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one can define cellular homology by letting $C_n(X)=\{\mathbb{S}^n, X^n/X^{n-1}\}$, where $X$ is a CW complex and the curly brackets mean stable homotopy classes of maps. Now the differential of the resulting complex is supposed to be given by a map $X^n/X^{n-1}\to\Sigma X^{n-1}/X^{n-2}$ and I struggle to understand this map. The only candidate I can think of would be the suspension of an attaching map. More precisely, on an $n$-cell, we use the inverse of the characteristic map to end up in $\mathbb{S}^n$, identify this with $\Sigma\mathbb{S}^{n-1}$ via a (canonical) homeomorphism, then apply the attaching map to end up in $\Sigma X^{n-1}$. If this is correct (is it?), I still do not see why this is indeed a differential, i.e. d^2=0.

Furthermore, I would like to do concrete calculations with this formulation, if necessary only in very low dimensions (say, 1 or 2). So in particular I would like to know how the ordinary cellular boundary operator can be recovered from the fancy one above. Does anyone know a detailed reference for this view on cellular homology or can provide any useful insights?

Thank you.

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You ask for a detailed reference. Do you know of any references that mention this perspective, maybe not in detail? I haven't seen it before. It seems to me that you have attaching maps $S^{n-1} \to X^{n-1}/X^{n-2}$ and you probably just want to suspend these and wedge them together (and identify the domain with $X^n/X^{n-1}$). I think this is the same as what you wrote? The fact that you get a differential should be checked just as in the case of the ordinary cellular boundary map, e.g. in Hatcher's book. Does this not work? –  Dan Ramras Jun 14 '12 at 18:29
    
This formulation is used in the paper "Equivariant ordinary homology and cohomology" by Costenoble and Waner, arxiv.org/pdf/math/0310237 to define equivariant cellular chains and is handled as if it were common knowledge. My motivation to use this boundary stems from the fact that I try to avoid orientation assumptions on the cells. In Bredons book, the cellular boundary is shown to be a boundary using mapping degrees and therefore orientation theory. I think (gonna check this), that Hatcher does the same thing. I am trying to understand this purely from the stable homotopy viewpoint. –  DanielW Jun 15 '12 at 9:32
    
Unfortunately I can't edit my comment any more... You find the definition on page 14 of the named paper. –  DanielW Jun 15 '12 at 9:40
    
Why are you using stable homotopy theory, though? The suspension homomorphisms $\pi_n(S^n) \rightarrow \pi_{n+1}(S^{n+1})$ are isomorphisms for $n\geq 1$, so really except for the bottom case there's no reason to introduce this language. –  Aaron Mazel-Gee Jun 15 '12 at 16:58
    
Of course, without equivariance one could replace stable homotopy groups by homotopy groups, but this does not help with the problem, does it? My main problem is to figure out why the map yields a boundary and how one can actually compute that boundary in terms of generators (without going back to the classical cellular description). –  DanielW Jun 16 '12 at 1:07
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1 Answer

I am not completely clear on the motivation of your question, but I think I can answer the question as-asked. In my opinion, this is easiest to see with a complicated diagram. Start with the inclusions of subcomplexes: $$\begin{array}{ccccccccccc}X^1 & \to & X^2 & \to & \cdots & \to & X^{n-1} & \to & X^n & \to & X^{n+1} & \to & \cdots\end{array},$$ and then throw in each quotient map dangling off: $$\begin{array}{ccccccccccc}X^1 & \to & X^2 & \to & \cdots & \to & X^{n-1} & \to & X^n & \to & X^{n+1} & \to & \cdots \\ \downarrow p_1 & & \downarrow & & & & \downarrow p_{n-1} & & \downarrow p_n & & \downarrow p_{n+1} & &\\ X^1 & & X^2 / X^1 & & & & X^{n-1} / X^{n-2} & & X^n / X^{n-1} & & X^{n+1} / X^n & & \cdots,\end{array}$$ where I've elected to name the downward maps for later use.

Each of these is what's called a cofiber sequence, where a cofiber sequence is a pair of sequential maps $A \to B \to C$ with the property that a test map $T \to B$ can be lifted to a commuting triangle $T \to A \to B$ if and only if the composite $T \to B \to C$ is null-homotopic. The way you get cofiber sequences in homotopy theory is by attaching cones to subspaces ($A \to B \to B \cup_A Cone(A)$), which for nice enough subspaces agrees with the quotient sequence $A \to B \to B/A$. An important lemma is that the maps in cofiber sequences satisfy the differential property:

Setting $T = A$, the map $A = T \to B$ lifts to a map $A = T \to A$ by taking the identity, and this forces (by the "only if") the long composite $A \to B \to C$ to be null.

An exceedingly useful fact about cofiber sequences is that they can be extended: in the sequence of maps $$\begin{array}{ccc} A & \to & B \\ & \to & B \cup_A Cone(A) \\ & \to & (B \cup_A Cone(A)) \cup_B Cone(B) \simeq \Sigma A \\ & \to & \Sigma A \cup_{B \cup_A Cone(A)} Cone(B \cup_A Cone(A)) \simeq \Sigma B \\ & \to & \cdots,\end{array}$$ any adjacent pair forms a cofiber sequence. Abbreviating all these messy cones, one says that $A \to B \to C$ extends to $$A \to B \to C \to \Sigma A \to \Sigma B \to \Sigma C \to \Sigma^2 A \to \cdots.$$

Now what does this have to do with you? Well, we have a whole bunch of cofiber sequences in that diagram with all the quotients of subcomplexes, and if we extend them, we find long sequences of the form $$X^{n-1} \to X^n \xrightarrow{p_n} X^n / X^{n-1} \xrightarrow{\partial_n} \Sigma X^{n-1} \to \Sigma X^n \to \cdots,$$ where I've again decided to give a name for later use to the "new map" that comes from extension. You'll notice that the subcomplexes $X^n$ themselves sit in a remarkable position in the diagram: they have a map to the right to $X^{n+1}$ and a map down to $X^n / X^{n-1}$, both of which sit in different cofiber sequences. Now, here's an important definitional assertion:

Your boundary map $X^{n+1} / X^n \to \Sigma(X^n / X^{n-1})$ is the same as the composite $(\Sigma p_n) \circ \partial_{n+1}$. That is: start at your favorite node on the bottom row, follow the map $\partial_{n+1}$ to move diagonally back up to (a suspension of) the top row, then follow $p_n$ down to move back to the bottom row.

Now that we've said all these things about cofiber sequences, the assertion that this gives a differential follows quickly: performing this pair of operations twice yields a four-fold composite $((\Sigma^2 p_{n-1}) \circ (\Sigma \partial_n)) \circ ((\Sigma p_n) \circ \partial_{n+1})$. Since we can associate composition, you find $\Sigma \partial_n$ and $\Sigma p_n$ right next to each other in the middle --- and these are two maps which appear adjacent to each other in the cofiber sequence $$X^{n-1} \to X^n \xrightarrow{p_n} X^n / X^{n-1} \xrightarrow{\partial_n} \Sigma X^{n-1} \to \cdots.$$ Hence, by the boxed lemma, their composite is zero, and hence the four-fold composite must also be zero, and that is exactly the differential condition on your boundary operator. Since the map of spaces $X^{n+1} / X^n \to \Sigma^2 X^{n-1} / X^{n-2}$ is itself null, applying any decent functor (such as homotopy groups) will also yield zero.

You can easily dress this up to fit into, e.g., the equivariant setting --- the important thing was just that cofiber sequences were around to work with. Moreover, if you know anything about spectral sequences, this is identical to the argument that the $d^1$-differential of the associated filtration spectral sequence is in fact a differential. Let me know if I've missed your point, and I'll revise the answer accordingly.

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@DanielW: You mention low dimensional cellular homology. For this I think you need crossed modules, which you can't get at by stable, i.e. abelian, methods. See a presentation I gave at Chicago in 2012 on my preprint page, or, for full details, Part I of our book "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" (pdf on my web page for the book). Note that this gives an approach to basic algebraic topology, including for example the Relative Hurewicz Theorem, using homotopically defined functors and without setting up singular homology. –  Ronnie Brown Aug 5 '13 at 9:51
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