Exercise B11 Prove that if A is an abelian category, then A^{op} is an abelian category too. We have to prove that A^{op} is an additive category which satisfies the following properties:

i) Every morphism in A^{op} has a kernel and a cokernel. ii) Every monic is the kernel of its cokernel. iii) Every epi is the cokernel of its kernel. We now prove that A^{op} is an additive category:

1. There is a zero object in the category, which is the zero object of A. In fact Hom_{A^{op}}(0, X) = Hom_{A}(X, 0) and Hom_{A^{op}}(X, O) = Hom_{A}(0, X).

2. For every B, C ∈ Obj(A^{op}) we have that Hom_{A^{op}}(B, C) is an abelian group and, moreover, the

composition

Hom_{A^{op}}(B, C) × Hom_{A^{op}}(C, D) ⟶ Hom_{A^{op}}(B, D)

is bilinear. This follows trivially by definition of additivity for A.

3. If B, C ∈ Obj(A^{op}), then B ∏ C exists in A^{op} and it coincides with B ∐ C in A. We now prove the remaining properties:

i) Every morphism f ∈ Hom_{A^{op}}(B, C) is a morphism f ∈ Hom_{A}(C, B). By hypotheses, every morphism f ∈ Hom_{A}(C, B) has a kernel and a cokernel. It is immediate to see that a kernel of f ∈ Hom_{A}(C, B) is a cokernel of f ∈ Hom_{A^{op}}(B, C) and a cokernel of f ∈ Hom_{A}(C, B) is a kernel of f ∈ Hom_{A^{op}}(B, C).

ii) Every monic f ∈ Hom_{A^{op}}(B, C) is an epimorphism f ∈ Hom_{A}(C, B), which is, by hypotheses, the cokernel of its kernel. Then, by previous, it follows that every monic f ∈ Hom_{A^{op}}(B, C) is the kernel of its cokernel.

iii) Every epi f ∈ Hom_{A^{op}}(B, C) is a monomorphism f ∈ Hom_{A}(C, B), which is, by hypotheses, the kernel of its cokernel. Then, as before, it follows that every epi f ∈ Hom_{A^{op}}(B, C) is the co kernel of its kernel.