## Obstruction Cocycles

Hey everyone, I was reading about obstruction theory, here's a bit of a summary. Take a cellular space $X$ and a fibre bundle $p:E \to X$ with fiber $F$; consider the problem of extending a section $s$, defined on the $(n-1)$-skeleton over to the $n$-skeleton. We work cell by cell pulling back the bundle via the characteristic map and the section via the restriction of the Char. Map to the boundary of our $n$-cell, since the cell is contractible, the bundle is isomorphic to $D^n \times F$ so the section defines a map from $S^{n-1} \to D^n \times F$ i.e. an element of $\pi_{n-1}(D^n \times F) \cong \pi_{n-1}(F)$. Define the obstruction cochain as the element in $C^n(X^n,\pi_{n-1}(F))$ taking each $n$-cell to the element in the $(n-1)$-homotopy group constructed before.

Here's what's bothering me, in Steenrod's book (The topology of Fibre Bundles) he proves that this cochain is actually a cocycle in a really weird way, it looks to me as if he makes no distinction between the homological boundary and the topological boundary of a cell. Roughly he writes the following composition: $C_{q+1}(X) \stackrel{\partial_*}\to Z_q(X) \stackrel{hurewicz}\to \pi_q(X^q) \stackrel{f=(p_2\circ \sigma)_*}\to \pi_q(F)$

And claims that this composition is the value of the obstruction cochain in an $n+1$ cell, how might one verify this?

Not being happy with this proof i went and looked at the one Kirk and Davis' book (Lecture notes on algebraic topology) and found it too complex (I know, I dont like anything sorry).

What I was wondering is if there was a way to prove this affirmation (the obstruction cochain is a cocycle) directly, i.e. denoting the cochain by $\Theta$ doing something like:

$\delta \Theta (e) = \Theta( \partial e) = \Theta (\sum [w_i;e]w_i) = \sum [w_i;e]\Theta(w_i) = \dots = 0$ (where $e$ is a $(n+1)$-cell, $w_i$ is a $n$-cell and $[w_i;e]$ is their incidence number, so that the third term is $\Theta$ evaluated on the cellular (homological) boundary).

Any help on the subject or a good refference is very very much appreciated! Thanks and have a great week.

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 Have you looked at Whitehead's book? He goes into obstruction theory for CW-complexes in quite a lot of detail. I haven't looked at it in a while but I remember being quite satisfied with the exposition. – Ryan Budney Sep 12 2010 at 20:09 Juan, could you re-read your question carefully? Presumably you're not interested in homotopy groups of $\mathbb{R}^n$, but that's what you have as things stand. You've indicated a composition - is it the value of the obstruction cochain or of its coboundary? Are $q$ and $n$ the same? Does the Hurewicz map really go from cycles to homotopy? What does "roughly" mean? – Tim Perutz Sep 12 2010 at 20:18 @Ryan, no I have not looked at Whiteheads book, I will try and find it, can you give me the title please? @Tim, I dont see where do the homotopy groups or $\mathbb{R}^n$ come in, theyre always 0 so there are no obstructions there... What i meant by "roughly" is that I wasn't going to write the whole of Steenrod's proof, if you read his book you'll see why I wrote the Hurewicz map that way (I may be bad at AT but Im not that bad) and about q and n, I did mess up there, but the meaning gets through I believe. Sorry for my bad english... – Juan OS Sep 12 2010 at 21:03 Whitehead, "Elements of Homotopy Theory" Springer-Verlag books.google.com/… – Ryan Budney Sep 12 2010 at 23:49 Juan, sorry, the meaning doesn't get through clearly enough for me. For instance, you say that $F$ is the fibre of a vector bundle. That makes it $\mathbb{R}^d$ for some $d$, and yet I don't think that's what you mean. – Tim Perutz Sep 13 2010 at 2:35
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You should have a look at the paper given in the answer to my earlier question on obstruction theory. It gives a very nice and direct proof that the obstruction cochain is a cocycle that also works in the case of non-simple spaces (a setting that most modern treatments gloss over). It's written in simplicial rather than cellular language, but I imagine the techniques could be carried over with a bit of effort.

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 Thanks a lot, I actually read your question and found the article, it's great! Id still like to see a direct proof in the sence of the cellular boundary operator so I'll leave the question open a while more, but that article is really helpful! Thanx – Juan OS Sep 12 2010 at 19:23

How are you defining the cellular chain complex and $[w_i;e]$? The usual way is to define $C_n(X):=H_n(X^n,X^{n-1})$ and the differential as the composite $H_n(X^n,X^{n-1})\to H_{n-1}(X^{n-1})\to H_{n-1}(X^{n-1},X^{n-2})$, where the first map is the connecting homomorphism for the pair. Steenrod's observation is then straightforward, and follows from the long exact sequence of the appropriate pair. After the fact, you can show that $C_n(X)$ is free on the $n$ cells and that $[w_i;e]$ is the degree of the composite of the attaching map $S^{n-1}\to X^{n-1}$ for $e$ composed with the projection $X^{n-1}\to X^{n-1}/X^{n-2}=\vee_i S^{n-1}\to w_i/\partial w_i\cong S^{n-1}$. That typically requires reconciling different notions of degree, e.g. homological and differential topological and perhaps this is where you are having difficulties.

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 You are correct about the definition of the cell complex, however my problem is that Steenrod appears to be stating that the composite of the maps in the sequence I wrote is the value of the obs. cochain, and I feel in doing so he is stating something like "the value of the map in the (topological) boundary is the same as its value in the (homological) boundary", I'll try to reread the proof from the first definition you gave of the cell complex. Thanks. – Juan OS Sep 12 2010 at 21:58 Juan: try this. Consider the map of pairs $(D^{q+1}, S^{q})\to (X^{q+1}, X^{q})$ given by the characteristic map for $e$ and consider the map of long exact sequences of these pairs. The way you define the obstruction cocycle in your question pulls back to $Ze=H_{q+1}(D^{q+1},S^q)\to H_q(S^q)\pi_q(S^q)\to \pi_q(F)$. This shows that the restriction of Steenrod's obstruction cochain to $e$ is the cochain you defined in your first paragraph. Now the argument I outlined shows that Steenrod's cochain is a cocycle. – Paul Sep 12 2010 at 23:09 PS. keep in mind that even showing that $\partial^2=0$ is not so easy if you use $\partial e=\sum_i[w_i:e]w_i$ as your definition of the differential, and the question about why the obstruction cochain is a cocycle is basically a variation on the same question. On the other hand, the with defn I outlined above $\partial^2=0$ is obvious. – Paul Sep 12 2010 at 23:15

You may find this and this by Tony Phillips useful.

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 Thank you for the links, I will look into them. – Juan OS Sep 12 2010 at 21:54