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.

Let $X$ be a finite CW complex, given singular cohomology with coefficents in $\hat{\mathbb{Z}}$, $H^i(X, \hat{\mathbb{Z}})$ can you recover singular cohomology with coefficents in $\mathbb{Z}$?

I was told that one cannot do this.

However, it seems to be true. The singular cohomology groups are finitely generated abelian groups.

1.) We can recover the rank of each of them, because in our case $H^i(X, \hat{\mathbb{Z}}) \cong H^i(X, \mathbb{Z}) \otimes_{\mathbb{Z}} \hat{\mathbb{Z}}$.

2.) Thus, the issue is to recover the torsion. However, for every integer $m$, $\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \hat{\mathbb{Z}} \cong \mathbb{Z}/m\mathbb{Z}$ as $\mathbb{Z}$-modules.

$\textbf{Question: }$ Am I making a terrible mistake somewhere? More generally, it seems like the same trivial proof above shows that for any CW complex with finitely generated cohomology groups, one can easily recover cohomology with $\mathbb{Z}$ coefficents from $\hat{\mathbb{Z}}$-coefficents.

I'd really like to see a proof or a counterexample.

share|improve this question
1  
clarification needed; you mean the profinite completion of $\mathbb{Z}$? –  Bombyx mori Jan 10 '13 at 6:28
    
Yes, $\hat{\mathbb{Z}} = \varprojlim \mathbb{Z}/n\mathbb{Z}$ –  user53932 Jan 10 '13 at 14:01

1 Answer 1

You are right. Since $\hat{\mathbb{Z}}$ is flat over $\mathbb{Z}$, the natural map $H^i(X,\mathbb{Z}) \otimes \hat{\mathbb{Z}} \longrightarrow H^i(X,\hat{\mathbb{Z}})$ is an isomophism (of $\hat{\mathbb{Z}}$-modules).

If $M = \mathbb{Z}^r \oplus M_{t}$ is a finitely generated group (where $M_t$ is the torsion subgroup), then $\hat{M} := M \otimes \hat{\mathbb{Z}} \cong \hat{\mathbb{Z}}^r \oplus M_t$ (because $M_t \otimes \hat{\mathbb{Z}} = M_t$). The torsion subgroup of $\hat{M}$ as $\mathbb{Z}$-module (NOT as $\hat{\mathbb{Z}}$-module) is $M_t$. The rank can be recovered in the following way : let $p$ be any prime and denote $\hat{M}_p$ a pro-$p$-Sylow of $M \otimes \hat{\mathbb{Z}}$ (it is a pro-$p$-subgroup with quotient of 'profinite order' prime to $p$), then $r$ is the rank of $\hat{M}_p$ as $\mathbb{Z}_p$-module (the rank is well defined for modules over PID).

share|improve this answer

Your Answer

 
discard

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

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