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Assume $H_n$ is a covariant homotopy functor on the category of locally compact Hausdorff spaces which has the Mayer-Vietoris property: whenever $X$ is the union of two closed subspaces $A$ and $B$ there is a long exact sequence

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

I want to prove an excision theorem in $H_n$. 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 $X$. But in case I made a mistake in the formulation, I'm really just looking for an explanation of one direction of the slogan "the Mayer-Vietoris sequence is equivalent to the excision theorem". I already know how to do this in ordinary homology theory (i.e. when the relevant long exact sequences come from short exact sequences of chain complexes), but I need to understand how it works when chain complexes are unavailable.

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I assume $H_n$ is a sequence of functors from the homotopy category of locally compact Hausdorff spaces to the category of abelian groups. Do you want the long exact sequence to be functorial in $X, A, B$? – Qiaochu Yuan May 19 '12 at 20:17
Yes on all accounts. – Paul Siegel May 19 '12 at 20:32
This may or may not be helpful, but the other direction (excision implies Mayer-Vietoris) is dealt with axiomatically in the original Eilenberg-Steenrod book. Pleasantly, you can find a searchable .pdf of the book here: – Dan Ramras May 21 '12 at 3:11
I have been able to find other references for "excision implies Mayer-Vietoris" (for example Hatcher does it), but the argument doesn't seem to me to be reversible. Maybe I'm just missing something. – Paul Siegel May 21 '12 at 12:07
I guess my point was that Eilenberg-Steenrod doesn't use chain complexes whereas Hatcher does. But I have no reason to think the E-S argument can be reversed... – Dan Ramras May 22 '12 at 7:03

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