Homological Algebra/Sequences

Lemma: In an -enriched category, if is a kernel, is a cokernel of , then is a kernel of . 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: is a cokernel of and is a kernel of . is a monomorphism and is a cokernel of . is an epimorphism and is a kernel of . short exact sequence We call a short exact sequence if it satisfies any of the equivalent conditions above. splitting lemma Let be an abelian category, and suppose that is a short exact sequence. Then the following are equivalent: There exists a morphism so that There exists a morphism so that The short exact sequence is isomorphic to the short exact sequence 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.