Homological Algebra/Printable version

Homological Algebra

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Defintion of abelian category

Definition ($Ab$-enriched category):

An $Ab$-enriched category is a category ${\mathcal {C}}$ such that:

1. $\forall a,b\in \operatorname {Obj} ({\mathcal {C}})$, ${\mathcal {C}}(a,b)$ is an abelian group.
2. $\forall a,b,c\in \operatorname {Obj} ({\mathcal {C}})$, $\circ _{a,b,c}:{\mathcal {C}}(b,c)\times {\mathcal {C}}(a,b)\to {\mathcal {C}}(a,c)$ is bilinear.

Definition (zero object):

A zero object is an object in an $Ab$-enriched category that is both initial and terminal. We usually denote it by $\mathbf {0}$.

Definition (biproduct):

Given an $Ab$-enriched category ${\mathcal {C}}$, a biproduct of $a,b\in \operatorname {Obj} ({\mathcal {C}})$ is a tuple $(c,i_{1}:a\to c,p_{1}:c\to a,i_{2}:b\to c,p_{2}:c\to b)$ such that:

1. $p_{1}\circ i_{1}=1_{a}.$
2. $p_{2}\circ i_{1}=0_{a,b}.$
3. $p_{1}\circ i_{2}=0_{b,a}.$
4. $p_{2}\circ i_{2}=1_{b}.$
5. $i_{1}\circ p_{1}+i_{2}\circ p_{2}=1_{c}.$

We usually denote $c$ by $a\oplus b$.

Definition (additive category):

An additive category is an $Ab$-enriched category ${\mathcal {C}}$ such that:

1. There is a zero product in ${\mathcal {C}}$.
2. Every $a,b\in \operatorname {Obj} ({\mathcal {C}})$ has a biproduct.

Definition ((co-)kernel):

Given $f:a\to b$ in an $Ab$-enriched category. A (co-)kernel of $f$ is a (co-)equalizer of $f$ and $0_{a,b}$.

Definition (abelian category):

An abelian category is an additive category where:

1. Every morphism has a kernel and cokernel.
2. Every monomorphism is a kernel and every epimorphism is a cokernel.

Example:

The category of all left $R$-modules of a ring $R$ is an abelian category.

Exercises

1. Given $f:a\to b$ in an $Ab$-enriched category with zero object. Prove that $f=0_{a,b}$ iff $f$ factors through $\mathbf {0}$.

1. Given a biproduct $(a\oplus b,i_{1},p_{1},i_{2},p_{2})$ of $a$ and $b$. Prove that $(a\oplus b,i_{1},i_{2})$ is a coproduct of $a$ and $b$ and $(a\oplus b,p_{1},p_{2})$ is a product of $a$ and $b$.

1. In an $Ab$-enriched category with zero object, a kernel of $f:a\to b$ can be equivalently be characterized as a pullback of $\mathbf {0} \to b$ along $f$.

Sequences

Lemma: In an $Ab$-enriched category, if $f$ is a kernel, $g$ is a cokernel of $f$, then $f$ is a kernel of $g$.

Proof: Let $f$ be a kernel of $h$. Since $hf=0$, and $g$ is a cokernel of $f$, there exists $k$ such that $h=kg$. For all $l$ such that $gl=0$, $hl=kgl=k0=0$. Since $f$ is a kernel of $h$, $l$ factorizes uniquely through $f$. $\Box$

Corollary: In an abelian category, consider a sequence $a{\xrightarrow {f}}b{\xrightarrow {g}}c$. The following conditions are equivalent:

1. $g$ is a cokernel of $f$ and $f$ is a kernel of $g$.
2. $f$ is a monomorphism and $g$ is a cokernel of $f$.
3. $g$ is an epimorphism and $f$ is a kernel of $g$.

Definition (short exact sequence):

We call $\cdot {\xrightarrow {f}}\cdot {\xrightarrow {g}}\cdot$ a short exact sequence if it satisfies any of the equivalent conditions above.

Proposition (splitting lemma):

Let ${\mathcal {C}}$ be an abelian category, and suppose that

$0\longrightarrow A\xrightarrow {\qquad f\qquad } B\xrightarrow {\qquad g\qquad } C\longrightarrow 0$

is a short exact sequence. Then the following are equivalent:

1. There exists a morphism $l:B\to A$ so that $l\circ f=\operatorname {id} _{A}$
2. There exists a morphism $r:C\to A$ so that $g\circ r=\operatorname {id} _{B}$
3. The short exact sequence $0\longrightarrow A\xrightarrow {\qquad f\qquad } B\xrightarrow {\qquad g\qquad } C\longrightarrow 0$ is isomorphic to the short exact sequence $0\longrightarrow A\xrightarrow {\qquad \iota _{A}\qquad } A\oplus C\xrightarrow {\qquad \pi _{C}\qquad } C\longrightarrow 0$

Proof: Suppose first that $0\longrightarrow A\xrightarrow {\qquad f\qquad } B\xrightarrow {\qquad g\qquad } C\longrightarrow 0$ is isomorphic to $0\longrightarrow A\xrightarrow {\qquad \iota _{A}\qquad } A\oplus C\xrightarrow {\qquad pi_{C}\qquad } C\longrightarrow 0$ via isomorphisms $\varphi :A\to A$, $\psi :B\to A\oplus C$ and $\chi :C\to C$. Then there exists $l$ as in 1., since we may just define

$l:=\varphi ^{-1}\circ \pi _{A}\circ \psi$;

$l$ has the required property since by definition of morphisms of chain complexes, $\psi \circ f=\iota _{A}\circ \phi$, and further $\pi _{A}\circ \iota _{A}=\operatorname {id} _{A}$. Suppose now that there does exist $l:B\to A$ so that $l\circ f=\operatorname {id} _{A}$. Since $A\oplus C$ is a biproduct, it is in particular a product, so that $l$ and $g$ define a unique morphism $\Phi :B\to A\oplus B$ such that

$l=\pi _{A}\circ \Phi$ and $g=\pi _{B}\circ \Phi$.

Then, $\Phi$, together with the identities on $A$ and $C$, yields an isomorphism of chain complexes. The equivalence of 2. and 3. is a dual statement to the equivalence of 1. and 3., whence it doesn't have to be proved separately. $\Box$