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

i) I'm asking about group morphisms

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 $J=\mathbb{N}$.

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

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  • $\begingroup$ 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) $\endgroup$ – Maxime Ramzi Apr 1 '19 at 21:19
  • $\begingroup$ What paper are you referring to? $\endgroup$ – Eric Wofsey Apr 1 '19 at 21:20
  • $\begingroup$ Of course I forgot to paste the link, here it is : lms.ac.uk/sites/lms.ac.uk/files/… $\endgroup$ – Maxime Ramzi Apr 1 '19 at 21:25
  • $\begingroup$ Actually the paper uses an axiom of "accessibility of ordinals", which forbids inaccessible cardinals hence measurable cardinals. $\endgroup$ – Maxime Ramzi Apr 1 '19 at 21:40
  • $\begingroup$ But I thought I had a proof of the result stated there so I'll have to find where it goes wrong $\endgroup$ – Maxime Ramzi Apr 1 '19 at 21:41

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