dual of Z^I for uncountable I - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T18:29:19Z http://mathoverflow.net/feeds/question/12586 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12586/dual-of-zi-for-uncountable-i dual of Z^I for uncountable I Martin Brandenburg 2010-01-22T00:04:29Z 2010-01-22T00:57:43Z <p>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 Specker<sup>1</sup> that this is actually an isomorphism. But what happens when $I$ is uncountable?</p> <p>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.</p> <p><sup>1</sup> Ernst Specker, Additive Gruppe von Folgen ganzer Zahlen, Portugaliae Math. 9 (1950), 131-140. <a href="http://www.ams.org/mathscinet-getitem?mr=MR0039719" rel="nofollow">MR0039719</a> (12,587b)</p> http://mathoverflow.net/questions/12586/dual-of-zi-for-uncountable-i/12588#12588 Answer by Mariano Suárez-Alvarez for dual of Z^I for uncountable I Mariano Suárez-Alvarez 2010-01-22T00:31:41Z 2010-01-22T00:46:52Z <p>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. <a href="http://www.ams.org/mathscinet-getitem?mr=MR1859179" rel="nofollow">MR1859179</a> (2002g:20049)]</p>