Timeline for Mayer-Vietoris implies Excision
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| May 23, 2012 at 2:34 | history | edited | Paul Siegel | CC BY-SA 3.0 |
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| May 22, 2012 at 13:08 | comment | added | Steven Landsburg | Given a pullback square of commutative rings, one has a Mayer-Vietoris sequence for lower K-theory that results from an excision isomorphism of relative K-groups. I once had occasion to ask whether there could be pullback squares with "Mayer-Vietoris" sequences that arise "accidentally" from some map not defined via excision. I started calling such a situation a "Mayer-Vietoris accident". Bob Thomason found this phrase hilarious and used it every chance he could get. If you find yourself in an analogous situation, I hope that in Bob's memory you'll call it a Mayer-Vietoris accident. | |
| May 22, 2012 at 10:39 | comment | added | Sergey Melikhov | @unknown (google): notice that Paul says that his $H_*$ is a functor on spaces, not pairs of spaces. In this context usual excision doesn't make sense, so it's reasonable to use "excision" to refer something else (which is not exactly the long exact sequence of the pair). I wonder however what he means when he assumes his $H_*$ satisfies usual Elienberg-Stenrod axioms except excision, because those do include the long exact sequence of the pair which needs $H_*$ to be a functor on pairs of spaces, not single spaces. | |
| May 22, 2012 at 10:22 | comment | added | Sergey Melikhov | The long exact sequence $H_n(A)\to H_n(X)\to H_n(X−A)\to H_{n−1}(A)$ does not hold for usual homology theories (say if $X$ is an $n$-ball, $A$ its boundary sphere and $H_*$ is ordinary homology). It does hold for locally-finite homology, which is not homotopy invariant (but is an invariant of proper homotopy). So you might want to edit the question, e.g. so as to eliminate homotopy invariance from your assumption of usual Eilenberg-Steenrod axioms apart from excision. | |
| May 22, 2012 at 9:08 | answer | added | Lennart Meier | timeline score: 9 | |
| May 22, 2012 at 7:09 | comment | added | Dan Ramras | I wrote something a little more detailed here: mathoverflow.net/questions/23175/… | |
| May 22, 2012 at 7:08 | comment | added | Dan Ramras | One thought: ideally your Mayer-Vietoris sequence ought to come from some sort of representability statement, and the same statement ought to give long exact sequences for pairs. I guess what I mean is that Mayer-Vietoris sequences come from homotopy pullback diagrams. When you have a (homotopy) pushout of spaces, and you map them all into something, you get a (homotopy) pullback and hence a M-V sequence. On the other hand, Puppe sequences give you a method for obtaining long exact sequences for pairs when dealing with a representable functor. No idea if this would help in your situation... | |
| May 22, 2012 at 4:29 | comment | added | John Pardon | On the other hand, many people (incorrectly) use "excision" to refer to the long exact sequence of the pair, and it's a bit more reasonable to replace this with the Mayer--Vietoris sequence than to replace excision with the Mayer--Vietoris sequence. So did you mean the long exact sequence of the pair instead of excision? | |
| May 22, 2012 at 4:28 | comment | added | John Pardon | Excision is the statement that $H_n(X,A)=H_n(X-U,A-U)$ for $U\subseteq A\subseteq X$ sufficiently nice (it doesn't involve any long exact sequence). So I'm not sure what you mean by the statement "excision takes the form of a long exact sequence . . .". | |
| May 22, 2012 at 4:25 | answer | added | Steven Landsburg | timeline score: 7 | |
| May 22, 2012 at 3:26 | history | asked | Paul Siegel | CC BY-SA 3.0 |