# Is there a surjective morphism from an infinite direct product of copies of $\mathbb{Z}$ to an infinite direct sum of copies of $\mathbb{Z}$?

Is there a surjective morphism $$\mathbb{Z}^I\to \mathbb{Z}^{(J)}$$ for some $$I,J$$?

ii) $$\mathbb{Z}^{(J)}$$ denotes the direct sum of $$J$$ copies of $$\mathbb Z$$

iii) Of course, I want $$J$$, and thus $$I$$ to be infinite, otherwise it's trivial.

I want to know if there is such a surjection for any $$J$$, so it suffices to find one $$J$$ with no such surjection and it'll be done. However it seems that the answer won't really depend on $$J$$.

Thoughts : I think there is no surjective morphism, and so was trying to find a contradiction. Of course such a morphism is split, so I was trying to analyze summands of $$\mathbb{Z}^I$$ but got essentially nowhere. I also tried studying maps $$\mathbb{Z}^I \to \mathbb{Z}$$, to study the projections. I know that if such a map vanishes on almost zero sequences, then it vanishes, so I was trying to prove that it would do that here, but with no success.

I don't really know how to proceed further.

EDIT: the link that was given in the comments can help. Indeed, if there were such a surjection, then by passing to hom's into $$\mathbb{Z}$$, we would have an injection $$\mathbb{Z}^J \to \hom (\mathbb{Z}^I, \mathbb{Z})$$. Now I don't know how general the quoted Baer's result is, but if it generalizes to any exponent, then this would provide an injection $$\mathbb{Z}^J\to \mathbb{Z}^{(I)}$$, which is clearly contradictory, by looking for instance at "almost" $$2^\infty$$-divisible elements of $$\mathbb{Z}^J$$ (if this isn't clear I can sketch the proof here, but it's not that complicated)

Alternatively, if Baer's result isn't that general, it might be possible to use some trickery to reduce to $$I= \mathbb{N}$$, at least when assuming that $$J=\mathbb{N}$$.

• Have you tried looking at the problem from a categorical perspective? – Shaun Apr 1 '19 at 16:03
• You defined $\mathbb{Z}^{(J)}$, but not $\mathbb{Z}^{I}$. Are they the same thing, or do your brackets have a meaning? – user1729 Apr 1 '19 at 16:07
• @Shaun it depends on what you mean, but I have tried to think categorically about it. I do think it's something specifically group-theoretic though – Maxime Ramzi Apr 1 '19 at 16:09
• Okay (but call it "the direct product" or something, rather than just "the power" :-p) – user1729 Apr 1 '19 at 16:13
• @user1729 : it is split because $\mathbb{Z}^{(I)}$ is free – Maxime Ramzi Apr 1 '19 at 16:27

By the Łoś–Eda Theorem, $$\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$$ is a free abelian group for any set $$I$$ (namely, it is free on the homomorphisms given by the countably complete ultrafilters on $$I$$). If a surjective homomorphism $$\mathbb{Z}^I\to\mathbb{Z}^{(J)}$$ existed, it would induce an injective homomorphism $$\operatorname{Hom}(\mathbb{Z}^{(J)},\mathbb{Z})\to\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$$, and so $$\operatorname{Hom}(\mathbb{Z}^{(J)},\mathbb{Z})$$ would be free since $$\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$$ is free. But $$\operatorname{Hom}(\mathbb{Z}^{(J)},\mathbb{Z})\cong\mathbb{Z}^J$$ is not free if $$J$$ is infinite, so this is a contradiction.
• I have a problem : this paper seems to state that $\hom (\mathbb{Z}^I, \mathbb{Z}) = \mathbb{Z}^{(I)}$, no matter what $I$ is, which seems to contradict the Los-Eda theorem you are quoting (though it is enough to conclude as well) – Maxime Ramzi Apr 1 '19 at 21:19