Chain maps of complex - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:56:39Z http://mathoverflow.net/feeds/question/36844 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36844/chain-maps-of-complex Chain maps of complex Amadeus 2010-08-27T06:08:50Z 2010-08-27T07:14:35Z <p>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).</p> <p>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.</p> <p>Is my reasoning correct? I really not confident on this stuff so can someone give me corrections or assurance. Thanks</p> http://mathoverflow.net/questions/36844/chain-maps-of-complex/36847#36847 Answer by babubba for Chain maps of complex babubba 2010-08-27T06:57:03Z 2010-08-27T07:14:35Z <p>Maybe I'm underestimating your problem, but it seems Mikael above is right.</p> <p>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.</p> <p>Categorically <a href="http://en.wikipedia.org/wiki/Kernel_(category_theory" rel="nofollow">http://en.wikipedia.org/wiki/Kernel_(category_theory</a>)</p> <p>Given a map $f: X \to Y$, a kernel is another map $k:K \to X$ satisfying blah blah.</p> <p>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).</p>