Homological Algebra/Sequences

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.