TITLE: Obstruction to homotopy, cohomology operations and Dold-Whitney theorem
QUESTION [8 upvotes]: I am reading the famous paper by Dold and Whitney "Classification of Oriented Sphere Bundles Over A 4-Complex".
I'll state their theorem for the case of SO(3) bundles
Classification Theorem:Let $B_1, B_2$ be principal $SO(3)$ bundles over a complex $K$ of dimension at most 4. let $h_i:K\to G_n\simeq BSO(3)$ be the classifying map for $B_i$.
Assume they have the same second Stiefel-Whitney class $w_2(B_1)=w_2(B_2)=: w_2$ (then we can assume that $h_1$ and $h_2$ agree on the 3-skeleton $K^3$). 
Let $d(h_1,h_2) \in H^4(K;\pi_4(G_n))$ be the difference cocycle cohomology class.
The bundles $B_1,B_2$ are equivalent iif
 there exists a cohomology class $x\in H^1(K,\mathbb{Z}/2)$ such that
$$(1) \quad \quad d(h_1,h_2) = \beta x \cup \beta x + \beta (x\cup w_2)$$
where $\beta$ is the Bockstein homomorphism associated with the coefficient sequence $0\to \mathbb{Z} \overset{2\cdot}{\to} \mathbb{Z}\to \mathbb{Z}/2\to 0$ and $\pi_3(G_n)$ has been identified with $\mathbb{Z}$.

Usually we require the cohomology class of the cocycle to $d(h_1,h_2)$ to be $0$, why in (1) they require that condition instead?

Explanation of the question
I have seen stated in others book on obstruction theory that $h_1,h_2:K^n\to Y$ are homotopic rel to $K^{n-2}$  iif $[d(h_1,h_2)]=0$ (under the hypothesis of $Y$ being simple, I don't know if this is the point) this seems to be very different from $(1)$.
Cohomology operations.  Dold & Whitney  appeal to a theorem from Eilenberg MacLane's paper "On the groups $H(\pi,n),\quad  III:$ Operations and Obstructions", theorem 14.2. 
I looked into it and indeed they don't require the difference cocycle to be null-cohomologous.
I only explained myself that the RHS of $(1)$ comes from the cohomology operation induced by the $k$-invariant  (I guess Postnikov invariant) of the target space.

Can someone explain better the setup and theorem 14.2 in Eilenberg MacLane's paper?

REPLY [9 votes]: Let me explain why the condition is not the triviality of the difference cocycle. Maybe an important point to note is that the condition with the difference cocycle is not about the vanishing of an obstruction, it is about the enumeration of possible lifts of maps from one Postnikov truncation to the next.
If we denote by $\tau_{\leq n} {\rm BSO}(3)$ the Postnikov truncation which kills all homotopy above degree $n$, then we have the usual fiber sequence of obstruction theory
$$
K(\pi_4{\rm BSO}(3),4)\to \tau_{\leq 4}{\rm BSO}(3) \to \tau_{\leq 3}{\rm BSO}(3).
$$
This gives rise to an exact sequence of pointed sets of homotopy classes
$$
[X_+,\Omega\tau_{\leq 3}{\rm BSO}(3)]\to [X_+,K(\pi_4{\rm BSO}(3),4)]\to [X_+,\tau_{\leq 4}{\rm BSO}(3)]\to [X_+,\tau_{\leq 3}{\rm BSO}(3)].
$$
Note that ${\rm BSO}(3)$ is simply connected (hence also simple), so there is no difference between pointed and unpointed maps here. Since $\tau_{\leq 3}{\rm BSO}(3)=K(\mathbb{Z}/2,2)$, we can rewrite that sequence
$$
{\rm H}^1(X,\mathbb{Z}/2)\to {\rm H}^4(X,\mathbb{Z})\to [X_+,{\rm BSO}(3)]\to {\rm H}^2(X,\mathbb{Z}/2).
$$
The set $[X,{\rm BSO}(3)]$ gives the isomorphism classes of oriented rank 3 bundles, the last map takes a bundle to its second Stiefel-Whitney class. The additional invariant lives in a quotient group of $H^4(X,\mathbb{Z})$ and the first map ${\rm H}^1(X,\mathbb{Z}/2)\to {\rm H}^4(X,\mathbb{Z})$ takes a class $x$ to $\beta x\cup \beta x+\beta(x\cup w_2)$. So the difference cocycle is a class in ${\rm H}^4$, but the bundle is still trivial if the cocycle lies in the image of ${\rm H}^1(X,\mathbb{Z}/2)$ which explains the condition.