Timeline for Künneth formula for cohomology
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 5, 2023 at 6:53 | history | edited | Mark Grant | CC BY-SA 4.0 |
Added boundedness hypothesis
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| May 26, 2015 at 9:17 | comment | added | Mark Grant | @Sigur: If you wanted to compute the cohomology of $A\otimes B\otimes C$, one way would be to apply the above Kunneth sequence twice: once to compute the cohomology of either $A\otimes B$ or $B\otimes C$, then again to compute the cohomology of $A\otimes B\otimes C$. | |
| May 26, 2015 at 9:04 | comment | added | Sigur | @MarkGrant, what about for 3? I mean, $A,B,C$? | |
| Sep 22, 2012 at 8:10 | vote | accept | Axel | ||
| Sep 21, 2012 at 12:45 | comment | added | Mark Grant | @Allen Hatcher: Thank you for taking the time to respond to my comment, which I hope you won't take as a slight against your wonderful book! | |
| Sep 21, 2012 at 12:29 | comment | added | Allen Hatcher | This was omitted from my book for two reasons. First, it is rarely needed for applications, where it usually suffices just to combine the homology version and the universal coefficient theorem. And second, the cohomology version has always seemed to me a little unnatural, as evidenced by the extra finiteness assumption and the presence of the Tor functor, which in the universal coefficient theorems appears in the homology version rather than the cohomology version. A lighter toolkit is always preferable, and specialized tools can always be added for specialized applications. | |
| Sep 21, 2012 at 8:21 | history | edited | Mark Grant | CC BY-SA 3.0 |
deleted 10 characters in body
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| Sep 21, 2012 at 8:21 | comment | added | Mark Grant | Yes, that's right. The finiteness assumptions are there exactly to make your unequality an equality. I wonder why the more modern textbooks (such as Hatcher and May) don't state this result? Dold's book has it also, as Proposition VI.12.16. | |
| Sep 21, 2012 at 7:37 | vote | accept | Axel | ||
| Sep 22, 2012 at 8:10 | |||||
| Sep 21, 2012 at 7:37 | comment | added | Axel | Thank you Mark! This is quite what I wanted to see. It seems to me that the finite type conditions you've mentioned somehow reflect the fact that $Hom(A, M)\otimes Hom(B, N)\neq Hom(A\otimes B, M\otimes N)$ so one has to impose some finiteness assumptions? | |
| Sep 21, 2012 at 7:06 | history | answered | Mark Grant | CC BY-SA 3.0 |