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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?

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  • $\begingroup$ 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. $\endgroup$ May 13 '19 at 1:41
  • $\begingroup$ 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. $\endgroup$
    – 0xd34df00d
    May 13 '19 at 1:44
  • $\begingroup$ 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 $\mathbb N+\mathbb N\cong\mathbb N$. $\endgroup$ May 13 '19 at 1:49
  • $\begingroup$ 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. $\endgroup$
    – 0xd34df00d
    May 13 '19 at 1:54
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    $\begingroup$ 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. $\endgroup$
    – logarithm
    May 13 '19 at 2:34
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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).

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    $\begingroup$ 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. $\endgroup$ May 13 '19 at 4:13
  • $\begingroup$ 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. $\endgroup$
    – 0xd34df00d
    May 13 '19 at 17:19

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