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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?

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 (12,587b)

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Related qn, when I is countable .. mathoverflow.net/questions/10239 –  Anweshi Jan 22 '10 at 0:14
To address the question in the title: The $\mathbb{Z}$-dual of $\mathbb{Z}^I$ is the free abelian group whose rank equals the cardinality of the set $D$ of all countably complete ultrafilters on $I$. Moreover, $|I| \le |D|$ and if the cardinality of $I$ is less than the first measurable cardinal, then $|I|=|D|$. For references see my answer to this question: mathoverflow.net/questions/132073/… –  Ralph May 29 '13 at 0:57
Why is Mariano's answer below still accepted as the answer? As YCor notes, it seems to have been based on a misunderstanding. –  Todd Trimble Feb 16 at 13:17

2 Answers 2

up vote 6 down vote accepted

Apparently the map is not an isomorphism: [Shelah, Saharon; Strüngmann, Lutz. The failure of the uncountable non-commutative Specker phenomenon. J. Group Theory 4 (2001), no. 4, 417--426. MR1859179 (2002g:20049)]

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Very interesting! –  Georges Elencwajg Jan 22 '10 at 0:51
This reference, which deals with noncommutative complete free products, does not seem relevant to the question. (And indeed $\phi$ is an isomorphism when $I$ has continuum cardinal, for instance.) –  YCor May 29 '13 at 8:46

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. 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) refer to this result as well, using the same source given by Ralph, just before Definition 2.1:

For generalizations to products of larger cardinalities and the resulting definition of slenderness for abelian groups we refer to [EM] or [F1]

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).

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