MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone tell me if there is a Mayer-Vietoris sequence for an arbitrary homotopy pushout (hence homotopy pullback) of spectra and an arbitrary (co)homology theory. If this comes from some easy way of writing down a pushout/pullback as a fiber sequence, it'd be really cool to see that spelled out (as if I were a baby). Also, it would also be really neat to know if there are general conditions on a model category or on the (co)homology theory to make this true.

Thanks in advance. :-)

share|cite|improve this question
Maybe I'm being silly... but having a homotopy cartesian square should maybe be the same as having a (co)fiber sequence like $A \rightarrow B \vee C \rightarrow D$ with appropriate maps... the associated long exact sequence would be what you wanted. But I think I'm overlooking something, because that seems too easy :) – Dylan Wilson Mar 1 '13 at 4:06
(if this argument works it would work for any stable model category/ stable infty-category/ nice enough triangulated category.) – Dylan Wilson Mar 1 '13 at 4:07
Thanks everyone for your answers. I understand this is an elementary question and perhaps MO was not an appropriate place to bring it up. But there seem to be a lot of simple facts in algebraic topology that are known about spaces and that people feel are obviously true about spectra, but struggle to completely and fully "prove" with all the relevant diagrams when pressed for it. Perhaps this is not one of those things, I don't know. As always, from a humble neophyte, thanks. – Jon Beardsley Mar 1 '13 at 19:50
up vote 3 down vote accepted

Mayer-Vietoris sequences can be obtained from excision isomorphisms.

Anything worthy of the name "homology theory" will give a long exact sequence $$\dots \to h_n(A)\to h_n(X)\to h_n(A\to X)\to h_{n-1}(A)\to \dots$$ for each morphism $A\to X$. And for a square, a.k.a. map of morphisms $(C\to B)\to (A\to X)$, it will give you a map from the long exact sequence of $C\to B$ to that of $A\to X$.

If the square happens to induce an isomorphism $h_n(C\to B)\to h_n(A\to X)$ for every $n$ (as it will if it is a homotopy pushout square) then the desired map $h_n(X)\to h_{n-1}(C)$ is given by composing $h_n(X)\to h_n(A\to X)\cong h_n(C\to B)\to h_{n-1}(C)$, and a little diagram chase gives you the exactness you want.

Depending on how one axiomatizes the notion of homology theory, the fact that homotopy pushout squares gives isomorphisms in relative homology is either an axiom or a consequence of the axioms.

There is no need to work with spectra or other stable objects here.

Note that if you had something like a homology theory except that it gave isomorphisms $h_n(C\to B)\to h_n(A\to X)$ for homotopy pullbacks instead of for homotopy pushouts then you would get a "Mayer-Vietoris sequence" for pullback squares. That's how it is for homotopy groups and based spaces, except that things get funny down around $\pi_1$ and relative $\pi_2$.

share|cite|improve this answer
Just to be fair, the OP was asking about homotopy push-pulls of spectra, so I didn't think it was too superfluous to work in that realm... Also does h(A --->X) mean something like the homology of the pair after replacing the map with a cofibration? (just making sure I understand) – Dylan Wilson Mar 1 '13 at 15:48
Yes. We think of homology of pairs, and sometimes we replace pairs by good pairs (where the inclusion is a cofibration), but we still define homology for all pairs (and a map of pairs that's a weak equivalence of both big spaces and subspaces induces an iso); so why not just define relative homology for all maps instead of just for inclusion maps? Sometimes "inclusion" doesn't make sense in abstract setting, and even if it does our habit is to define relative homology for all pairs not just nice ones, so why not for all maps rather than just cofibrations? – Tom Goodwillie Mar 1 '13 at 17:21
Excellent point :) – Dylan Wilson Mar 1 '13 at 18:14

I didn't want to answer this because the question seemed too elementary to spend time on. But to see quasicategories invoked for something so classically elementary is truly painful. (Forgive me Dylan). Jon asked "to see that spelled out (as if I were a baby)". Here goes. The question was about spectra, and you place yourself in any halfway reasonable category of such. If you are really old-fashioned your objects are CW spectra (alias cofibrant), but to have a homotopically meaningful construction we may as well use CW approximation or model theoretic cofibrant approximation to assume that $A$, $B$, and $C$ are CW or cofibrant. We have standard cylinders $A\wedge I_+$ and form the double mapping cylinder (explicit homotopy pushout) $$D = B \cup (A\wedge I_+) \cup C$$ with cofibration (yes, I know, that requires just a smidgen of trivial verbiage) $B\vee C \longrightarrow D$ with quotient (yes, there is no problem with quotients) $D\longrightarrow \Sigma A$. You are in a stable situation so the cofiber sequence extends to the left, but you can also just look at its extension $\Sigma A\longrightarrow \Sigma B\vee \Sigma C$ to the right to identify the maps. This is all exactly as if you were just in based spaces, except that in a stable situation cofiber sequences are fiber sequences. Nothing at all fancy, and the argument adapts very easily to any stable model category.

Dylan, you asked about classical excision; one way that goes is to use CW approximation to replace an excisive triad $(X;A,B)$ by a CW triad; then, with $C=A\cap B$, $X/C \cong A/C\vee B/C$. (Concise, page 110).

share|cite|improve this answer
No apology necessary :) I was just being lazy. Thank you for giving the model-theoretic proof. – Dylan Wilson Mar 1 '13 at 15:13

Ok, I'll state this a bit more confidently... (but I'm still worried I'm missing something).

Any time we're in, say, a stable model category where every object is cofibrant (or stable $\infty$-category), then given an object $X$ and a homotopy push-pull diagram (which I won't draw), involving $A \rightarrow B$, $A \rightarrow C$ and $C, B \rightarrow D$ we get a long exact sequence like $$ \cdots \rightarrow [D, X] \rightarrow [B, X]\oplus [C, X] \rightarrow [A, X] \rightarrow [\Sigma^{-1} D, X] \rightarrow\cdots $$ Indeed, we have a homotopy push-pull as above precisely if we have a (co)fiber sequence $$ A \rightarrow B \oplus C \rightarrow D $$ where the first map is the difference of the two obvious ones. There's probably a good way to do this without being fancy, but the easiest way I see to do this is in the setting of $\infty$-categories: the pushout and cofiber displayed above both have manifestly the same universal property. (I think I'm using cofibrancy here to say that the coproduct and homotopy coproduct should agree... maybe I don't need this- I'm bad with model categories, someone should correct me.)

Anyway, this must be in one of the obvious references. Adams? Neeman's book on triangulated categories? Something like that.

It's not immediately obvious to me that this specializes to Mayer-Vietoris, but that's me revealing too much of my ignorance. Inclusions of open subsets don't seem to be cofibrations, so why should the usual square we write down be a homotopy pullback/pushout (after taking suspension spectra)?

share|cite|improve this answer
Hey Dylan! The square for the Mayer-Vietoris sequence is in fact a homotopy pushout even before stabilizing. That's not a completely trivial fact, however, as it immediately implies the standard Seifert-van Kampen theorem. For an elementary proof see Theorem A.3.1 in Higher Algebra. – Marc Hoyois Mar 1 '13 at 5:19
Aha! Thanks Marc :) – Dylan Wilson Mar 1 '13 at 6:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.