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.

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    $\begingroup$ Is this in the context of topological structures? Rings? Groups? Sets? Brain surgery? $\endgroup$ – Asaf Karagila Jun 1 '13 at 23:53
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    $\begingroup$ @Asaf Karagila It's only for groups, but it could be for brain surgeries if you want. $\endgroup$ – Diego Silvera Jun 1 '13 at 23:57
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    $\begingroup$ 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. $\endgroup$ – Berci Jun 2 '13 at 0:52
  • $\begingroup$ @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. $\endgroup$ – Diego Silvera Jun 2 '13 at 0:58
  • $\begingroup$ Ah, ok... $\,\,\!$ $\endgroup$ – Berci Jun 2 '13 at 1:02

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 $\mathbb Z^A$.

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  • $\begingroup$ 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. $\endgroup$ – Diego Silvera Jun 2 '13 at 1:35
  • $\begingroup$ 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 $\mathbb{Z}^A$. $\endgroup$ – Diego Silvera Jun 2 '13 at 13:46

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