Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm working on (yet) an(other) exercise from Mosher & Tangora's "Cohomology Operations and Applications to Homotopy Theory". This one is about the homotopy exact couple, which is defined for a complex $K$ by $D_{p,q}=\pi_{p+q}(K^p)$ and $E_{p,q}=\pi_{p+q}(K^p,K^{p-1})$. So that we have relative Hurewicz, we assume K to be simply connected. As stated in the title, the object of the exercise is to show that this is not a homotopy invariant but that its derived couple is.

The motivating example I've got in my head (let me know if you've got a better one) is $S^2$ realized either with 1 vertex, 1 edge, and 2 faces, or with 1 edge and 1 face. This already easily proves that the homotopy exact couple itself is not an invariant. For the harder part, I've drawn the (presumably standard) grid with rows like $\cdots \rightarrow D_{p,q} \rightarrow E_{p,q} \rightarrow D_{p-1,q} \rightarrow \cdots$ connected by vertical inclusion maps $D_{p,q} \rightarrow D_{p+1,q-1}$, and I can see how these both give the same derived couple, but I'm having trouble figuring out exactly how to make this into a general argument. I begin with a homotopy equivalence $f:K \rightarrow L$, $g:L \rightarrow K$, and I can assume these maps are cellular so I get induced maps between all corresponding groups of the homotopy exact couples associated to $K$ and $L$. But what can I say about these maps? Clearly from my motivating example the restrictions to skeleta need not be homotopy equivalences, or even anything close. I'm pretty sure they commute with the intra-couple maps, but I haven't had any success pushing through the commutative algebra with that fact alone. It smells like obstruction theory should be involved here since in general you'll need to move $K^p$ through $K^{p+1}$ to realize the homotopy $gf\simeq 1_K$ (consider it as a map $K\times I \rightarrow K$, which can be assumed to be cellular), but I don't think I understand it well enough to see how (or if that's even true, I guess). Am I headed in the right direction?


P.S. I'm camping right now so I typed all of this on my phone. Might this be a first for MO? Or have people been asking math questions from their phones since before I was born...

share|improve this question
    
People used to write letters before you were born. –  supercooldave Jul 5 '10 at 8:58
4  
Obligatory xkcd link: xkcd.com/378 –  Mariano Suárez-Alvarez Jul 5 '10 at 8:59

1 Answer 1

up vote 3 down vote accepted

Let me start by making a definition: an $n$-skeleton of a space $X$ is an $n$-equivalence $X_n \to X$, where $X_n$ is an $n$-dimensional (at most) CW complex ($X$ itself need not be a CW complex). Obviously, $n$-skeleta are not unique, but any two $n$-skeleta for the same space factor through one another: there are compositions $X_n' \to X_n \to X$ and $X_n \to X_n' \to X$.

Let's concentrate on the $D$s. By definition, $D^2_{p,q} = \mathrm{im}( \pi_{p+q} (K_q) \to \pi_{p+q}( K_{q+1}))$. Any two $q$- and $(q+1)$-skeleta $K_q\to K_{q+1}\to K$ and ${K_q}'\to K_{q+1}' \to K$ factor through one another, so
$\mathrm{im}( \pi_{p+q} (K_q) \to \pi_{p+q}( K_{q+1})) \cong \mathrm{im}( \pi_{p+q} ({K_q}') \to \pi_{p+q}( K_{q+1}'))$.

This shows that $D^2_{p,q}$ is independent of the choice of CW decomposition. The isomorphism of the $E$-groups follows by the Five lemma.

EDIT: Of course the last bit of the second paragraph was ridiculous, and unnecessary; fixed now.

share|improve this answer
    
Thanks! So I assume that when you say that $n$-skeleta factor through each other, you mean up to homotopy -- this is basically the same as saying that the hom-eq maps can be chosen to be cellular, right? Also, I think at the end of your second paragraph you need $p\leq 0$ otherwise that's not an iso. But in any case, since as I said you only need the $(p+1)$-skeleton to get the $p$-skeleton in the right place, then we can get a commutative-up-to-homotopy square with $K^p \leftrightarrow L^p$ included into $K^{p+1}\leftrightarrow L^{p+1}$, which (I think) should prove it... –  Aaron Mazel-Gee Jul 5 '10 at 18:39

Your Answer

 
discard

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.