Timeline for Are there 'cohomology' functors that respect all Eilenberg-Steenrod axioms except homotopy invariance?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2014 at 12:46 | answer | added | Dennis Sullivan | timeline score: 13 | |
| Dec 31, 2011 at 1:14 | comment | added | Minhyong Kim | Fernando: I'm sure you know all this, but perhaps I should be a bit more careful so as not to mislead anyone. The covariant functoriality I was referring to does indeed use duality, and hence, changes degree. Thus, I suppose any given $H^n_c$ is not even covariant functorial. As you mention, the preservation of degrees only happens when one identifies with homology. On the other hand, for these reasons, any fixed $H^n_c$ is not a homotopy invariant, in spite of duality. Anyways, all these subtleties was the reason for my original qualification that $H^n_c$ is only a 'casual' cohomology. | |
| Dec 30, 2011 at 14:51 | comment | added | Fernando Muro | @Minhyong For oriented manifolds the covariant version is just homology by Poicaré duality, which is functorial and also homotopy invariant, so it does not produce a counterexample. And the contravariant version is not functorial, just think of $\mathbb{R}\rightarrow S^1$. | |
| Dec 30, 2011 at 10:49 | comment | added | Minhyong Kim | Fernando: I should have written oriented manifolds, I guess. Is this your objection? Or am I missing something even more basic? | |
| Dec 30, 2011 at 9:08 | comment | added | Fernando Muro | @Minhyong It's not functorial for manifolds either | |
| Dec 30, 2011 at 6:31 | answer | added | Tyler Lawson | timeline score: 8 | |
| Dec 30, 2011 at 1:02 | comment | added | Minhyong Kim | Well, I guess it's covariant functorial for manifolds. Compact support cohomology should be thought of as closer to homology than cohomology, I think. This reminds me: intersection (co)homology is not a homotopy invariant either. | |
| Dec 29, 2011 at 23:01 | comment | added | Fernando Muro | Cohomology with compact support is not functorial either... unless you consider proper maps, but then it's proper homotopy invariant! | |
| Dec 29, 2011 at 22:39 | comment | added | Qfwfq | Yes, that is the kind of example I was expecting. | |
| Dec 29, 2011 at 21:57 | comment | added | Minhyong Kim | I don't think I've ever checked how it relates to the Eilenberg-Steenrod axioms, but one natural cohomology theory (in the slightly more casual sense) that is not homotopy invariant is cohomology with compact support. This lack of invariance has nice applications to proofs of natural facts, for example, that the tangent bundle of the sphere is not homeomorphic to the trivial bundle. | |
| Dec 29, 2011 at 20:28 | answer | added | Steven Landsburg | timeline score: 8 | |
| Dec 29, 2011 at 18:44 | history | edited | Qfwfq | CC BY-SA 3.0 |
added 98 characters in body; added 11 characters in body
|
| Dec 29, 2011 at 18:35 | history | asked | Qfwfq | CC BY-SA 3.0 |