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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?

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    $\begingroup$ (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'). $\endgroup$
    – Dylan Wilson
    Oct 8, 2017 at 18:57
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    $\begingroup$ (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.) $\endgroup$
    – Dylan Wilson
    Oct 8, 2017 at 18:59
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    $\begingroup$ 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 $\endgroup$
    – Dan Ramras
    Oct 9, 2017 at 15:33
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    $\begingroup$ There is an article by Richard Steiner, similar in spirit to your approach, written exactly for this purpose. $\endgroup$
    – Ken
    Jul 19, 2021 at 5:48

1 Answer 1

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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.)

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  • $\begingroup$ PS: This is basically the archaic formulation of Dylan Wilson's comments from October 2017. $\endgroup$
    – John Rognes
    Dec 17, 2018 at 1:51
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    $\begingroup$ 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." $\endgroup$
    – Johannes Hahn
    Dec 17, 2018 at 19:00
  • $\begingroup$ You could take the second lemma to be the definition of what it means for a triple to be excisive. The third lemma is simpler than the five-lemma; it just uses the long exact sequence of (D',D). Regarding the first lemma(ta) you may object that forming mapping cones might take you out of an "admissible category for homology theory", as on page 5 of Eilenberg-Steenrod. Are you looking for a proof that refers to only the pairs of spaces you can form from A_{12}, A_1, A_2, A, X_{12}, X_1, X_2, X and the empty space? $\endgroup$
    – John Rognes
    Dec 17, 2018 at 20:12
  • $\begingroup$ I'm quite happy with a proof that works with mapping cones. Do you know of any proof that relies solely on the long exact sequences for pairs? $\endgroup$
    – Johannes Hahn
    Dec 18, 2018 at 0:36

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