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As the title suggests, is it possible to deduce the Mayer-Vietoris sequence from the Eilenberg-Steenrod axioms alone? By this I mean without appealing some kind of barycentric subdivision argument. Most of the statements of the Mayer-Vietoris sequence I have seen are for a space $X$ which is covered by the interiors $A^{\circ}$ and $B^{\circ}$ of some subsets $A$ and $B$, so I expect you would need to restate the excision axiom to say that the inclusion $(B, A \cap B) \hookrightarrow(X, A)$ induces isomorphisms $H_{n}(B, A \cap B) \rightarrow H_{n}(X,A)$ for all $n$. Even with this, I am not able to get Mayer-Vietoris without repeating the long tedious arguments found in, for example, Hatcher. I was under the impression that an axiomatic approach to Homology was developed for precisely this kind of purpose.

Thanks

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This is a corollary of exercise $2.2.38$ of Hatcher (I'll write down a proof of that if the OP wishes so).

Note that if a CW compelx $X$ can be decomposed as $A \cup B$ in terms of two of it's subcomplexes, then the inclusion of pairs $(A, A \cap B) \to (X, B)$ gives a morphism of long exact sequences by naturality of the snake map:

$$\require{AMScd} \begin{CD} H_{n+1}(X, B) @>>> H_n(B) @>>> H_n(X) @>>> H_n(X, B) \\ @AAA @AAA @AAA @AAA \\ H_{n+1}(A, A \cap B) @>>> H_n(A \cap B) @>>> H_n(A) @>>> H_n(A, A \cap B) \end{CD}$$

The first and the third map are isomorphisms as $X/B \cong A/A \cap B$. By the homological algebra exercise in $38$, you'd be done.

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