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Alexander Gelbukh's user avatar
Alexander Gelbukh's user avatar
Alexander Gelbukh's user avatar
Alexander Gelbukh
  • Member for 10 years, 1 month
  • Last seen more than 4 years ago
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Closed orientable 4-manifold with $H^1(M;\Bbb Z_2)=\Bbb Z_2$ and non-zero cup product $H^1\times H^1\to H^2$
@abx Could you please clarify why $y$ is 4-torsion? Also, in the last line you say $H^2(N,\mathbb Z)$, shouldn't it me $M$ instead of $N$? Sorry for not getting something very basic...
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Every free abelian group is slender, why?
Thank you very much for your contribution! Still I think advising on useful direction to look for an answer should be made as a comment, not as an answer. Then we can upvote the comment if it proved to be useful. An answer is as a proof as a theorem: you don't say "PROOF. I think one could look for a proof along these lines... QED"
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Every free abelian group is slender, why?
Sorry, this is not quite an answer... this might be, or might not be, a useful direction to look for an answer.
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Every homomorphism from the Baer–Specker group into a slender group factors through ${\bf Z}^n$, why?
@AlexDegtyarev Sorry I can't see how the fact follows from the definition. Could you please provide more details?
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