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.