# How can I prove that the derived couple of the homotopy exact couple is an invariant?

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...

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

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.
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