The group $\mathbb{Z}^\mathbb{N}/\mathbb{Z}^{(\mathbb{N})}$ can't be embedded in a product $\mathbb{Z}^A$ for any $A$ 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.
 A: 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 $\mathbb Z^A$.
