Timeline for Does homology have a coproduct?
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Oct 14, 2009 at 6:45 | comment | added | Andrew Stacey | I'd need to look it up to be sure, but the slogan that I've absorbed from algebraic topology is that to have "good" behaviour for all spaces then you need the coefficient ring to be a (graded) field. But "good" generally means Kunneth formula and cohomology dual to homology, rather than just one of them. I don't know enough about homological algebra to state the conditions precisely, though, without looking them up (the Boardman et al papers are a good reference, btw). But that's why the Morava K-theories are so popular: they are the only ones where the coefficient ring is a graded field. | |
Oct 13, 2009 at 18:41 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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Oct 13, 2009 at 18:39 | comment | added | Ben Webster♦ | Well, semi-simple assures that all modules are flat. As this question of Anton's shows, any other examples would be pretty pathological. mathoverflow.net/questions/208/… | |
Oct 13, 2009 at 18:06 | comment | added | Mikael Vejdemo-Johansson | @ Andrew Does it have to be a field? Or is it enough that all Tor vanish? And for that matter, does 'is a field' follow from 'all Tor vanish'? | |
Oct 13, 2009 at 18:05 | comment | added | Mikael Vejdemo-Johansson | This is the answer I'd have brought - with the added comment that the reason I hear most often /why/ we don't study the coalgebra of homology is that coalgebras are so much less studied than algebras, and we don't have a firm intuition formed for coalgebras in the same way. | |
Oct 13, 2009 at 17:59 | comment | added | Andrew Stacey | I know this is included in my answer above, but so that answers are self-contained: for this to be always correct everything needs to be over a field right from the start. | |
Oct 13, 2009 at 12:56 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |