So let $G = \mathbb{Z}^{\oplus\mathbb{N}}$. We need to prove $G \cong G \times G$ (Aluffi ex. II.5.9). Here's my stab at it.
First, denote the set-function from $\mathbb{N}$ to $G$ that's the part of the free group construction as $j$.
Then, consider $G \times G$ as the (categorical) product of $G$ with itself (along with the projection functions $\pi_1,\pi_2$). Then there exists an unique morphism $\sigma$ in $\mathbf{Set}$ such that $j = \pi_1 \sigma = \pi_2 \sigma$.
Then, let's fix some arbitrary abelian group $H$ along with a set-function $f : \mathbb{N} \rightarrow H$. By the corresponding universal property for $G$ being the free abelian group for $\mathbb{N}$, there exists an unique morphism $\varphi : G \rightarrow H$ such that $f = \varphi j$.
Now note that in $\mathbf{Ab}$ products coincide with coproducts, so there are injection functions $\iota_1, \iota_2 : G \rightarrow G \times G$. Also, by the universal property for coproducts, for the $H$ fixed above, there exists an unique $\sigma' : G \times G \rightarrow H$ such that $\varphi = \sigma' \iota_1 = \sigma' \iota_2$.
Combining all of the above, $f = \sigma' \iota_i \pi_j \sigma$ for $i, j \in \{ 1, 2 \}$. Now, further note that $\sigma' \iota_i \pi_j$ define a morphism $G \times G \rightarrow H$, which must coincide with $\sigma'$ by the universal property, so $f = \sigma' \sigma$.
And that's it! We've proven that there exists $\sigma : \mathbb{N} \rightarrow G \times G$ such that for every abelian $H, f : \mathbb{N} \rightarrow H$ there exists an unique $\sigma' : G \times G \rightarrow H$ such that $f = \sigma' \sigma$, which is precisely the universal property that the free abelian group for $\mathbb{N}$ shall satisfy, which shows that $G \times G$ is also a free abelian group for $\mathbb{N}$.
Does the above seem reasonable? If so, does it generalize to free abelian groups for arbitrary sets? I don't think I've used any particular properties specific to $\mathbb{N}$. And are there better proofs?
Best Answer
No, this is wrong: you have not proved that $\sigma'$ is unique (and indeed it is not, for your choice of $\sigma$). Your $\sigma'$ is unique with the property that $\varphi=\sigma'\iota_1=\sigma'\iota_2$ but there is no reason to believe it is also unique with the property that $f=\sigma'\sigma$. In particular, note that the image of $\sigma$ generates only the diagonal subgroup of $G\times G$, not all of $G\times G$. So, $\sigma'$ could behave in all sorts of ways on the elements of $G\times G$ that are not in the diagonal, and that will not disturb the equation $f=\sigma'\sigma$.
As mentioned in the comments, you do need to use something special about $\mathbb{N}$, namely that $\mathbb{N}\sqcup\mathbb{N}\cong\mathbb{N}$. The idea is that $G\times G$, being also a coproduct of two copies of $G$, will be free on a coproduct of two copies of $\mathbb{N}$.
Alternatively, I would strongly encourage you to try to prove this just by concretely looking at what $G$ is as a set. If you take $G$ to be the set of finite support sequences of elements of $\mathbb{Z}$, it's quite easy to write down an explicit isomorphism $G\cong G\times G$ (again, the key idea is to use $\mathbb{N}\sqcup\mathbb{N}\cong\mathbb{N}$, where this time the $\mathbb{N}$ shows up as the index set of the sequences). Categorical proofs are valuable but it's also extremely valuable to have a concrete picture of what's going on (and that picture can help you find a categorical argument, if you want one).