# Retract of a product

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 $$\mathbb Z$$?

• The introduction to R.J. Nunke’s On direct products of infinite cyclic groups (Proc. Amer. Math. Soc. 13 1962 (66-71)) 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. Jun 11 at 11:15

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 $$i and $$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, and then ensure that additionally $$|a_{n+1}| > 2|f(x_n)|$$ 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.

• 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). Jun 9 at 14:55
• 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?
– Echo
Jun 9 at 15:08
• 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?) Jun 9 at 15:39
• @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.
– YCor
Jun 11 at 7:10
• If $\eta$ is a $\sigma$-complete ultrafilter on $I$, then $f\mapsto \lim_\eta f$ is a well-defined homomorphism $\mathbf{Z}^I\to\mathbf{Z}$. If $\eta$ is not principal, then this has "infinite support".
– YCor
Jun 11 at 7:12