Proving that the free abelian group $G$ for the set $\mathbb{N}$ is isomorphic to its product with itself $G \times G$

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?

• If you replace $N$ with a singleton set, the F.A.G on $N$ looks like $Z$, but $G \times G$ is then $Z \times Z$, which is not $Z$. So either something about your argument uses the fact that $N$ is different from a singleton, or something about your argument is wrong. May 13 '19 at 1:41
• Yeah, finite sets (and singletons in particular) made me worry, but I was not able to quickly disprove the existence of group isomorphism between $\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{Z}$, nor I was able to find a hole in my argument, hence the question. May 13 '19 at 1:44
• A "better" proof that will become the evident way to handle this in the future is to note that the free abelian group functor, call it $F$, is a left adjoint and all left adjoints preserve colimits including coproducts. (Proving this is actually really easy if everything is formulated in terms of representability.) As you note, $\oplus$/$\times$/$+$ is a biproduct (i.e. simultaneously a product and a coproduct). Given $G=F(\mathbb N)$ we have $F(\mathbb N)+F(\mathbb N)\cong F(\mathbb N+\mathbb N)$. Then we use the set theoretic fact that $\mathbb N+\mathbb N\cong\mathbb N$. May 13 '19 at 1:49
• So you're relying on $\mathbb{N} + \mathbb{N} \cong \mathbb{N}$, while I'm not (or at least I don't see where I use it if I do). Now I'm pretty much convinced my proof is incomplete, but I don't know where. May 13 '19 at 1:54
• The problem is then when you say that $\sigma'\iota_i\pi_i$ is $\sigma'$. $f$ and $\sigma$ are not in the same category in which $\sigma'$ is the unique morphism. May 13 '19 at 2:34

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).
• Conversely, it is pretty mechanical to produce a concrete argument/construction given concrete representations of certain categorical structures, most particularly (co)limits in $\mathbf{Set}$. Given a concrete representation of $F(1)$, e.g. $\mathbb Z$, the arguments I gave can be mechanically unfolded to produce an explicit group isomorphism. Extracting the concrete content of "abstract nonsense" proofs is also a valuable exercise. This is probably more work than constructing it directly in this case but in more complex cases a direct solution may not be as forthcoming. May 13 '19 at 4:13
• In addition to that, I've just realized I've proven $F^{Ab}(A \amalg B) \cong F^{Ab}(A) * F^{Ab}(B)$ earlier (shame on me for forgetting that!), and proving $F(S_1) \cong F(S_2)$ for $S_1 \cong S_2$ is trivial. The result follows immediately, and I guess that's what Derek mentioned in one of his comments. May 13 '19 at 17:19