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

Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0}

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 Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0} is isomorphic to the short exact sequence Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{\iota_A} A \oplus C \xlongrightarrow{\pi_C} C \longrightarrow 0}

Proof: Suppose first that Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{f} B \xlongrightarrow{g} C \longrightarrow 0} is isomorphic to Failed to parse (unknown function "\xlongrightarrow"): {\displaystyle 0 \longrightarrow A \xlongrightarrow{\iota_A} A \oplus C \xlongrightarrow{\pi_C} C \longrightarrow 0} 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.