Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I don't remember if I've already seen this question even here or in MO or in my mind. This is partly related to questions arose about differences between homology and cohomology; I'm wondering if some kind of difference can appear before computing $H^n(-)$, $H_n(-)$, looking at the bare complex of (co)chains...

My question: Given a chain complex $\mathcal C=\{C_n,\partial_n\}$, consider the image complex $\hom(\mathcal C,X)=\{\hom(C_n,X),\hom(\partial_n,X)\}$; is it always a cochain complex? Can I always define some kind of ``equivalence'' between chain complexes and cochain complexes using a suitable contrav. functor? Can I conversely choose carefully $X$ in order not to have such an equivalence?

share|improve this question
2  
No matter what $X$ is and what sort of chain complexes you're looking at, applying the contravariant additive functor $\hom{({-},X)}$ will always result in a cochain complex (of abelian groups) since the relation $\partial_{n}\partial_{n+1} = 0$ is sent to the relation $\delta^{n+1} \delta^{n} = 0$, where I write $\delta^n = \hom{(\partial_n},X)$ for short. However, you might e.g. kill torsion: suppose for instance each $C_n$ is a torsion abelian group. If you take $X = \mathbb{Z}$ then $\hom{(C_n,X)} = 0$ for all $n$. –  t.b. Apr 30 '11 at 15:50
    
I don't see if it is true just because I'm taking $\hom$ or if it is true with any contrav. functor... I suppose there's anything peculiar in that functor. Good example that with torsion! –  tetrapharmakon Apr 30 '11 at 15:55
2  
If $F$ is contravariant then $F(\partial_n \partial_{+1}) = F(\partial_{n+1})F(\partial_n)$. So if you want to send chain complexes to cochain complexes, you need contravariance and sending zero to zero. Since you're also interested in chain maps and the additive notion of (co)chain homotopy it seems best to restrict attention to additive functors. –  t.b. Apr 30 '11 at 16:00
2  
On the other hand, there are not that many contravariant additive endofunctors of abelian groups by the contravariant version of the Eilenberg-Watts theorem that says in particular that a contravariant additive endofunctor $F:\mathfrak{Ab} \to \mathfrak{Ab}$ has to be of the form $F({-}) = \hom{({-},F(\mathbb{Z}))}$, as soon as it transforms colimits into limits. –  t.b. Apr 30 '11 at 16:07
    
@t.b: If we knew $H_n$ was finite dimensional, and we were to apply $hom(-,X)$ twice, would we recover $H_n$? –  user99680 Jun 12 at 1:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.