0
$\begingroup$

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

Improve this question
$\endgroup$
2
  • $\begingroup$ 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
    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
    Aug 27, 2010 at 9:13

1 Answer 1

Reset to default
2
$\begingroup$

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).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.