Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is going to basically be a cross-post of a MSE question: http://math.stackexchange.com/questions/147146/mayer-vietoris-implies-excision. I suspect that the answer to this question will turn out to be fairly straightforward, but (a) I can't figure it out myself, (b) it's not in any of the obvious references, (c) it hasn't gotten any answers on MSE, and (d) the question honestly is relevant to my research.

Let $H_n$ be a sequence of functors from the category of locally compact Hausdorff spaces to the category of abelian groups which satisfy all of the usual Eilenberg-Steenrod axioms for a generalized homology theory, except the excision axiom is replaced by a Mayer-Vietoris axiom: whenever $X$ is the union of two closed subspaces $A$ and $B$ there is a (functorial) exact sequence

$$\to H_n(A \cap B) \to H_n(A) \oplus H_n(B) \to H_n(X) \to H_{n-1}(A \cap B) \to$$

According to the slogan "Mayer-Vietoris is equivalent to excision", one ought to be able to prove an excision theorem for $H_n$. This probably takes the form of a long exact sequence

$$\to H_n(A) \to H_n(X) \to H_n(X - A) \to H_{n-1}(A) \to$$

where $A$ is a closed subset of $X$, though if this isn't the right formulation I welcome corrections. I know how to go the other direction (excision implies Mayer-Vietoris), but this direction eludes me. Can anyone help?

EDIT: I guess I was a bit sloppy about my use of the term "Eilenberg-Steenrod Axioms" which require that $H_n$ be a functor of pairs. So let me be more precise. We'll work in the category of pointed locally compact (second countable if desired) Hausdorff spaces where the morphisms between $X$ and $Y$ are basepoint preserving continuous maps from the one point compactification of $X$ to the one point compactification of $Y$. Define a homotopy of maps between $X$ and $Y$ to be a morphism between $X \times [0,1]$ and $Y$.

I will declare that a sequence of covariant functors $H_n$ from this category to the category of abelian groups defines a generalized homology theory if:

  • Each $H_n$ is a homotopy functor

  • There is a functorial long exact sequence of the form above associated to any closed subset $A \subseteq X$.

Under these circumstances there is a Mayer-Vietoris sequence associated to any decomposition of $X$ into closed subspaces. A precise statement of my question is: if we replace the second axiom with a Mayer-Vietoris axiom, can we reconstruct the long exact sequence associated to a single closed subset?

However, I am flexible. In the setting that motivates this question, I really do have a notion of homology of pairs, so I would really be happy if somebody could deduce any version of the excision theorem from the existence of a Mayer-Vietoris sequence (perhaps with some extra conditions on the boundary map if it helps).

share|improve this question
1  
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 . . .". –  John Pardon May 22 '12 at 4:28
2  
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? –  John Pardon May 22 '12 at 4:29
1  
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... –  Dan Ramras May 22 '12 at 7:08
    
I wrote something a little more detailed here: mathoverflow.net/questions/23175/… –  Dan Ramras May 22 '12 at 7:09
1  
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. –  Sergey Melikhov May 22 '12 at 10:22
show 2 more comments

2 Answers 2

I'd have taken "excision" to mean that the map $H_n(A,A\cap B)\rightarrow H_n(X,B)$ is an isomorphism.

I'd have taken "Mayer-Vietoris" to mean not just the exactness of some sequence $$\cdots\rightarrow H_n(A\cap B)\rightarrow H_n(A)\oplus H_n(B)\rightarrow H_n(X)\rightarrow H_{n-1}(A\cap B)\rightarrow\cdots$$ but the exactness of the particular sequence you get when the map $H_n(X)\rightarrow H_{n-1}(A\cap B)$ is defined by composing first the natural map from $H_n(X)\rightarrow H_n(X,B)$, then the inverse of the excision isomorphism, and then the natural map $H_{n}(A,A\cap B)\rightarrow H_{n-1}(A\cap B)$.

Given that, Mayer-Vietoris implies excision in the sense that the very statement of the Mayer-Vietoris property makes no sense in the absence of the excision isomorphism.

This does raise the question of whether a "Mayer-Vietoris type exact sequence" could occur in the absence of excision, with some other functorial map $H_n(X)\rightarrow H_{n-1}(A\cap B)$ replacing the one defined via excision. I'm not sure whether we'd still want to call this a Mayer-Vietoris sequence.

share|improve this answer
    
I see your point. Interestingly enough, in the setting that motivated this question I have a functorial map $H_n(X) \to H_{n-1}(A \cap B)$ which fits into a Mayer-Vietoris-like sequence but I am having trouble proving the excision theorem directly and I was hoping there were some formal manipulations that would do the job. I'm working in a category a bit different from LCH, and probably I just need to think harder about that category. Thanks for the insight. –  Paul Siegel May 22 '12 at 4:56
1  
Are you in the end interested in arbitrary spaces or only in CW-complexes? –  Fabian Lenhardt May 22 '12 at 7:54
    
In the category that really interests me, the objects are proper separable metric spaces and the morphisms are basically continuous quasi-isometries. I'm happy to assume that the proper separable metric spaces are also CW-complexes, but this might not help as much as usual since the morphisms are a little weird. –  Paul Siegel May 23 '12 at 2:40
add comment

I must admit, I don't know whether you can just replace excision by Mayer-Vietoris in the usual Eilenberg-Steenrod axioms. But you can do something slightly different:

While the Eilenberg-Steenrod axioms speak of homology of pairs of spaces, you can define a homology theory on spaces also as a sequence of homotopy invariant functors $h_n$ fulfilling a Mayer-Vietoris sequence (as Steven pointed out, you have just some functorial map $h_n(X) \to h_{n-1}(A\cap B)$, which replace as a datum the boundary maps of the long exact sequence of a pair). Then you define the homology of a pair $(X,A)$ as $h_n(X,A) := \ker(h_n(C_AX) \to h_n(pt))$. Here, $C_A X$ stands for the mapping cone of the inclusion $A\hookrightarrow X$. The Mayer-Vietoris sequence (based on the covering $CA \cup X = C_AX$) implies the long exact sequence of the pair. Excision (i.e. $h_n(X,A) \cong h_n(X-B, A-B)$ for $\overline{B} \subset A^\circ$) is a homotopy fact in the sense that the inclusion $C_{A-B}(X-B) \to C_A X$ is a homotopy equivalence. I must admit, I didn't check the last point in detail, but it should be at least true if $B\to A$ is a closed cofibration and in general you might replace the inclusion via a cylinder construction by a closed cofibration.

Thus, you get a different axiomatization of a homology theory using the Mayer-Vietoris sequence [There is also a third axiomatization of a homology theory, using pointed spaces as input and having a sequence for the cone and a suspension isomorphism as axioms]. Of course, if you have already a functor defined on pairs of spaces, you have to check that something like $h_n(X,A) = h_n(C_AX, pt)$ holds.

I want to comment, I know this "absolute" axiomatization primarly from bordism theory, where you can show Mayer-Vietoris quite easily.

share|improve this answer
    
This is very promising! The specific functor I care about actually is defined on pairs, and I think it is even true that $h_n(X,A) = h_n(C_A X, pt)$ as you stipulate. I'll try this idea... –  Paul Siegel May 23 '12 at 2:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.