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Fix a cardinal $\kappa$ and consider $\mathbb{Z}^\kappa$ with componentwise addition and the subgroup $$F_\kappa :=\{g:\kappa \to \mathbb{Z}: \{\alpha\in \kappa: g(\alpha)\neq 0\} \text{ is finite}\}.$$

Questions:

  1. Is $\text{Hom}(F_\kappa,\mathbb{Z})\cong \mathbb{Z}^\kappa$?
  2. Is $\text{Hom}(\mathbb{Z}^\kappa,\mathbb{Z})\cong F_\kappa$?
  3. A positive answer to 1. and 2. would imly $\text{Hom}(\text{Hom}(F_\kappa,\mathbb{Z}), \mathbb{Z}) \cong F_\kappa$, but maybe this statment is correct even if 1. and 2. are false?
Dominic van der Zypen
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    A standard notation for $F_\kappa$ is $\mathbb{Z}^{(\kappa)}$, and has the advantage to carry over other rings. 1) is a trivial exercise. 2) is more subtle, it holds iff $\kappa$ is smaller than the first "measurable cardinal". This problem is classical and well-referenced in internet. An reference is Laszlo Fuchs's book on abelian groups. – YCor Feb 16 '15 at 13:20
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    The answer to 1. is yes (trivially); this $F_\kappa$ is a free abelian group generated by the set $\kappa$. The answer to 2. has been treated in several places around MO, for example http://mathoverflow.net/questions/12586/dual-of-zi-for-uncountable-i/12588#12588 and http://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z. Unfortunately, the first of these has an accepted answer which seems incorrect. – Todd Trimble Feb 16 '15 at 13:21
  • Thanks YCor and Todd for your comments and references - could you post it as an answer? – Dominic van der Zypen Feb 16 '15 at 13:22
  • Dominic, what I'm saying is that your question is a duplicate. I'm afraid I'm going to have to close it as such. – Todd Trimble Feb 16 '15 at 13:27
  • OK that's fine. – Dominic van der Zypen Feb 16 '15 at 13:28

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