Let $G$ be a retract of a product $\prod_I\mathbb{Z}$ of copies of $\mathbb Z$. This means there are group homomorphisms $\pi:\prod_I\mathbb{Z}\to G$ and $\sigma:G\to\prod_I\mathbb Z$ such that $\pi\circ\sigma=Id_G$. Is it true that $G$ must be a product of copies of $\mathbb Z$?

$\begingroup$ The introduction to R.J. Nunke’s On direct products of infinite cyclic groups (Proc. Amer. Math. Soc. 13 1962 (6671)) says that the paper proves that the answer to the OP’s question is “yes”. However, on a quick look through the paper I don’t see where this is proved in full generality. Probably I’m missing something. $\endgroup$– Jeremy RickardJun 11 at 11:15
1 Answer
EDIT: In an earlier version, I claimed erroneously that the following statement holds for general $I$, and can be formally reduced to the case of countable $I$. I currently do not see how to perform that reduction. The argument also had some gaps, which I hopefully fixed now. So here's an argument for countable $I$:
For countable $I$, we have $\operatorname{Hom}(\prod_I \mathbb{Z}, \mathbb{Z})\cong \bigoplus_I \mathbb{Z}$. The key statement is that a homomorphism $\prod_I \mathbb{Z} \to \mathbb{Z} $ necessarily has "finite support", i.e. there is a finite subset $I'\subseteq I$ such that $f$ factors through the projection $\prod_I \mathbb{Z}\to \prod_{I'}\mathbb{Z}$.
If that is not the case, then (identifying $I$ with $\mathbb{N}$) for any $n$, we find $v_n\in \prod_{\mathbb{N}} \mathbb{Z}$ with $v_n(i)=0$ for $i<n$ and $f(v_n)\neq 0$. But then we may consider sequences $a_i$ with $a_i\mid a_{i+1}$, and form $x_n = \sum_{i=0}^n a_i v_i$. Note this also makes sense for $n=\infty$. Now observe that $f(x_n) = f(x_\infty)$ modulo $a_{n+1}$, but if we inductively fix the elements $a_n$, we may arrange for the absolute value of the minimal representative of $f(x_n)$ modulo $a_{n+1}$ to diverge, leading to a contradiction. Simply pick $a_n$ large enough so that $f(x_n)$ is larger than $f(x_i)$ for all $i<n$, and then ensure that additionally $a_{n+1} > 2f(x_n)$ when we pick $a_{n+1}$.
Why does this help? If $M$ is a retract of $\prod_I \mathbb{Z}$, then $\operatorname{Hom}(M,\mathbb{Z})$ is a retract of $\bigoplus_I \mathbb{Z}$, hence free, and so $\operatorname{Hom}(\operatorname{Hom}(M,\mathbb{Z}),\mathbb{Z})$ is a product of copies of $\mathbb{Z}$. In general, this does not agree with $M$, however, in this situation it does! We always have a natural transformation $M\to \operatorname{Hom}(\operatorname{Hom}(M,\mathbb{Z}),\mathbb{Z})$, let us call an abelian group "reflexive" if this is an isomorphism. The above argument shows that $\prod_I \mathbb{Z}$ is reflexive, and since retracts of isomorphisms are retracts, it follows more generally that every retract of $\prod_I \mathbb{Z}$ is reflexive.

$\begingroup$ For the statement in your first paragraph, I think you need $I$ to be smaller than the first measurable cardinal (if such a thing exists). $\endgroup$ Jun 9 at 14:55

$\begingroup$ What do you mean by "dualize"? If you mean Potryagin Duality, I don't see how that helps. If your dualizing object is $\mathbb Z$, I see that you get that $Hom(G,\mathbb{Z})$ is a free abelian group. But how does that prove the original statement? $\endgroup$– EchoJun 9 at 15:08

1$\begingroup$ I just edited my answer: I addressed the question how the dualizing argument works, but unfortunately I also observed that the countable case does not formally imply the case of general $I$. So for now this is only an answer for countable $I$ (and maybe one can extend it somehow to the generality addressed by @JeremyRickard?) $\endgroup$ Jun 9 at 15:39

$\begingroup$ @AchimKrause (To complement Jeremy's comment) if you precisely need that the canonical map $\mathbf{Z}^{(I)}\to\mathrm{Hom}(\mathbf{Z}^I,\mathbf{Z})$ is an isomorphism, then you precisely need the assumption that $I$ has cardinal $<$ than the smallest measurable cardinal (if it exists) — equivalently that every ultrafilter on $I$, $\sigma$complete (=stable under countable intersections) is principal. This is true for $I$ countable, but also $I$ continuum, power of the continuum, and so on. $\endgroup$– YCorJun 11 at 7:10

$\begingroup$ If $\eta$ is a $\sigma$complete ultrafilter on $I$, then $f\mapsto \lim_\eta f$ is a welldefined homomorphism $\mathbf{Z}^I\to\mathbf{Z}$. If $\eta$ is not principal, then this has "infinite support". $\endgroup$– YCorJun 11 at 7:12