Timeline for Does the cohomology Bockstein homomorphism map to the homology Bockstein homomorphism under Poincarè duality?
Current License: CC BY-SA 4.0
5 events
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| May 17, 2023 at 9:38 | comment | added | Andrea Antinucci | @DaveBenson in this case I am not sure I understand your answer. Does your answer refer to Poincarè duality? My question was specifically about Poincarè duality and I was trying to unpack the Poincarè dual of the Bockstein in homology and compare it with the Bockstein in cohomology. This makes sense since both are homomorphisms between the same two cohomology group, so they either are equal or not equal. This was my question. | |
| May 16, 2023 at 16:41 | vote | accept | Andrea Antinucci | ||
| May 23, 2023 at 7:55 | |||||
| May 15, 2023 at 12:13 | comment | added | Fernando Muro | Dave, up to a $(-1)^d$ sign as per your argument. | |
| May 15, 2023 at 11:40 | comment | added | Andrea Antinucci | About the section is was a typo (I copied and pasted so it appeared twice), which I now corrected: the section is $s:A_3\rightarrow A_2$. Thank you for the answer. So if I understand correctly this implies that, denoting by $\Phi _A :C_p(X,A)\rightarrow C^{d-p}(X,A)$ Poincarè duality on chains (which I understand is somehow a meaningful thing), this commutes with the section, namely $\Phi_{A_2}(s(c))=s(\Phi_{A_3}(c))$, where $c\in C_p(X,A_3)$. Is this tautological from the definition of Poincarè duality? If not, is there a simple way to prove it directly? | |
| May 15, 2023 at 9:52 | history | answered | Dave Benson | CC BY-SA 4.0 |