By Tomi Pannila

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**Additional resources for An Introduction to Homological Algebra**

**Example text**

If A is an additive category, then CpAq is additive. 39 Proof. Abelian group: Let f, g : K ‚ Ñ L‚ be two morphisms in CpAq. The following diagram commutes di´2 A‚ ... Ai´1 di´1 A‚ Ai ... B i´1 ai`2 f i `g i f i´1 `g i´1 i´2 dB ‚ diA‚ di´i B‚ Bi di`1 A‚ ... f i`1 `g i`1 diB ‚ B i`1 di`1 B‚ ... so f ` g :“ pf i ` g i qiPZ is a morphism of complexes. If α, β : L‚ Ñ M ‚ are two morphisms in CpAq we have ppf ` gq ˝ pα ` βqqi “ pf i ` g i q ˝ pαi ` β i q “ pf i ˝ αi q ` pf i ˝ β i q ` pg i ˝ αi q ` pg i ˝ β i q “ pf ˝ αqi ` pf ˝ βqi ` pg ˝ αqi ` pg ˝ βqi .

Let A be an abelian category and f “ me a morphism of A where m is a monomorphism and e is an epimorphism. Then ker f – ker e and coker f – coker m. Proof. By duality it suffices to show that ker f – ker e. Let k1 : ker f Ñ f and k2 : ker e Ñ e be the corresponding morphisms. Then f k1 “ 0 implies ek1 “ 0 and ek2 “ 0 implies f k2 “ 0. From the universal properties if follows that ker f – ker e. The rest of this section is devoted for a quick introduction to exact sequences in abelian categories.

Exactness at coker h: It suffices to show that g˜2 is an epimorphism. By commutativity g˜2 c2 “ c3 n where c2 , c3 , and n are epimorphisms. Hence g˜2 is an epimorphism. 9) is not in general the zero morphism, consider the following morphism of short exact sequences over Ab. 9) the last morphisms between kernels 0 – kerpZ Ñ Zq Ñ kerpZ Ñ Z{2q – 2Z – Z is not Id surjective and Z – cokerp0 Ñ Zq Ñ cokerpZ Ñ Zq – 0 is not injective. 9) is exact when δ˜ “ IdZ , which is not the zero morphism. If one embeddes this diagram to complexes over Ab in degree 0, one gets ˜ now between complexes, is not the zero morphism.