I read in my book a chain map q is a kernel of p iff each q(n) is a kernel of p(n). I think there's something wrong with this, it has to do with domain and codomain. Instead of using chain complexes I will give an example of a functor category where I think there's something wrong (after all, chain maps are like natural transformations).
Take J to be the category with 2 objects and 1 non identity between them, and Ab be the abelian groups, so every functor J->Ab is like an arrow in Ab, so I'm just gonna call the functors arrows in Ab. Now let f1 be the 0 arrow Z->Z, where Z is the integers, and f2 be the 0 arrow R->R, R the real numbers. Then define the two components of a natural transformation q: f1->f2 to be the inclusion Z->R, this is clearly natural. Next let g1 be the identity R->R, and g2 be the identity R/Z->R/Z, and define the two components of a natural transformation p:g1->g2 to be the projection onto quotient. Clearly q1 is a kernel of p1 and the same for q2,p2, but the codomain of q is f2 the 0 arrow R->R not the identity R->R (the domain of p), so q can't possibly be a kernel of p. I fiddled around a bit and think that if we assume to codomain of q to equal the domain of p, then it works out. I think natural transformations are not just determined by their components; domain and range also matter.
Is my reasoning correct? I really not confident on this stuff so can someone give me corrections or assurance. Thanks