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Concept
Category of abelian groups
Summary
In mathematics, the Ab has the abelian groups as and group homomorphisms as morphisms. This is the prototype of an : indeed, every can be embedded in Ab.
The zero object of Ab is the trivial group {0} which consists only of its neutral element.
The monomorphisms in Ab are the injective group homomorphisms, the epimorphisms are the surjective group homomorphisms, and the isomorphisms are the bijective group homomorphisms.
Ab is a of Grp, the . The main difference between Ab and Grp is that the sum of two homomorphisms f and g between abelian groups is again a group homomorphism:
(f+g)(x+y) = f(x+y) + g(x+y) = f(x) + f(y) + g(x) + g(y)
= f(x) + g(x) + f(y) + g(y) = (f+g)(x) + (f+g)(y)
The third equality requires the group to be abelian. This addition of morphism turns Ab into a , and because the direct sum of finitely many abelian groups yields a biproduct, we indeed have an .
In Ab, the notion of coincides with kernel in the algebraic sense, i.e. the categorical kernel of the morphism f : A → B is the subgroup K of A defined by K = {x ∈ A : f(x) = 0}, together with the inclusion homomorphism i : K → A. The same is true for cokernels; the cokernel of f is the quotient group C = B / f(A) together with the natural projection p : B → C. (Note a further crucial difference between Ab and Grp: in Grp it can happen that f(A) is not a normal subgroup of B, and that therefore the quotient group B / f(A) cannot be formed.) With these concrete descriptions of kernels and cokernels, it is quite easy to check that Ab is indeed an .
The in Ab is given by the product of groups, formed by taking the cartesian product of the underlying sets and performing the group operation componentwise. Because Ab has kernels, one can then show that Ab is a . The coproduct in Ab is given by the direct sum; since Ab has cokernels, it follows that Ab is also cocomplete.
We have a forgetful functor Ab → which assigns to each abelian group the underlying set, and to each group homomorphism the underlying function. This functor is faithful, and therefore Ab is a .
Official source
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