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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 2010 at 0:14

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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 2010 at 0:51

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