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.

How the tittle says I need to prove that:

There isn't a group monomorphism $\psi: \mathbb{Z}^\mathbb{N}/\mathbb{Z}^{(\mathbb{N})} \to \mathbb{Z}^A$ for any $A$

and, of course, this is equivalent to prove that there isn't any $\psi: \mathbb{Z}^\mathbb{N} \to \mathbb{Z}^A$ such that $\ker (\psi) = \mathbb{Z}^{(\mathbb{N})}$.

For this purpose I have tried to put the discrete topology $\tau_D$ on $\mathbb{Z}$ and the product topology $\tau$ on $\mathbb{Z}^\mathbb{N}$ which turn out to be Hausdorff and $\mathbb{Z}^{(\mathbb{N})}$ is dense. So I just need to put a Hausdorff topology on $\mathbb{Z}^A$ for which all linear maps such that $\mathbb{Z}^{(\mathbb{N})} \subset \ker (\psi)$ are continuous to conclude that $\psi$ must to be constant.

I have tried with the product topology as above on $\mathbb{Z}^A$, but I'm stuck proving that linear maps are continuous.

Please don't spoil my question with a different proof if it's possible, because this is my homework. Thank you very much.


I come with a new approach, I'm trying to prove that the topology $$\{B \subset \mathbb{Z}^A: \psi^{-1}(B) \text{ is open for } \psi: \mathbb{Z}^\mathbb{N} \to \mathbb{Z}^A \text{ linear such that } \mathbb{Z}^{(\mathbb{N})} \subset \operatorname{ker}(\psi)\}$$ is Hausdorff, can you help me? Sorry if I'm being too annoying with this.

share|improve this question
5  
Is this in the context of topological structures? Rings? Groups? Sets? Brain surgery? –  Asaf Karagila Jun 1 '13 at 23:53
1  
@Asaf Karagila It's only for groups, but it could be for brain surgeries if you want. –  Diego Silvera Jun 1 '13 at 23:57
1  
Well, not all $f:\Bbb Z^{\Bbb N}\to\Bbb Z^A$ homomorphisms are constant, so you cannot conclude that. You should try to prove that $\ker f$ is closed. –  Berci Jun 2 '13 at 0:52
    
@Berci I'm assuming that $\mathbb{Z}^{(\mathbb{N})} \subset \operatorname{ker}(\psi)$ and since $\mathbb{Z}^{(\mathbb{N})}$ is dense then $\psi$ is constant if I can prove that $\psi$ es continuous. –  Diego Silvera Jun 2 '13 at 0:58
    
Ah, ok... $\,\,\!$ –  Berci Jun 2 '13 at 1:02

1 Answer 1

up vote 7 down vote accepted

Remember that a function into a topological product space is continuous if and only if each of its components (i.e., its compositions with the projection maps f the product to the factors) is continuous. So to prove that all homomorphisms $\mathbb Z^{\mathbb N}\to\mathbb Z^A$ are continuous, it would suffice to prove this for homomorphisms $\mathbb Z^{\mathbb N}\to\mathbb Z$. The good news is that this continuity result is true; the bad news is that it's a nontrivial theorem of Specker. Specifically, for every homomorphism $h:\mathbb Z^{\mathbb N}\to\mathbb Z$, there is a finite $n$ such that $h(x_1,x_2,\dots)$ depends only on the first $n$ components $x_1,\dots,x_n$ of the input $(x_1,x_2,\dots)\in\mathbb Z^{\mathbb N}$. Proofs of this can be found in textbooks on abelian groups, for example Fuchs's "Infinite Abelian Groups" or Eklof and Mekler's "Almost Free Modules", but, as I said, it's not trivial and probably not what was intended by the person assigning this homework.

If you're willing to deviate from the topological approach, I suggest showing that $\mathbb Z^{\mathbb N}/\mathbb Z^{(\mathbb N)}$ has a non-trivial divisible subgroup and that such a subgroup cannot have a monomorphism into $\mathbb Z^A$.

share|improve this answer
    
I'll keep trying with the original approach a bit more, but essentially to prove the continuity of $h$, as you say, is to prove that $h(x_1,x_2,\ldots)$ depends only on a finite number of $x_i$'s. –  Diego Silvera Jun 2 '13 at 1:35
    
Only a little comment, when you say "it would suffice to prove this for homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}^A$" you are assuming I want a product topology on $\mathbb{Z}^A$. –  Diego Silvera Jun 2 '13 at 13:46

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.