Dual of $\mathbb Z^I$ for uncountable $I$

Question

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)

Answer 1

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).