Timeline for Does homology have a coproduct?
Current License: CC BY-SA 4.0
12 events
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| Dec 17, 2020 at 22:13 | comment | added | ziggurism | On the one hand, the answers here say that, at least over field coefficients, homology has a comultiplication which is formally dual to cup product of cohomology. On the other hand, I am under the impression that the dual of cup product of cohomology classes is intersection product of homology classes. So which is it? Is there some connection between intersection product and comultiplication of homology classes? Maybe this would be better as a new question | |
| Oct 1, 2020 at 23:09 | comment | added | Theo Johnson-Freyd | I know this post is (more than a decade) old, but I only now found it, and it is awesome. "Whether this is more because we have a greater history and more experience, or whether they are inherently simpler is something I shall leave for another to answer." I won't answer this, but I will mention something that I think I first learned from Dylan Wilson: The category $SET$ is fundamentally qualititively different from the category $SET^{op}$, and this is one of the reasons why algebras and coalgebras end up being so different. | |
| S Dec 18, 2019 at 15:30 | history | edited | Andrew Stacey | CC BY-SA 4.0 |
Rewrite formulas using latex
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| S Dec 18, 2019 at 15:30 | history | suggested | lisyarus | CC BY-SA 4.0 |
Rewrite formulas using latex
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| Dec 18, 2019 at 14:19 | review | Suggested edits | |||
| S Dec 18, 2019 at 15:30 | |||||
| Jul 4, 2014 at 6:05 | comment | added | Jim Conant | Nice answer! I am amused by the term cooperation. | |
| Oct 16, 2009 at 8:23 | history | edited | Andrew Stacey | CC BY-SA 2.5 |
added 334 characters in body
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| Oct 13, 2009 at 18:03 | history | edited | Andrew Stacey | CC BY-SA 2.5 |
added 592 characters in body
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| Oct 13, 2009 at 17:58 | comment | added | Andrew Stacey | The short answer is that I don't see why not, but you'd need every term in the spectral sequence to be flat in order to get this. I'm also not so sure how much help it would be. The point about rings is that once you know where x goes to then you know where x^2 goes to. But knowing where x goes to doesn't obviously tell you where everything in the comultiplication of x goes to. | |
| Oct 13, 2009 at 15:36 | comment | added | Ben Webster♦ | Is there any reason one couldn't have spectral sequences of coalgebras? | |
| Oct 13, 2009 at 14:05 | history | edited | Andrew Stacey | CC BY-SA 2.5 |
minor correction to splitting conditions
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| Oct 13, 2009 at 13:08 | history | answered | Andrew Stacey | CC BY-SA 2.5 |