
> 
> Let $I$ be an infinite set. There is a homomorphism of abelian groups $\mathbb{Z}^{(I)} \to \hom(\mathbb{Z}^I,\mathbb{Z})$ which sends the basis element $e\_i$ to the projection $p\_i$. If $I$ is countable, it's a famous result of Specker1 that this is actually an isomorphism. But what happens when $I$ is uncountable?
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Clearly it is injective. Surjectivity means that $\phi \in \hom(\mathbb{Z}^I,\mathbb{Z})$ is determined by the values $\phi(e\_i)$ and that these values vanisch for almost all $i$. I can't copy the proof for the countable case.


1 Ernst Specker, Additive Gruppe von Folgen ganzer Zahlen, Portugaliae Math. 9 (1950), 131-140. [MR0039719](http://www.ams.org/mathscinet-getitem?mr=MR0039719) (12,587b)



Dual of $\mathbb Z^I$ for uncountable $I$
With regard to Mariano's answer, I believe some clarification is in order. A closely related question was asked by Michael Barr and answered by user Ralph [here](https://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z/132083#132083). In brief, the homomorphism named in Martin's question is in fact an isomorphism, provided that $I$ has cardinality less than the first measurable cardinal. 


Shelah and Strüngmann (accessible [here](http://arxiv.org/pdf/math/0009045v1.pdf)) refer to this result as well, using the same source given by Ralph, just before Definition 2.1: 



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> For generalizations to products of larger cardinalities and the resulting definition of slenderness for abelian groups we refer to [EM] or [F1] 
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where [EM] is the text by Eklof and Mekler. It seems that Shelah and Strüngmann are talking about something slightly different: homomorphisms out of *free complete products* (but using a notation which could unfortunately suggest direct products). 


