# Homological Algebra/Printable version

Homological Algebra

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# Defintion of abelian category

Definition (${\displaystyle Ab}$ -enriched category):

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

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

Definition (zero object):

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

Definition (biproduct):

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

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

We usually denote ${\displaystyle c}$  by ${\displaystyle a\oplus b}$ .

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

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

Definition ((co-)kernel):

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

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 ${\displaystyle R}$ -modules of a ring ${\displaystyle R}$  is an abelian category.

## Exercises

1. Given ${\displaystyle f:a\to b}$  in an ${\displaystyle Ab}$ -enriched category with zero object. Prove that ${\displaystyle f=0_{a,b}}$  iff ${\displaystyle f}$  factors through ${\displaystyle \mathbf {0} }$ .
1. Given a biproduct ${\displaystyle (a\oplus b,i_{1},p_{1},i_{2},p_{2})}$  of ${\displaystyle a}$  and ${\displaystyle b}$ . Prove that ${\displaystyle (a\oplus b,i_{1},i_{2})}$  is a coproduct of ${\displaystyle a}$  and ${\displaystyle b}$  and ${\displaystyle (a\oplus b,p_{1},p_{2})}$  is a product of ${\displaystyle a}$  and ${\displaystyle b}$ .
1. In an ${\displaystyle Ab}$ -enriched category with zero object, a kernel of ${\displaystyle f:a\to b}$  can be equivalently be characterized as a pullback of ${\displaystyle \mathbf {0} \to b}$  along ${\displaystyle f}$ .

# Sequences

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

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

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

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

Definition (short exact sequence):

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

Proposition (splitting lemma):

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

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

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

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

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

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

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

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

Then, ${\displaystyle \Phi }$ , together with the identities on ${\displaystyle A}$  and ${\displaystyle C}$ , 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. ${\displaystyle \Box }$