dual of Z^I for uncountable I

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 '15 at 13:17

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.