Proof that $\mathbb{Z}^{\oplus \mathbb{N}}$ is isomorphic to its product with itself I am trying to prove $\mathbb{Z}^{\oplus \mathbb{N}} \times \mathbb{Z}^{\oplus \mathbb{N}} \cong \mathbb{Z}^{\oplus \mathbb{N}}$ and will appreciate hints to approach the question.

Background:
This question is taken from Aluffi. 
$\mathbb{Z}^{\oplus \mathbb{N}}$ is defined as 
$$\{ \alpha \colon \mathbb{N} \to \mathbb{Z} \mid \alpha(n) \neq 0 \textrm{ for only finitely many elements } n \in \mathbb{N}\}$$
with the group operation $(\alpha + \beta) (n) = \alpha(n) + \beta(n)$, and it forms the free group of $\mathbb{N}$.
I have tried to show the isomorphism by defining group homomorphisms
$\phi \colon \mathbb{Z}^{\oplus \mathbb{N}} \times \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ by $\phi(\alpha,\beta) = \alpha + \beta$ and $\psi \colon \mathbb{Z}^{\oplus \mathbb{N}} \times \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ by $\psi(\alpha) = (\alpha, \alpha)$. However, the first homomorphism is not injective while the second isn't surjective.
 A: $\newcommand{\NN}{\mathbb{N}}$
$\newcommand{\ZZ}{\mathbb{Z}}$
The key is to notice that $\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ is just $\NN$-many copies of $\ZZ$, and $\ZZ^{\oplus\NN}$ is also just $\NN$-many copies of $\ZZ$. Concretely, $\ZZ^{\oplus\NN}$ is indexed by $\NN$, whereas $\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ is indexed by two copies of $\NN$, so it's indexed by $\NN\sqcup\NN$, where $\sqcup$ denotes disjoint union. 
The fact that "$\NN$-many copies of $\ZZ$ is isomorphic to $\NN\sqcup\NN$-copies of $\ZZ$" relies on the fact that $\NN$ and $\NN\sqcup\NN$ have the same cardinality, which in turn relies on the fact that there is a bijection between $\NN$ and $\NN\sqcup\NN$. Let $i_1,i_2 : \NN\hookrightarrow\NN\sqcup\NN$ denote the two injections of $\NN$ into each copy of $\NN$ inside $\NN\sqcup\NN$. Then, for example, the map $f : \NN\rightarrow\NN\sqcup\NN$ given by
$$f(n)\ = \left\{\begin{array}{ll}
i_1\left(\frac{n}{2}\right) & \text{if $n$ is even} \\
i_2\left(\frac{n+1}{2}\right) & \text{if $n$ is odd}
\end{array}\right.$$
is a bijection.
Then, you can define $\phi : \ZZ^{\oplus\NN}\rightarrow\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ as follows:
$$\phi(a_1,a_2,a_3,\ldots) = \phi(a_{f(1)},a_{f(2)},a_{f(3)},\ldots)$$
where recall that $\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$ is indexed by $\NN\sqcup\NN$.
Here's another way to think about $\phi$. Let $j_1,j_2$ denote the natural injections
$$j_k : \ZZ^{\oplus\NN}\hookrightarrow\ZZ^{\oplus\NN}\times\ZZ^{\oplus\NN}$$
(For any group $A$, there are two "obvious" natural injections $A\rightarrow A\times A$). Let $e_i\in\ZZ^{\oplus\NN}$ denote the usual basis vector with a 1 in the $i$th position and 0's everywhere else. Then
$$\phi(e_i)\ = \left\{\begin{array}{ll}
j_1\left(e_{\frac{n}{2}}\right) & \text{if $n$ is even} \\
j_2\left(e_{\frac{n+1}{2}}\right) & \text{if $n$ is odd}
\end{array}\right.$$
