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Oct 9, 2021 at 19:18 comment added John Baez Is this really an example of Brown's representability theorem in action? Maybe I'm misunderstanding, but this argument seems to rely only on $H^1(X,\mathbb{Z}) = [X, S^1]$, and thus $S^1 = K(\mathbb{Z},1)$. We don't really need Brown representability to see that.
Jan 14, 2010 at 19:01 comment added Ryan Budney This argument went unnoticed until 1963 when Kervaire pointed it out. It's a good example of how some pretty significant ideas can take time to filter across fields, even though knot theory and cobordism are so close. Thom and Serre had set up the machinery by 1954.
Jan 14, 2010 at 18:57 history edited Ryan Budney CC BY-SA 2.5
reference correction -- the other article in the same book
Jan 14, 2010 at 18:48 history edited Ryan Budney CC BY-SA 2.5
qualify a statement
Jan 12, 2010 at 15:24 comment added Daniel Moskovich That's a lovely proof!
Jan 11, 2010 at 21:28 history answered Ryan Budney CC BY-SA 2.5