Homological Algebra/Printable version

Homological Algebra

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Defintion of abelian category

Definition ( -enriched category):

An  -enriched category is a category   such that:

  1.  ,   is an abelian group.
  2.  ,   is bilinear.

Definition (zero object):

A zero object is an object in an  -enriched category that is both initial and terminal. We usually denote it by  .

Definition (biproduct):

Given an  -enriched category  , a biproduct of   is a tuple   such that:


We usually denote   by  .

Definition (additive category):

An additive category is an  -enriched category   such that:

  1. There is a zero product in  .
  2. Every   has a biproduct.

Definition ((co-)kernel):

Given   in an  -enriched category. A (co-)kernel of   is a (co-)equalizer of   and  .

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.


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


  1. Given   in an  -enriched category with zero object. Prove that   iff   factors through  .
  1. Given a biproduct   of   and  . Prove that   is a coproduct of   and   and   is a product of   and  .
  1. In an  -enriched category with zero object, a kernel of   can be equivalently be characterized as a pullback of   along  .


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

Proof: Let   be a kernel of  . Since  , and   is a cokernel of  , there exists   such that  . For all   such that  ,  . Since   is a kernel of  ,   factorizes uniquely through  .  

Corollary: In an abelian category, consider a sequence  . The following conditions are equivalent:

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

Definition (short exact sequence):

We call   a short exact sequence if it satisfies any of the equivalent conditions above.

Proposition (splitting lemma):

Let   be an abelian category, and suppose that


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

  1. There exists a morphism   so that  
  2. There exists a morphism   so that  
  3. The short exact sequence   is isomorphic to the short exact sequence  

Proof: Suppose first that   is isomorphic to   via isomorphisms  ,   and  . Then there exists   as in 1., since we may just define


  has the required property since by definition of morphisms of chain complexes,  , and further  . Suppose now that there does exist   so that  . Since   is a biproduct, it is in particular a product, so that   and   define a unique morphism   such that

  and  .

Then,  , together with the identities on   and  , 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.