Timeline for Request: intermediate-level proof: every 2-homology class of a 4-manifold is generated by a surface.
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2010 at 16:53 | comment | added | Torsten Ekedahl | Note that this is analogous to what I did for $H^1(M)$ where one has $H^1(M)=[M,S^1]$. | |
| Apr 25, 2010 at 9:08 | comment | added | Lennart Meier | If one does not like sheaf cohomology, one can circumvent its use like follows: it holds that $H^2(M) = [M, \mathbb{CP}^\infty]$ since $\mathbb{CP}^\infty$ is an Eilenberg-MacLane space. Since $M$ is four-dimensional, you can homotope it via cellular approximation into $\mathbb{CP}^2$ and you can make it transverse to $\mathbb{CP}^1\subset \mathbb{CP}^2$. Take the preimage and this is then your wished-for surface. - But this may just be one of the terse proofs Herb has found ;). | |
| Apr 25, 2010 at 5:32 | comment | added | Somnath Basu | Just beat me to it! :) | |
| Apr 25, 2010 at 5:27 | history | answered | Torsten Ekedahl | CC BY-SA 2.5 |