# Getting the most general form of Mayer-Vietoris from the Eilenberg-Steenrod axioms

I asked this question a while ago on MSE, 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...

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?

• (Homotopy) colimits commute with (homotopy) colimits so the answer should be 'yes' (modulo me not knowing point-set topology enough to confidently assert anything is an 'excisive triad'). – Dylan Wilson Oct 8 '17 at 18:57
• (You have a pushout of (* <--- * ---> *) shaped diagrams and you'd like to know if the colimit of the pushout of diagrams is the pushout of the colimits, and the answer is yes.) – Dylan Wilson Oct 8 '17 at 18:59
• I think what you want is actually in the original Eilenberg-Steenrod book. Look at the discussion of Mayer-Vietoris in Section 15. You can find the pdf here: www.maths.ed.ac.uk/~aar/papers/eilestee.pdf – Dan Ramras Oct 9 '17 at 15:33

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.)
• So let me get this straight: First new (to me) lemma is "The two possible iterated mapping cones of a commutative square are homoemorphic" and similarly, "the various iterated mapping cones of a commutative cube are homoemorphic". The second lemma is a special case of the five-lemma: "$(Y,Y_1,Y_2)$ is excisive iff the double mapping cone of the square $\begin{smallmatrix}Y_1&\to&Y\\\uparrow&&\uparrow\\Y_{12}&\to&Y_2\end{smallmatrix}$ is acyclic." The third lemma is also an application of the five-lemma: If $D\subseteq D'$ are two acyclic spaces, the mapping cone $C_D^{D'}$ is also acyclic." – Johannes Hahn Dec 17 '18 at 19:00