I asked this question a while ago [on MSE](https://math.stackexchange.com/questions/1596348/getting-the-most-general-form-of-mayer-vietoris-from-the-axioms-of-homology), got no answer, put a bounty on it, still got no answer, was advised to ask here instead, hesitated, forgot about the question for a while and now remembered it. I'm still not sure if it really qualifies as research level. I let you decide...




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I'd like to derive the most general form of the Mayer-Vietoris sequence from the Eilenberg-Steenrod axioms for homology (in particular: I do not want to use the definition of $H\_\ast(X)$ in terms of simplices). By that I mean the existence of an exact sequence of the form
$\cdots \to H\_n(X\_{12},A\_{12})\to H\_n(X\_1,A\_1)\oplus H\_n(X\_2,A\_2)\to H\_n(X,A) \to H\_{n-1}(X\_{12},A\_{12})\to\cdots$
whenever $(A,A\_1,A\_2)\subseteq(X,X\_1,X\_2)$ are two excisive triads and $A\_{12}:=A\_1\cap A\_2$, $X\_{12}:=X\_1\cap X\_2$.


It is easy to do that in the special case $A=A\_1=A\_2$ by looking at the two long exact sequences for pairs from the inclusions $A\subseteq X\_{12}\subseteq X\_1$ and $A\subseteq X\_2\subseteq X$ respectively. These two sequences form a Barratt-Whitehead ladder and the lemma of Mayer-Vietoris applies.


Basically the same approach works in the special case $X=X\_1=X\_2$.


Both of these proofs are well known in the literatur, but so far I was unable to find a proof for the general version either in the books or myself that did not go through the realisation of $H\_\ast$ as the homology of some chain-complex generated by simplices.


All my previous attempts consisted of doodling one diagram after the other, but may be there is a simpler solution. After realising that $H\_\ast(X,A) = \tilde{H}\_\ast(C\_A^X)$, where $C\_A^X$ is the mapping cone of the inclusion $A\to X$, one could also ask whether $(C\_A^X, C\_{A\_1}^{X\_1}, C\_{A\_2}^{X\_2})$ is an excisive triad. So I'm asking: Is it?



Getting the most general form of Mayer-Vietoris from the Eilenberg-Steenrod axioms
If you are willing to work with mapping cones, then this follows from looking at the triple (= threefold iterated) mapping cone for the cube with vertices $A\_{12} = A\_1 \cap A\_2$, $A\_1$, $A\_2$, $A$, $X\_{12} = X\_1 \cap X\_2$, $X\_1$, $X\_2$ and $X$ in two different ways.


Let us use your notation $C\_A^X = X \cup\_A CA$, so that there is a natural isomorphism $H\_\*(X, A) \cong \tilde H\_\*(C\_A^X)$. If $(A, A\_1, A\_2)$ is excisive, then the double (= twofold iterated) mapping cone for the square with vertices $A\_{12}$, $A\_1$, $A\_2$ and $A$ has the homology of a point. Likewise, if $(X, X\_1, X\_2)$ is excisive, then the double mapping cone for the square with vertices $X\_{12}$, $X\_1$, $X\_2$ and $X$ has the homology of a point. Thus the triple mapping cone for the cube has the homology of a point. This is homeomorphic to the double mapping cone for the square with vertices $C\_{A\_{12}}^{X\_{12}}$, $C\_{A\_1}^{X\_1}$, $C\_{A\_2}^{X\_2}$ and $C\_A^X$. Since $C\_{A\_{12}}^{X\_{12}} = C\_{A\_1}^{X\_1} \cap C\_{A\_2}^{X\_2}$, this shows that $(C\_A^X, C\_{A\_1}^{X\_1}, C\_{A\_2}^{X\_2})$ is excisive, and gives you the exact Mayer-Vietoris sequence
$$
\dots \overset{\partial}\to H\_n(X\_{12}, A\_{12}) \to H\_n(X\_1, A\_1) \oplus H\_n(X\_2, A\_2) \to H\_n(X, A) \overset{\partial}\to \dots
$$
by the Barratt-Whitehead lemma. (Note the spelling of Michael Barratt's name.)


