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 \input{preamble} % OK, start here. % \begin{document} \title{Homological Algebra} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent Basic homological algebra will be explained in this document. We add as needed in the other parts, since there is clearly an infinite amount of this stuff around. A reference is \cite{Maclane}. \section{Basic notions} \label{section-topology-basic} \noindent The following notions are considered basic and will not be defined, and or proved. This does not mean they are all necessarily easy or well known. \begin{enumerate} \item Nothing yet. \end{enumerate} \section{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}