Timeline for Coboundary of a cup-product
Current License: CC BY-SA 3.0
14 events
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| Feb 23, 2015 at 8:34 | comment | added | abx | @Chris Gerig: Thanks for your efforts! Unfortunately in the situation I have $H^p(X)$ is zero, so Dold's stability result says nothing. I still hope there is some way to express this cup-product. | |
| Feb 23, 2015 at 8:32 | history | rollback | Chris Gerig |
Rollback to Revision 6
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| Feb 23, 2015 at 8:31 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Feb 23, 2015 at 8:26 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Feb 23, 2015 at 8:14 | comment | added | Nerses Aramian | Here is a kind of an example that kind of shows we need some other structure than derivations (maybe): the coboundary map $H^n(X)\arrow H^{n+1}(CX,X)\simeq H^{n+1}(\Sigma X)$ is the suspension isomorphism, where $CX$ is the cone on $X$ and $\Sigma X$ is the suspension. Any reasonable action of the cohomology of X on cohomology of $(CX,X)$ should be up to sign the same if acted on the right or on the left. If we pick a class $x\in H^n(X)$, then we would get the $\Sigma(x\smile x)$ is $\Sigma(x)\smile x \pm x\smile \Sigma(x)$, which 0 module 2. Yuck! | |
| Feb 23, 2015 at 7:52 | comment | added | Chris Gerig | True, but an exercise in that book gives this as a special case to a more general "stability" result, which I was hopeful would be of use here. And I now agree with Neil's comment. I was originally equating $H^\ast(X\times A,A\times A)$ with $H^\ast(X,A)$ and I no longer hold that in my mind. | |
| Feb 23, 2015 at 7:49 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Feb 23, 2015 at 2:01 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Feb 23, 2015 at 1:54 | history | edited | Chris Gerig | CC BY-SA 3.0 |
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| Feb 23, 2015 at 1:50 | comment | added | Tyler Lawson | @ChrisGerig In that situation, Dold's book has $A_1 = A$ and $A_2 = \emptyset$, so this diagram actually states that $\partial$ is map of $H^*(X)$-modules. (It does show that you are correct in the case where $\partial_q \beta = 0$.) | |
| Feb 23, 2015 at 1:34 | comment | added | Chris Gerig | Sure there is, see chapter VII section 8 of Dold's Lectures on Algebraic Topology, where $(X,A,\varnothing)$ is an excisive triad. The "stability" property 8.10 seems highly relevant. | |
| Feb 23, 2015 at 1:18 | comment | added | Neil Strickland | I don't think that this makes sense. Note that $\partial_p\alpha$ lies in $H^*(X,A)$ and $\beta$ lies in $H^*(A)$, and there is no natural product $H^*(X,A)\otimes H^*(A)\to H^*(X,A)$ so $\partial_p\alpha\cup\beta$ is not defined. | |
| Feb 23, 2015 at 1:15 | history | edited | Chris Gerig | CC BY-SA 3.0 |
mainly removed a redundant sentence
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| Feb 22, 2015 at 22:38 | history | answered | Chris Gerig | CC BY-SA 3.0 |