Timeline for Equivalence of different cohomology groups
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 29, 2016 at 8:02 | comment | added | Sebastian Goette | The typical counterexample for (1) and (2) is the Polnish circle $X\subset\mathbb R^2$, also known as "topological chicken" (draw a picture and you will understand the name). It is compact Hausdorff, but not locally pathconnected. Therefore $0=\pi_1(X)=H_1(X)=H^1_{\mathrm{sing}}(X)$. On the other hand, $X$ separates $\mathbb R^2$, so one can deduce that $H_{\mathrm{sh}}(X;\underline{\mathbb Z})\cong\mathbb Z$. Or using Neil's argument, there is a nontrivial map $X\to S^1$ generating $[X,S^1]\cong\mathbb Z$ (which would be "representable cohomology" in an adequat model structure). | |
| Apr 28, 2016 at 19:55 | comment | added | Denis Nardin | I believe that the equivalence between (2) and (3) is done in Godement's Topologie Algébrique et Théorie de Faisceaux for paracompact Hausdorff spaces | |
| Apr 28, 2016 at 19:24 | answer | added | Neil Strickland | timeline score: 7 | |
| Apr 28, 2016 at 17:34 | comment | added | Gregory Arone | This paper arxiv.org/abs/1602.06674 says that it is a classical result for locally contractible paracompact $X$, and it purports to remove the paracompactness assumption. | |
| Apr 28, 2016 at 17:08 | answer | added | user19475 | timeline score: 2 | |
| Apr 28, 2016 at 16:49 | history | asked | asv | CC BY-SA 3.0 |