Isomorphism between Z-modules I am having trouble to prove that $\mathbb{Z}^{(\mathbb{N})}$ is isomorphic as a $\mathbb{Z}$-module to $\text{Hom}_\mathbb{Z}(\mathbb{Z}^\mathbb{N},\mathbb{Z})$, where the isomorphism $\varphi$ is given by $\varphi(e_n)=\text{pr}_n$. 
I have read in this other link that the key step to prove this is the fact that any morphism from $\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$ that vanishes in $\mathbb{Z}^{(\mathbb{N})}$ must be the zero morphism, and I am able to prove this fact, but I do not know how to use it to prove that $\varphi$ is an epimorphism.
 A: I will post a proof that uses the idea provided by Max. I hope this is correct.
Consider a $\mathbb{Z}$-module morphism $f:\mathbb{Z}^\mathbb{N} \rightarrow \mathbb{Z}$, and suppose that $f(e_n) \neq 0$ for infinitely many $n$. Without loss of generality, we can assume that $f(e_n)>0$ for infinitely many $n$. By considering the projections onto those coordinates, we may assume that $f(e_n)>0$ for all $n$.
Let's consider a sequence $(\alpha_n)_{n \in \mathbb{N}}$ such that:


*

*$f(2^{\alpha_n}e_n)>2\sum_{i=1}^{n-1}f(2^{\alpha_i}e_i)$

*$\alpha_{n+1} > \alpha_n$
Let $a_n = 2^{\alpha_n}$, and consider $X=f((a_n)_{n\in\mathbb{N}})$.
Then, for every $n\in \mathbb{N}$:
$X \equiv \sum_{i=1}^{n-1}f(a_ie_i) \pmod{a_n}$
Let's take $k$ sufficiently large such that $\sum_{i=1}^{k-1}f(a_ie_i) > |X|$, $a_k - \sum_{i=1}^{k-1}f(a_ie_i) > |X|$ and $a_k > |X|$, and this is possible because $f(e_n)>0$ for all $n$ and the first condition imposed to the sequence.
Then, considering $X$ modulo $a_k$:
$X \equiv \sum_{i=1}^{k-1}f(a_ie_i) \pmod{a_k}$
But because $k$ is large enough, we obtain that $X = \sum_{i=1}^{k-1}f(a_ie_i)$, or $X = \sum_{i=1}^{n-1}f(a_ie_i) - a_k$, but this contradicts our choice for $k$.
Hence our initial assumption was wrong.
This means that $f=a_1\text{pr}_{x_1}+...+a_n\text{pr}_{x_n}$ in $\mathbb{Z}^{(\mathbb{N})}$, but using the fact that if a $\mathbb{Z}$ module morphsim $g:\mathbb{Z}^\mathbb{N} \rightarrow \mathbb{Z}$ that vanishes in $\mathbb{Z}^{(\mathbb{N})}$ must vanish everywhere, we obtain that $f=a_1\text{pr}_{x_1}+...+a_n\text{pr}_{x_n}$ everywhere, as desired.
