcal{A}, \mathcal{B}$ be abelian categories.
Let $F = (F^n, \delta_F)$ be a $\delta$-functor
from $\mathcal{A}$ to $\mathcal{B}$.
We say $F$ is a {\it universal $\delta$-functor} if an only
if for every $\delta$-functor $G = (G^n, \delta_G)$ and any
morphism of functors $t : F^0 \to G^0$ there exists
a unique morphism of $\delta$-functors $\{t^n\}_{n \geq 0} : F \to G$
such that $t = t^0$.
\end{definition}
\begin{lemma}
\label{lemma-efface-implies-universal}
Let $\mathcal{A}, \mathcal{B}$ be abelian categories.
Let $F = (F^n, \delta_F)$ be a $\delta$-functor
from $\mathcal{A}$ to $\mathcal{B}$.
Suppose that for every $n > 0$ and any $A \in \Ob(\mathcal{A})$
there exists an injective morphism $u : A \to B$ (depending on $A$ and $n$)
such that $F^n(u) : F^n(A) \to F^n(B)$ is zero. Then $F$ is a universal
$\delta$-functor.
\end{lemma}
\begin{proof}
Let $G = (G^n, \delta_G)$ be a $\delta$-functor
from $\mathcal{A}$ to $\mathcal{B}$ and let $t : F^0 \to G^0$
be a morphism of functors. We have to show there exists
a unique morphism of $\delta$-functors $\{t^n\}_{n \geq 0} : F \to G$
such that $t = t^0$. We construct $t^n$ by induction on $n$.
For $n = 0$ we set $t^0 = t$.
Suppose we have already constructed a unique sequence of
transformation of functors $t^i$ for $i \leq n$ compatible with
the maps $\delta$ in degrees $\leq n$.
\medskip\noindent
Let $A \in \Ob(\mathcal{A})$. By assumption we may choose
a embedding $u : A \to B$ such that $F^{n + 1}(u) = 0$.
Let $C = B/u(A)$. The long exact cohomology sequence for
the short exact sequence $0 \to A \to B \to C \to 0$ and the
$\delta$-functor $F$ gives that
$F^{n + 1}(A) = \Coker(F^n(B) \to F^n(C))$ by our choice of $u$.
Since we have already defined $t^n$ we can set
$$
t^{n + 1}_A : F^{n + 1}(A) \to G^{n + 1}(A)
$$
equal to the unique map such that
$$
\xymatrix{
\Coker(F^n(B) \to F^n(C)) \ar[r]_{t^n}
\ar[d]_{\delta_{F, A \to B \to C}} &
\Coker(G^n(B) \to G^n(C))
\ar[d]^{\delta_{G, A \to B \to C}} \\
F^{n + 1}(A) \ar[r]^{t^{n + 1}_A} &
G^{n + 1}(A)
}
$$
commutes. This is clearly uniquely determined by the requirements
imposed. We omit the verification that this defines a transformation
of functors.
\end{proof}
\begin{lemma}
\label{lemma-uniqueness-universal-delta-functor}
Let $\mathcal{A}, \mathcal{B}$ be abelian categories.
Let $F : \mathcal{A} \to \mathcal{B}$ be a functor.
If there exists a universal $\delta$-functor
$(F^n, \delta_F)$ from $\mathcal{A}$ to $\mathcal{B}$
with $F^0 = F$, then it is determined up to unique isomorphism
of $\delta$-functors.
\end{lemma}
\begin{proof}
Immediate from the definitions.
\end{proof}
\section{Complexes}
\label{section-complexes}
\noindent
Of course the notions of a chain complex and a cochain complex
are dual and you only have to read one of the two parts of
this section. So pick the one you like. (Actually, this doesn't
quite work right since the conventions on numbering things
are not adapted to an easy transition between chain and cochain
complexes.)
\medskip\noindent
A {\it chain complex $A_\bullet$} in an additive category $\mathcal{A}$
is a complex
$$
\ldots \to
A_{n + 1} \xrightarrow{d_{n + 1}}
A_n \xrightarrow{d_n}
A_{n - 1} \to
\ldots
$$
of $\mathcal{A}$. In other words, we are given an object $A_i$ of
$\mathcal{A}$ for all $i \in \mathbf{Z}$ and for
all $i \in \mathbf{Z}$ a morphism $d_i : A_i \to A_{i - 1}$ such that
$d_{i - 1} \circ d_i = 0$ for all $i$. A {\it morphism of chain
complexes $f : A_\bullet \to B_\bullet$} is given by a
family of morphisms $f_i : A_i \to B_i$ such that all
the diagrams
$$
\xymatrix{
A_i \ar[r]_{d_i} \ar[d]_{f_i} & A_{i - 1} \ar[d]^{f_{i - 1}} \\
B_i \ar[r]^{d_i} & B_{i - 1}
}
$$
commute. The {\it category of chain complexes of $\mathcal{A}$}
is denoted $\text{Ch}(\mathcal{A})$. The full subcategory consisting
of objects of the form
$$
\ldots \to A_2 \to A_1 \to A_0 \to 0 \to 0 \to \ldots
$$
is denoted $\text{Ch}_{\geq 0}(\mathcal{A})$.
In other words, a chain complex $A_\bullet$ belongs to
$\text{Ch}_{\geq 0}(\mathcal{A})$ if and only if
$A_i = 0$ for all $i < 0$.
A {\it homotopy $h$} between a pair of morphisms
of chain complexes $f, g : A_\bullet \to B_\bullet$ is
is a collection of morphisms $h_i : A_i \to B_{i + 1}$
such that we have
$$
f_i - g_i = d_{i + 1} \circ h_i + h_{i - 1} \circ d_i
$$
for all $i$. Clearly, the notions of chain complex, morphism of
chain complexes, and homotopies between morphisms of chain complexes
makes sense even in a preadditive category.
\begin{lemma}
\label{lemma-compose-homotopy}
Let $\mathcal{A}$ be an additive category.
Let $f, g : B_\bullet \to C_\bullet$ be morphisms
of chain complexes. Suppose given morphisms of chain
complexes $a : A_\bullet \to B_\bullet$, and
$c : C_\bullet \to D_\bullet$.
If $\{h_i : B_i \to C_{i + 1}\}$ defines a homotopy
between $f$ and $g$, then $\{c_{i + 1} \circ h_i \circ a_i\}$
defines a homotopy between $c \circ f \circ a$ and
$c \circ g \circ a$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
In particular this means that it makes sense to define
the category of chain complexes with maps up to homotopy.
We'll return to this later.
\begin{definition}
\label{definition-homotopy-equivalent}
Let $\mathcal{A}$ be an additive category.
We say a morphism $a : A_\bullet \to B_\bullet$
is a {\it homotopy equivalence} if there exists
a morphism $b : B_\bullet \to A_\bullet$
such that there exists a homotopy between
$a \circ b$ and $\text{id}_A$
and there exists a homotopy between $b \circ a$ and $\text{id}_B$.
If there exists such a morphism between $A_\bullet$ and $B_\bullet$, then
we say that $A_\bullet$ and $B_\bullet$ are {\it homotopy equivalent}.
\end{definition}
\noindent
In other words, two complexes are homotopy equivalent if they become
isomorphic in the category of complexes up to homotopy.
\begin{lemma}
\label{lemma-cat-chain-abelian}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item The category of chain complexes in $\mathcal{A}$ is
abelian.
\item A morphism of complexes
$f : A_\bullet \to B_\bullet$ is injective
if and only if each $f_n : A_n \to B_n$ is injective.
\item A morphism of complexes
$f : A_\bullet \to B_\bullet$ is surjective
if and only if each $f_n : A_n \to B_n$ is surjective.
\item A sequence of chain complexes
$$
A_\bullet \xrightarrow{f} B_\bullet \xrightarrow{g} C_\bullet
$$
is exact at $B_\bullet$ if and only if each sequence
$$
A_i \xrightarrow{f_i} B_i \xrightarrow{g_i} C_i
$$
is exact at $B_i$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
For any $i \in \mathbf{Z}$ the $i$th {\it homology group}
of a chain complex $A_\bullet$ in an abelian category is defined by
the following formula
$$
H_i(A_\bullet) = \Ker(d_i)/\Im(d_{i + 1}).
$$
If $f : A_\bullet \to B_\bullet$ is a morphism of chain
complexes of $\mathcal{A}$ then we get an induced
morphism $H_i(f) : H_i(A_\bullet) \to H_i(B_\bullet)$
because clearly
$f_i(\Ker(d_i : A_i \to A_{i - 1})) \subset
\Ker(d_i : B_i \to B_{i - 1})$, and similarly
for $\Im(d_{i + 1})$.
Thus we obtain a functor
$$
H_i : \text{Ch}(\mathcal{A}) \longrightarrow \mathcal{A}.
$$
\begin{definition}
\label{definition-quasi-isomorphism}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item A morphism of chain complexes $f : A_\bullet \to B_\bullet$
is called a {\it quasi-isomorphism} if the induced
map $H_i(f) : H_i(A_\bullet) \to H_i(B_\bullet)$
is an isomorphism for all $i \in \mathbf{Z}$.
\item A chain complex $A_\bullet$ is called
{\it acyclic} if all of its homology objects
$H_i(A_\bullet)$ are zero.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-map-homology-homotopy}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item If the maps $f, g : A_\bullet \to B_\bullet$ are
homotopic, then the induced maps $H_i(f)$ and $H_i(g)$
are equal.
\item If the map $f : A_\bullet \to B_\bullet$ is a homotopy
equivalence, then $f$ is a quasi-isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-long-exact-sequence-chain}
Let $\mathcal{A}$ be an abelian category.
Suppose that
$$
0 \to
A_\bullet \to
B_\bullet \to
C_\bullet \to
0
$$
is a short exact sequence of chain complexes of $\mathcal{A}$.
Then there is a canonical long exact homology sequence
$$
\xymatrix{
\ldots & \ldots & \ldots \ar[lld] \\
H_i(A_\bullet) \ar[r] & H_i(B_\bullet) \ar[r] & H_i(C_\bullet) \ar[lld] \\
H_{i - 1}(A_\bullet) \ar[r] &
H_{i - 1}(B_\bullet) \ar[r] &
H_{i - 1}(C_\bullet) \ar[lld] \\
\ldots & \ldots & \ldots \\
}
$$
\end{lemma}
\begin{proof}
Omitted. The maps come from the Snake Lemma \ref{lemma-snake}
applied to the diagrams
$$
\xymatrix{
&
A_i/\Im(d_{A, i + 1}) \ar[r] \ar[d]^{d_{A, i}} &
B_i/\Im(d_{B, i + 1}) \ar[r] \ar[d]^{d_{B, i}} &
C_i/\Im(d_{C, i + 1}) \ar[r] \ar[d]^{d_{C, i}} &
0 \\
0 \ar[r] &
\Ker(d_{A, i - 1}) \ar[r] &
\Ker(d_{B, i - 1}) \ar[r] &
\Ker(d_{C, i - 1}) &
}
$$
\end{proof}
\noindent
A {\it cochain complex $A^\bullet$} in an additive category $\mathcal{A}$
is a complex
$$
\ldots \to
A^{n - 1} \xrightarrow{d^{n - 1}}
A^n \xrightarrow{d^n}
A^{n + 1} \to
\ldots
$$
of $\mathcal{A}$. In other words, we are given an object $A^i$ of
$\mathcal{A}$ for all $i \in \mathbf{Z}$ and for
all $i \in \mathbf{Z}$ a morphism $d^i : A^i \to A^{i + 1}$ such that
$d^{i + 1} \circ d^i = 0$ for all $i$. A {\it morphism of cochain
complexes $f : A^\bullet \to B^\bullet$} is given by a
family of morphisms $f^i : A^i \to B^i$ such that all
the diagrams
$$
\xymatrix{
A^i \ar[r]_{d^i} \ar[d]_{f^i} & A^{i + 1} \ar[d]^{f^{i + 1}} \\
B^i \ar[r]^{d^i} & B^{i + 1}
}
$$
commute. The {\it category of cochain complexes of $\mathcal{A}$}
is denoted $\text{CoCh}(\mathcal{A})$. The full subcategory consisting
of objects of the form
$$
\ldots \to 0 \to 0 \to A^0 \to A^1 \to A^2 \to \ldots
$$
is denoted $\text{CoCh}_{\geq 0}(\mathcal{A})$.
In other words, a cochain complex $A^\bullet$ belongs to the subcategory
$\text{CoCh}_{\geq 0}(\mathcal{A})$ if and only if
$A^i = 0$ for all $i < 0$.
A {\it homotopy $h$} between a pair of morphisms
of cochain complexes $f, g : A^\bullet \to B^\bullet$ is
is a collection of morphisms $h^i : A^i \to B^{i - 1}$
such that we have
$$
f^i - g^i = d^{i - 1} \circ h^i + h^{i + 1} \circ d^i
$$
for all $i$. Clearly, the notions of cochain complex, morphism of
cochain complexes, and homotopies between morphisms of cochain complexes
makes sense even in a preadditive category.
\begin{lemma}
\label{lemma-compose-homotopy-cochain}
Let $\mathcal{A}$ be an additive category.
Let $f, g : B^\bullet \to C^\bullet$ be morphisms
of cochain complexes. Suppose given morphisms of cochain
complexes $a : A^\bullet \to B^\bullet$, and
$c : C^\bullet \to D^\bullet$.
If $\{h^i : B^i \to C^{i - 1}\}$ defines a homotopy
between $f$ and $g$, then $\{c^{i - 1} \circ h^i \circ a^i\}$
defines a homotopy between $c \circ f \circ a$ and
$c \circ g \circ a$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
In particular this means that it makes sense to define
the category of cochain complexes with maps up to homotopy.
We'll return to this later.
\begin{definition}
\label{definition-homotopy-equivalent-cochain}
Let $\mathcal{A}$ be an additive category.
We say a morphism $a : A^\bullet \to B^\bullet$
is a {\it homotopy equivalence} if there exists
a morphism $b : B^\bullet \to A^\bullet$
such that there exists a homotopy between
$a \circ b$ and $\text{id}_A$
and there exists a homotopy between $b \circ a$ and $\text{id}_B$.
If there exists such a morphism between $A^\bullet$ and $B^\bullet$, then
we say that $A^\bullet$ and $B^\bullet$ are {\it homotopy equivalent}.
\end{definition}
\noindent
In other words, two complexes are homotopy equivalent if they become
isomorphic in the category of complexes up to homotopy.
\begin{lemma}
\label{lemma-cat-cochain-abelian}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item The category of cochain complexes in $\mathcal{A}$ is
abelian.
\item A morphism of cochain complexes
$f : A^\bullet \to B^\bullet$ is injective
if and only if each $f^n : A^n \to B^n$ is injective.
\item A morphism of cochain complexes
$f : A^\bullet \to B^\bullet$ is surjective
if and only if each $f^n : A^n \to B^n$ is surjective.
\item A sequence of cochain complexes
$$
A^\bullet \xrightarrow{f} B^\bullet \xrightarrow{g} C^\bullet
$$
is exact at $B^\bullet$ if and only if each sequence
$$
A^i \xrightarrow{f^i} B^i \xrightarrow{g^i} C^i
$$
is exact at $B^i$.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\noindent
For any $i \in \mathbf{Z}$ the $i$th {\it cohomology group}
of a cochain complex $A^\bullet$ is defined by
the following formula
$$
H^i(A^\bullet) = \Ker(d^i)/\Im(d^{i - 1}).
$$
If $f : A^\bullet \to B^\bullet$ is a morphism of cochain
complexes of $\mathcal{A}$ then we get an induced
morphism $H^i(f) : H^i(A^\bullet) \to H^i(B^\bullet)$
because clearly
$f^i(\Ker(d^i : A^i \to A^{i + 1})) \subset
\Ker(d^i : B^i \to B^{i + 1})$, and similarly
for $\Im(d^{i - 1})$.
Thus we obtain a functor
$$
H^i : \text{CoCh}(\mathcal{A}) \longrightarrow \mathcal{A}.
$$
\begin{definition}
\label{definition-quasi-isomorphism-cochain}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item A morphism of cochain complexes $f : A^\bullet \to B^\bullet$
of $\mathcal{A}$ is called a {\it quasi-isomorphism} if the induced
maps $H^i(f) : H^i(A^\bullet) \to H^i(B^\bullet)$
is an isomorphism for all $i \in \mathbf{Z}$.
\item A cochain complex $A^\bullet$ is called
{\it acyclic} if all of its cohomology objects
$H^i(A^\bullet)$ are zero.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma-map-cohomology-homotopy-cochain}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item If the maps $f, g : A^\bullet \to B^\bullet$ are
homotopic, then the induced maps $H^i(f)$ and $H^i(g)$
are equal.
\item If $f : A^\bullet \to B^\bullet$ is a homotopy equivalence,
then $f$ is a quasi-isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-long-exact-sequence-cochain}
\begin{slogan}
Short exact sequences of complexes give rise to long exact sequences
of (co)homology.
\end{slogan}
Let $\mathcal{A}$ be an abelian category.
Suppose that
$$
0 \to
A^\bullet \to
B^\bullet \to
C^\bullet \to
0
$$
is a short exact sequence of chain complexes of $\mathcal{A}$.
Then there is a canonical long exact cohomology sequence
$$
\xymatrix{
\ldots & \ldots & \ldots \ar[lld] \\
H^i(A^\bullet) \ar[r] &
H^i(B^\bullet) \ar[r] &
H^i(C^\bullet) \ar[lld] \\
H^{i + 1}(A^\bullet) \ar[r] &
H^{i + 1}(B^\bullet) \ar[r] &
H^{i + 1}(C^\bullet) \ar[lld] \\
\ldots & \ldots & \ldots \\
}
$$
\end{lemma}
\begin{proof}
Omitted. The maps come from the Snake Lemma \ref{lemma-snake}
applied to the diagrams
$$
\xymatrix{
&
A^i/\Im(d_A^{i - 1}) \ar[r] \ar[d]^{d_A^i} &
B^i/\Im(d_B^{i - 1}) \ar[r] \ar[d]^{d_B^i} &
C^i/\Im(d_C^{i - 1}) \ar[r] \ar[d]^{d_C^i} &
0 \\
0 \ar[r] &
\Ker(d_A^{i + 1}) \ar[r] &
\Ker(d_B^{i + 1}) \ar[r] &
\Ker(d_C^{i + 1}) &
}
$$
\end{proof}
\section{Truncation of complexes}
\label{section-truncations}
\noindent
Let $\mathcal{A}$ be an abelian category.
Let $A_\bullet$ be a chain complex. There
are several ways to {\it truncate} the complex $A_\bullet$.
\begin{enumerate}
\item The {\it ``stupid'' truncation $\sigma_{\leq n}$}
is the subcomplex $\sigma_{\leq n} A_\bullet$ defined
by the rule $(\sigma_{\leq n} A_\bullet)_i = 0$ if
$i > n$ and $(\sigma_{\leq n} A_\bullet)_i = A_i$ if
$i \leq n$. In a picture
$$
\xymatrix{
\sigma_{\leq n}A_\bullet \ar[d] &
\ldots \ar[r] &
0 \ar[r] \ar[d] &
A_n \ar[r] \ar[d] &
A_{n - 1} \ar[r] \ar[d] &
\ldots \\
A_\bullet &
\ldots \ar[r] &
A_{n + 1} \ar[r] &
A_n \ar[r] &
A_{n - 1} \ar[r] &
\ldots
}
$$
Note the property
$\sigma_{\leq n}A_\bullet / \sigma_{\leq n - 1}A_\bullet = A_n[-n]$.
\item The {\it ``stupid'' truncation $\sigma_{\geq n}$}
is the quotient complex $\sigma_{\geq n} A_\bullet$ defined
by the rule $(\sigma_{\geq n} A_\bullet)_i = A_i$ if
$i \geq n$ and $(\sigma_{\geq n} A_\bullet)_i = 0$ if
$i < n$. In a picture
$$
\xymatrix{
A_\bullet \ar[d] &
\ldots \ar[r] &
A_{n + 1} \ar[r] \ar[d] &
A_n \ar[r] \ar[d] &
A_{n - 1} \ar[r] \ar[d] &
\ldots \\
\sigma_{\geq n}A_\bullet &
\ldots \ar[r] &
A_{n + 1} \ar[r] &
A_n \ar[r] &
0 \ar[r] &
\ldots
}
$$
The map of complexes
$\sigma_{\geq n}A_\bullet \to \sigma_{\geq n + 1}A_\bullet$ is surjective
with kernel $A_n[-n]$.
\item The {\it canonical truncation} $\tau_{\geq n}A_\bullet$
is defined by the picture
$$
\xymatrix{
\tau_{\geq n}A_\bullet \ar[d] &
\ldots \ar[r] &
A_{n + 1} \ar[r] \ar[d] &
\Ker(d_n) \ar[r] \ar[d] &
0 \ar[r] \ar[d] &
\ldots \\
A_\bullet &
\ldots \ar[r] &
A_{n + 1} \ar[r] &
A_n \ar[r] &
A_{n - 1} \ar[r] &
\ldots
}
$$
Note that these complexes have the property that
$$
H_i(\tau_{\geq n}A_\bullet) =
\left\{
\begin{matrix}
H_i(A_\bullet) & \text{if} & i \geq n \\
0 & \text{if} & i < n
\end{matrix}
\right.
$$
\item The {\it canonical truncation} $\tau_{\leq n}A_\bullet$
is defined by the picture
$$
\xymatrix{
A_\bullet \ar[d] &
\ldots \ar[r] &
A_{n + 1} \ar[r] \ar[d] &
A_n \ar[r] \ar[d] &
A_{n - 1} \ar[r] \ar[d] &
\ldots \\
\tau_{\leq n}A_\bullet &
\ldots \ar[r] &
0 \ar[r] &
\Coker(d_{n + 1}) \ar[r] &
A_{n - 1} \ar[r] &
\ldots
}
$$
Note that these complexes have the property that
$$
H_i(\tau_{\leq n}A_\bullet) =
\left\{
\begin{matrix}
H_i(A_\bullet) & \text{if} & i \leq n \\
0 & \text{if} & i > n
\end{matrix}
\right.
$$
\end{enumerate}
\noindent
Let $\mathcal{A}$ be an abelian category.
Let $A^\bullet$ be a cochain complex. There
are four ways to truncate the complex $A^\bullet$.
\begin{enumerate}
\item The {\it ``stupid'' truncation $\sigma_{\geq n}$} is the subcomplex
$\sigma_{\geq n} A^\bullet$ defined by the rule
$(\sigma_{\geq n} A^\bullet)^i = 0$ if
$i < n$ and $(\sigma_{\geq n} A^\bullet)^i = A_i$ if
$i \geq n$. In a picture
$$
\xymatrix{
\sigma_{\geq n}A^\bullet \ar[d] &
\ldots \ar[r] &
0 \ar[r] \ar[d] &
A^n \ar[r] \ar[d] &
A^{n + 1} \ar[r] \ar[d] &
\ldots \\
A^\bullet &
\ldots \ar[r] &
A^{n - 1} \ar[r] &
A^n \ar[r] &
A^{n + 1} \ar[r] &
\ldots
}
$$
Note the property
$\sigma_{\geq n}A^\bullet / \sigma_{\geq n + 1}A^\bullet
= A^n[-n]$.
\item The {\it ``stupid'' truncation $\sigma_{\leq n}$}
is the quotient complex $\sigma_{\leq n} A^\bullet$ defined
by the rule $(\sigma_{\leq n} A^\bullet)^i = 0$ if
$i > n$ and $(\sigma_{\leq n} A^\bullet)^i = A^i$ if
$i \leq n$. In a picture
$$
\xymatrix{
A^\bullet \ar[d] &
\ldots \ar[r] &
A^{n - 1} \ar[r] \ar[d] &
A^n \ar[r] \ar[d] &
A^{n + 1} \ar[r] \ar[d] &
\ldots \\
\sigma_{\leq n}A^\bullet &
\ldots \ar[r] &
A^{n - 1} \ar[r] &
A^n \ar[r] &
0 \ar[r] &
\ldots \\
}
$$
The map of complexes
$\sigma_{\leq n}A^\bullet \to \sigma_{\leq n - 1}A^\bullet$ is surjective
with kernel $A^n[-n]$.
\item The {\it canonical truncation} $\tau_{\leq n}A^\bullet$
is defined by the picture
$$
\xymatrix{
\tau_{\leq n}A^\bullet \ar[d] &
\ldots \ar[r] &
A^{n - 1} \ar[r] \ar[d] &
\Ker(d^n) \ar[r] \ar[d] &
0 \ar[r] \ar[d] &
\ldots \\
A^\bullet &
\ldots \ar[r] &
A^{n - 1} \ar[r] &
A^n \ar[r] &
A^{n + 1} \ar[r] &
\ldots
}
$$
Note that these complexes have the property that
$$
H^i(\tau_{\leq n}A^\bullet) =
\left\{
\begin{matrix}
H^i(A^\bullet) & \text{if} & i \leq n \\
0 & \text{if} & i > n
\end{matrix}
\right.
$$
\item The {\it canonical truncation} $\tau_{\geq n}A^\bullet$
is defined by the picture
$$
\xymatrix{
A^\bullet \ar[d] &
\ldots \ar[r] &
A^{n - 1} \ar[r] \ar[d] &
A^n \ar[r] \ar[d] &
A^{n + 1} \ar[r] \ar[d] &
\ldots \\
\tau_{\geq n}A^\bullet &
\ldots \ar[r] &
0 \ar[r] &
\Coker(d^{n - 1}) \ar[r] &
A^{n + 1} \ar[r] &
\ldots
}
$$
Note that these complexes have the property that
$$
H^i(\tau_{\geq n}A^\bullet) =
\left\{
\begin{matrix}
0 & \text{if} & i < n \\
H^i(A^\bullet) & \text{if} & i \geq n
\end{matrix}
\right.
$$
\end{enumerate}
\section{Homotopy and the shift functor}
\label{section-homotopy-shift}
\noindent
It is an annoying feature that signs and indices
have to be part of any discussion of homological
algebra\footnote{I am sure you think that my conventions
are wrong. If so and if you feel strongly about it
then drop me an email with an explanation.}.
\begin{definition}
\label{definition-shift}
Let $\mathcal{A}$ be an additive category.
Let $A_\bullet$ be a chain complex
with boundary maps $d_{A, n} : A_n \to A_{n - 1}$.
For any $k \in \mathbf{Z}$ we define the
{\it $k$-shifted chain complex $A[k]_\bullet$}
as follows:
\begin{enumerate}
\item we set $A[k]_n = A_{n + k}$, and
\item we set $d_{A[k], n} : A[k]_n \to A[k]_{n - 1}$
equal to $d_{A[k], n} = (-1)^k d_{A, n + k}$.
\end{enumerate}
If $f : A_\bullet \to B_\bullet$ is a morphism of
chain complexes, then we let
$f[k] : A[k]_\bullet \to B[k]_\bullet$ be the
morphism of chain complexes with
$f[k]_n = f_{k + n}$.
\end{definition}
\noindent
Of course this means we have functors
$[k] : \text{Ch}(\mathcal{A}) \to \text{Ch}(\mathcal{A})$
which mutually commute (on the nose, without
any intervening isomorphisms of functors),
such that $A[k][l]_\bullet = A[k + l]_\bullet$ and
with $[0] = \text{id}_{\text{Ch}(\mathcal{A})}$.
\begin{definition}
\label{definition-homology-shift}
Let $\mathcal{A}$ be an abelian category.
Let $A_\bullet$ be a chain complex
with boundary maps $d_{A, n} : A_n \to A_{n - 1}$.
For any $k \in \mathbf{Z}$ we identify
{\it $H_{i + k}(A_\bullet) \rightarrow H_i(A[k]_\bullet)$}
via the identification
$A_{i + k} = A[k]_i$.
\end{definition}
\noindent
This identification is functorial in $A_\bullet$.
Note that since no signs are involved in this
definition we actually get a compatible system
of identifications of all the homology
objects $H_{i - k}(A[k]_\bullet)$, which are
further compatible with the identifications
$A[k][l]_\bullet = A[k + l]_\bullet$ and
with $[0] = \text{id}_{\text{Ch}(\mathcal{A})}$.
\medskip\noindent
Let $\mathcal{A}$ be an additive category.
Suppose that $A_\bullet$ and $B_\bullet$ are
chain complexes, $a, b : A_\bullet \to B_\bullet$ are
morphisms of chain complexes, and $\{h_i : A_i \to B_{i + 1}\}$
is a homotopy between $a$ and $b$. Recall that this means
that
$a_i - b_i = d_{i + 1} \circ h_i + h_{i - 1} \circ d_i$.
What if $a = b$? Then we obtain the formula
$0 = d_{i + 1} \circ h_i + h_{i - 1} \circ d_i$,
in other words, $ - d_{i + 1} \circ h_i = h_{i - 1} \circ d_i$.
By definition above this means the collection $\{h_i\}$
above defines a morphism of chain complexes
$$
A_\bullet \longrightarrow B[1]_\bullet.
$$
Such a thing is the same as a morphism $A[-1]_\bullet \to B_\bullet$
by our remarks above. This proves the following lemma.
\begin{lemma}
\label{lemma-homotopy-shift}
Let $\mathcal{A}$ be an additive category.
Suppose that $A_\bullet$ and $B_\bullet$ are
chain complexes. Given any morphism of chain
complexes $a : A_\bullet \to B_\bullet$ there
is a bijection between the set of homotopies
from $a$ to $a$ and
$\Mor_{\text{Ch}(\mathcal{A})}(A_\bullet, B[1]_\bullet)$.
More generally, the set of homotopies between
$a$ and $b$ is either empty or a principal homogeneous
space under the group
$\Mor_{\text{Ch}(\mathcal{A})}(A_\bullet, B[1]_\bullet)$.
\end{lemma}
\begin{proof}
See above.
\end{proof}
\begin{lemma}
\label{lemma-ses-termwise-split}
Let $\mathcal{A}$ be an abelian category.
Let
$$
0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0
$$
be a sort exact sequence of complexes.
Suppose that $\{s_n : C_n \to B_n\}$ is a family
of morphisms which split the short exact sequences
$0 \to A_n \to B_n \to C_n \to 0$. Let
$\pi_n : B_n \to A_n$ be the associated
projections, see Lemma \ref{lemma-ses-split}.
Then the family of morphisms
$$
\pi_{n - 1} \circ d_{B, n} \circ s_n
:
C_n \to A_{n - 1}
$$
define a morphism of complexes $\delta(s) : C_\bullet \to A[-1]_\bullet$.
\end{lemma}
\begin{proof}
Denote $i : A_\bullet \to B_\bullet$ and $q : B_\bullet \to C_\bullet$
the maps of complexes in the short exact sequence. Then
$i_{n - 1} \circ \pi_{n - 1} \circ d_{B, n} \circ s_n =
d_{B, n} \circ s_n - s_{n - 1} \circ d_{C, n}$. Hence
$i_{n - 2} \circ d_{A, n - 1} \circ \pi_{n - 1} \circ d_{B, n} \circ s_n =
d_{B, n - 1} \circ (d_{B, n} \circ s_n - s_{n - 1} \circ d_{C, n}) =
- d_{B, n - 1} \circ s_{n - 1} \circ d_{C, n}$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-ses-termwise-split-long}
Notation and assumptions as in Lemma \ref{lemma-ses-termwise-split} above.
The morphism of complexes $\delta(s) : C_\bullet \to A[-1]_\bullet$
induces the maps
$$
H_i(\delta(s)) :
H_i(C_\bullet) \longrightarrow H_i(A[-1]_\bullet) = H_{i - 1}(A_\bullet)
$$
which occur in the long exact homology sequence associated
to the short exact sequence of chain complexes by
Lemma \ref{lemma-long-exact-sequence-chain}.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-ses-termwise-split-homotopy}
Notation and assumptions as in Lemma \ref{lemma-ses-termwise-split} above.
Suppose $\{s'_n : C_n \to B_n\}$ is a second choice of splittings.
Write $s'_n = s_n + i_n \circ h_n$ for some unique
morphisms $h_n : C_n \to A_n$. The family of maps
$\{h_n : C_n \to A[-1]_{n + 1}\}$ is a homotopy between
the associated morphisms
$\delta(s), \delta(s') : C_\bullet \to A[-1]_\bullet$.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{definition}
\label{definition-shift-cochain}
Let $\mathcal{A}$ be an additive category.
Let $A^\bullet$ be a cochain complex
with boundary maps $d_A^n : A^n \to A^{n + 1}$.
For any $k \in \mathbf{Z}$ we define the
{\it $k$-shifted cochain complex $A[k]^\bullet$}
as follows:
\begin{enumerate}
\item we set $A[k]^n = A^{n + k}$, and
\item we set $d_{A[k]}^n : A[k]^n \to A[k]^{n + 1}$
equal to $d_{A[k]}^n = (-1)^k d_A^{n + k}$.
\end{enumerate}
If $f : A^\bullet \to B^\bullet$ is a morphism of
cochain complexes, then we let
$f[k] : A[k]^\bullet \to B[k]^\bullet$ be the
morphism of cochain complexes with
$f[k]^n = f^{k + n}$.
\end{definition}
\noindent
Of course this means we have functors
$[k] : \text{CoCh}(\mathcal{A}) \to \text{CoCh}(\mathcal{A})$
which mutually commute (on the nose, without
any intervening isomorphisms of functors) and
such that $A[k][l]^\bullet = A[k + l]^\bullet$ and
with $[0] = \text{id}_{\text{CoCh}(\mathcal{A})}$.
\begin{definition}
\label{definition-cohomology-shift}
Let $\mathcal{A}$ be an abelian category.
Let $A^\bullet$ be a cochain complex
with boundary maps $d_A^n : A^n \to A^{n + 1}$.
For any $k \in \mathbf{Z}$ we identify
{\it $H^{i + k}(A^\bullet) \longrightarrow H^i(A[k]^\bullet)$}
via the identification $A^{i + k} = A[k]^i$.
\end{definition}
\noindent
This identification is functorial in $A^\bullet$.
Note that since no signs are involved in this
definition we actually get a compatible system
of identifications of all the homology
objects $H^{i - k}(A[k]^\bullet)$, which are
further compatible with the identifications
$A[k][l]^\bullet = A[k + l]^\bullet$ and
with $[0] = \text{id}_{\text{CoCh}(\mathcal{A})}$.
\medskip\noindent
Let $\mathcal{A}$ be an additive category.
Suppose that $A^\bullet$ and $B^\bullet$ are
cochain complexes, $a, b : A^\bullet \to B^\bullet$ are
morphisms of cochain complexes, and $\{h^i : A^i \to B^{i - 1}\}$
is a homotopy between $a$ and $b$. Recall that this means
that
$a^i - b^i = d^{i - 1} \circ h^i + h^{i + 1} \circ d^i$.
What if $a = b$? Then we obtain the formula
$0 = d^{i - 1} \circ h^i + h^{i + 1} \circ d^i$,
in other words, $ - d^{i - 1} \circ h^i = h^{i + 1} \circ d^i$.
By definition above this means the collection $\{h^i\}$
above defines a morphism of cochain complexes
$$
A^\bullet \longrightarrow B[-1]^\bullet.
$$
Such a thing is the same as a morphism $A[1]^\bullet \to B^\bullet$
by our remarks above. This proves the following lemma.
\begin{lemma}
\label{lemma-homotopy-shift-cochain}
Let $\mathcal{A}$ be an additive category.
Suppose that $A^\bullet$ and $B^\bullet$ are
cochain complexes. Given any morphism of cochain
complexes $a : A^\bullet \to B^\bullet$ there
is a bijection between the set of homotopies
from $a$ to $a$ and
$\Mor_{\text{CoCh}(\mathcal{A})}(A^\bullet, B[-1]^\bullet)$.
More generally, the set of homotopies between
$a$ and $b$ is either empty or a principal homogeneous
space under the group
$\Mor_{\text{CoCh}(\mathcal{A})}(A^\bullet, B[-1]^\bullet)$.
\end{lemma}
\begin{proof}
See above.
\end{proof}
\begin{lemma}
\label{lemma-ses-termwise-split-cochain}
Let $\mathcal{A}$ be an additive category.
Let
$$
0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0
$$
be a complex (!) of complexes.
Suppose that we are given splittings $B^n = A^n \oplus C^n$
compatible with the maps in the displayed sequence.
Let $s^n : C^n \to B^n$ and $\pi^n : B^n \to A^n$ be the
corresponding maps. Then the family of morphisms
$$
\pi^{n + 1} \circ d_B^n \circ s^n
:
C^n \to A^{n + 1}
$$
define a morphism of complexes $\delta : C^\bullet \to A[1]^\bullet$.
\end{lemma}
\begin{proof}
Denote $i : A^\bullet \to B^\bullet$ and $q : B^\bullet \to C^\bullet$
the maps of complexes in the short exact sequence. Then
$i^{n + 1} \circ \pi^{n + 1} \circ d_B^n \circ s^n =
d_B^n \circ s^n - s^{n + 1} \circ d_C^n$. Hence
$i^{n + 2} \circ d_A^{n + 1} \circ \pi^{n + 1} \circ d_B^n \circ s^n =
d_B^{n + 1} \circ (d_B^n \circ s^n - s^{n + 1} \circ d_C^n) =
- d_B^{n + 1} \circ s^{n + 1} \circ d_C^n$ as desired.
\end{proof}
\begin{lemma}
\label{lemma-ses-termwise-split-long-cochain}
Notation and assumptions as in
Lemma \ref{lemma-ses-termwise-split-cochain} above.
Assume in addition that $\mathcal{A}$ is abelian.
The morphism of complexes $\delta : C^\bullet \to A[1]^\bullet$
induces the maps
$$
H^i(\delta) :
H^i(C^\bullet) \longrightarrow H^i(A[1]^\bullet) = H^{i + 1}(A^\bullet)
$$
which occur in the long exact homology sequence associated
to the short exact sequence of cochain complexes by
Lemma \ref{lemma-long-exact-sequence-cochain}.
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
\begin{lemma}
\label{lemma-ses-termwise-split-homotopy-cochain}
Notation and assumptions as in
Lemma \ref{lemma-ses-termwise-split-cochain}.
Let $\alpha : A^\bullet \to B^\bullet$,
$\beta : B^\bullet \to C^\bullet$ be the given
morphisms of complexes.
Suppose $(s')^n : C^n \to B^n$ and $(\pi')^n : B^n \to A^n$
is a second choice of splittings.
Write $(s')^n = s^n + \alpha^n \circ h^n$ and
$(\pi')^n = \pi^n + g^n \circ \beta^n$ for some unique
morphisms $h^n : C^n \to A^n$ and $g^n : C^n \to A^n$. Then
\begin{enumerate}
\item $g^n = - h^n$, and
\item the family of maps $\{g^n : C^n \to A[1]^{n - 1}\}$ is a homotopy
between $\delta, \delta' : C^\bullet \to A[1]^\bullet$, more precisely
$(\delta')^n = \delta^n + g^{n + 1} \circ d_C^n + d_{A[1]}^{n - 1} \circ g^n$.
\end{enumerate}
\end{lemma}
\begin{proof}
As $(s')^n$ and $(\pi')^n$ are splittings we have $(\pi')^n \circ (s')^n = 0$.
Hence
$$
0 = ( \pi^n + g^n \circ \beta^n ) \circ ( s^n + \alpha^n \circ h^n ) =
g^n \circ \beta^n \circ s^n + \pi^n \circ \alpha^n \circ h^n =
g^n + h^n
$$
which proves (1). We compute $(\delta')^n$ as follows
$$
( \pi^{n + 1} + g^{n + 1} \circ \beta^{n + 1} )
\circ d_B^n \circ
( s^n + \alpha^n \circ h^n )
= \delta^n + g^{n + 1} \circ d_C^n + d_A^n \circ h^n
$$
Since $h^n = -g^n$ and since $d_{A[1]}^{n - 1} = -d_A^n$ we conclude that (2)
holds.
\end{proof}
\section{Graded objects}
\label{section-graded}
\noindent
We make the following definition.
\begin{definition}
\label{definition-graded}
Let $\mathcal{A}$ be an additive category. The {\it category of graded
objects of $\mathcal{A}$}, denoted $\text{Gr}(\mathcal{A})$, is
the category with
\begin{enumerate}
\item objects $A = (A^i)$ are families of objects $A^i$, $i \in \mathbf{Z}$
of objects of $\mathcal{A}$, and
\item morphisms $f : A = (A^i) \to B = (B^i)$ are families of
morphisms $f^i : A^i \to B^i$ of $\mathcal{A}$.
\end{enumerate}
\end{definition}
\noindent
If $\mathcal{A}$ has countable direct sums, then we can associate to
an object $A = (A^i)$ of $\text{Gr}(\mathcal{A})$ the object
$$
A = \bigoplus\nolimits_{i \in \mathbf{Z}} A^i
$$
and set $k^iA = A^i$. In this case $\text{Gr}(\mathcal{A})$ is equivalent
to the category of pairs $(A, k)$ consisting of an object $A$ of
$\mathcal{A}$ and a direct sum decomposition
$$
A = \bigoplus\nolimits_{i \in \mathbf{Z}} k^iA
$$
by direct summands indexed by $\mathbf{Z}$ and a morphism $(A, k) \to (B, k)$
of such objects is given by a morphism $\varphi : A \to B$ of $\mathcal{A}$
such that $\varphi(k^iA) \subset k^iB$ for all $i \in \mathbf{Z}$. Whenever
our additive category $\mathcal{A}$ has countable direct sums we will
use this equivalence without further mention.
\medskip\noindent
However, with our definitions an additive or abelian category does not
necessarily have all (countable) direct sums. In this case our definition
still makes sense. For example, if $\mathcal{A} = \text{Vect}_k$ is the
category of finite dimensional vector spaces over a field $k$, then
$\text{Gr}(\text{Vect}_k)$ is the category of vector
spaces with a given gradation all of whose graded pieces are finite
dimensional, and not the category of finite dimensional vector
spaces with a given graduation.
\begin{lemma}
\label{lemma-graded}
Let $\mathcal{A}$ be an abelian category. The category of graded objects
$\text{Gr}(\mathcal{A})$ is abelian.
\end{lemma}
\begin{proof}
Let $f : A = (A^i) \to B = (B^i)$ be a morphism of graded objects
of $\mathcal{A}$ given by
morphisms $f^i : A^i \to B^i$ of $\mathcal{A}$.
Then we have $\Ker(f) = (\Ker(f^i))$ and $\Coker(f) = (\Coker(f^i))$
in the category $\text{Gr}(\mathcal{A})$.
Since we have $\Im = \text{Coim}$ in $\mathcal{A}$
we see the same thing holds in $\text{Gr}(\mathcal{A})$.
\end{proof}
\begin{remark}[Warning]
\label{remark-direct-sums-not-exact}
There are abelian categories $\mathcal{A}$ having countable direct sums
but where countable direct sums are not exact. An example
is the opposite of the category of abelian sheaves on $\mathbf{R}$.
Namely, the category of abelian sheaves on $\mathbf{R}$ has
countable products, but countable products are not exact.
For such a category the functor $\text{Gr}(\mathcal{A}) \to \mathcal{A}$,
$(A^i) \mapsto \bigoplus A^i$
described above is not exact. It is still true that
$\text{Gr}(\mathcal{A})$ is equivalent to the category of
graded objects $(A, k)$ of $\mathcal{A}$, but the kernel in the category
of graded objects of a map $\varphi : (A, k) \to (B, k)$ is not equal to
$\Ker(\varphi)$ endowed with a direct sum decomposition, but rather it is
the direct sum of the kernels of the maps $k^iA \to k^iB$.
\end{remark}
\begin{definition}
\label{definition-graded-shift}
Let $\mathcal{A}$ be an additive category. If $A = (A^i)$ is a graded object,
then the $k$th {\it shift} $A[k]$ is the graded object with
$A[k]^i = A^{k + i}$.
\end{definition}
\noindent
If $A$ and $B$ are graded objects of $\mathcal{A}$, then we have
\begin{equation}
\label{equation-hom-into-shift}
\Hom_{\text{Gr}(\mathcal{A})}(A, B[k]) =
\Hom_{\text{Gr}(\mathcal{A})}(A[-k], B)
\end{equation}
and an element of this group is sometimes called a map of graded
objects {\it homogeneous of degree $k$}.
\medskip\noindent
Given any set $G$ we can define $G$-graded objects of $\mathcal{A}$
as the category whose objects are $A = (A^g)_{g \in G}$
families of objects parametrized by elements of $G$. Morphisms
$f : A \to B$ are defined as families of maps $f^g : A^g \to B^g$
where $g$ runs over the elements of $G$. If $G$ is an abelian group,
then we can (unambiguously) define shift functors $[g]$ on the category
of $G$-graded objects by the rule $(A[g])^{g_0} = A^{g + g_0}$.
A particular case of this type of construction is when
$G = \mathbf{Z} \times \mathbf{Z}$. In this case the objects of
the category are called {\it bigraded} objects of $\mathcal{A}$.
The $(p, q)$ component of a bigraded object $A$ is usually denoted
$A^{p, q}$. For $(a, b) \in \mathbf{Z} \times \mathbf{Z}$ we write
$A[a, b]$ in stead of $A[(a, b)]$.
A morphism $A \to A[a, b]$ is sometimes called a
{\it map of bidegree $(a, b)$}.
\section{Filtrations}
\label{section-filtrations}
\noindent
A nice reference for this material is \cite[Section 1]{HodgeII}.
(Note that our conventions regarding abelian categories are different.)
\begin{definition}
\label{definition-filtered}
Let $\mathcal{A}$ be an abelian category.
\begin{enumerate}
\item A {\it decreasing filtration} $F$ on an object $A$
is a family $(F^nA)_{n \in \mathbf{Z}}$ of subobjects of $A$ such that
$$
A \supset \ldots \supset F^nA \supset F^{n + 1}A \supset \ldots \supset 0
$$
\item A {\it filtered object of $\mathcal{A}$} is
pair $(A, F)$ consisting of an object $A$ of $\mathcal{A}$
and a decreasing filtration $F$ on $A$.
\item A {\it morphism $(A, F) \to (B, F)$ of filtered objects}
is given by a morphism $\varphi : A \to B$ of $\mathcal{A}$
such that $\varphi(F^iA) \subset F^iB$ for all $i \in \mathbf{Z}$.
\item The category of filtered objects is denoted $\text{Fil}(\mathcal{A})$.
\item Given a filtered object $(A, F)$ and a subobject $X \subset A$ the
{\it induced filtration} on $X$ is the filtration with $F^nX = X \cap F^nA$.
\item Given a filtered object $(A, F)$ and a surjection
$\pi : A \to Y$ the {\it quotient filtration} is the filtration with
$F^nY = \pi(F^nA)$.
\item A filtration $F$ on an object $A$ is said to be {\it finite}
if there exist $n, m$ such that $F^nA = A$ and $F^mA = 0$.
\item Given a filtered object $(A, F)$ we say $\bigcap F^iA$ exists
if there exists a biggest subobject of $A$ contained in all $F^iA$.
We say $\bigcup F^iA$ exists if there exists a smallest subobject
of $A$ containing all $F^iA$.
\item The filtration on a filtered object $(A, F)$ is said to be
{\it separated} if $\bigcap_i F^iA = 0$ and
{\it exhaustive} if $\bigcup F^iA = A$.
\end{enumerate}
\end{definition}
\noindent
By abuse of notation we say that a morphism $f : (A, F) \to (B, F)$
of filtered objects is {\it injective} if $f : A \to B$ is injective
in the abelian category $\mathcal{A}$. Similarly we say $f$ is
{\it surjective} if $f : A \to B$ is surjective in the category
$\mathcal{A}$. Being injective (resp.\ surjective)
is equivalent to being a monomorphism (resp.\ epimorphism)
in $\text{Fil}(\mathcal{A})$. By
Lemma \ref{lemma-filtered}