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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

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  • $\begingroup$ Your book might be somewhat terse on this, but I would expect the assumption for domain and codomain to agree to be fundamental. Possible fundamental to the point where the book doesn't bother stating it explicitly. Specifically, a function - and thus more so a chain map - carries its domain and codomain in its definition, and all of it has to match where functions meet, so it's not actually enough for q(n) to be a kernel of p(n): you need the modules you use to fit into the right chain complex as well. $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Aug 27, 2010 at 6:35
  • $\begingroup$ Thanks. I think my problem is the way I visualize the quantification: I think of two random arrows q and p (not necessarily composable), and the statement "q is a kernel of p" is like a binary predicate on the arrows of the category. So my above example goes through if I interpret "q is a kernel of p iff each q(n) is a kernel of p(n)" in my way $\endgroup$
    – Amadeus
    Commented Aug 27, 2010 at 9:13

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Maybe I'm underestimating your problem, but it seems Mikael above is right.

In your example you define $q:f_1\to f_2$, so if it's a kernel of some other map r, then $r$ must have $f_2$ for domain. $q$ can't possibly be a kernel of $p$, as the composition $pq$ does not make sense.

Categorically http://en.wikipedia.org/wiki/Kernel_(category_theory)

Given a map $f: X \to Y$, a kernel is another map $k:K \to X$ satisfying blah blah.

Now, if X and Y are complexes you have a criterion to check wether $k$ is a kernel: checking the components $k^n$ (but you already must have a chain map $k$ to begin with).

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