Euler Class on Oriented Even Rank Sphere Bundles I'm trying to solve Exercise 11.19 from Bott & Tu.

Show that the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, at least when the sphere bundle comes from a vector bundle.

I figured that the key to solving this would be to show that since the antipodal map $a: S^{2n} \to S^{2n}$ is orientation reversing, I could do some kind of manipulation with the integrals $\int_M e = \sum_i \int_{\partial \overline{D_i}} s^* \psi$ where $\psi$ is the global angular form on the sphere bundle. We also have $\int_{\partial \overline{D_r}} s^* \psi = \int_{\partial \overline{D_r}} s^* \rho^* \sigma =$ local degree of the section $s$ at $x_i$.
Suppose I have a local section $S: U \to E|_U = U \times S^{2n}$, $S(x) = (x,s(x)v)$, perhaps I could augment it with the antipodal map, $S'(x) = (x,a(s(x)v)) = (x,-s(x)v)$. Since the Euler class is independent of the section, and $S,S'$ both have the same zeros, this would mean that $\int_M e = \sum_i (\text{local degree of $S$ at $x_i$}) = \sum_i (\text{local degree of $S'$ at $x_i$})$.
However I believe that $\sum_i (\text{local degree of $S'$ at $x_i$}) = - \sum_i (\text{local degree of $S$ at $x_i$})$, which would solve the problem. Is this correct? Are there other sources with more information on this way of working with the Euler class?
 A: For $M$ to be of any (finite) dimension, we could consider the pullback of fibrewise antipodal map acting as an automorphism of the C$\check{e}$ch-de Rham (double) complex of $E\rightarrow M$. Then the pullback of orientation changes sign but the pullback of Euler class does not. That suffices to show the Euler class vanishes.
Let $\pi: E\rightarrow M$ be an oriented sphere bundle with even-dimensional fibres.
First of all, we define the fibre-wise antipodal map
\begin{align*}
    A: E&\rightarrow E\\
    (p,v) &\mapsto  (p,-v)\quad.
\end{align*}
Note that it is a bundle morphism: $\pi\circ A = \pi$ commutes. Moreover, since $A\circ A = \mathrm{id}_E$, we know $A$ is a bundle isomorphism.
Let $\mathscr{U}=\{U_\alpha\}$ be a good cover of trivializations of M. Since $E$ is orientable, we can find a collection of $(0,n)$-forms $\{\sigma_\alpha\}$ such that $[\sigma_\alpha]=[\sigma_\beta]$ on $U_{\alpha\beta}$ and the restriction of $\sigma$ to each fibre is a generator of $H^n(S^n)$.
Define the pullback of A on the C$\check{e}$ch-de Rham complexes as follows:
\begin{align*}
A^*: \mathcal{C}(\pi^{-1}\mathscr{U},\Omega^*) &\rightarrow \mathcal{C}(\pi^{-1}\mathscr{U}, \Omega^*)\\
\omega_{\alpha_0\cdots\alpha_p} \in \Omega^q(U_{\alpha_0\cdots\alpha_p})&\mapsto (A|_{U_{\alpha_0\cdots\alpha_p}})^*\omega_{\alpha_0\cdots\alpha_p}\in \Omega^q(U_{\alpha_0\cdots\alpha_p})
\end{align*}
This then extends to $C^p(\pi^{-1}\mathscr{U}, \Omega^q) = \prod_{\alpha_0<...<\alpha_p} \Omega^q(U_{\alpha_0\cdots\alpha_p}) \rightarrow C^p(\pi^{-1}\mathscr{U}, \Omega^q) $ natually. Hence the pullback $A^*$ is well-defined in a natural way.
Now, we claim $A$ induces double complex morphism $A^*$ between two C$\check{e}$ch-de Rham complexes $C^*(\pi^{-1}\mathscr{U}, \Omega^*) \rightarrow C^*(\pi^{-1}\mathscr{U},\Omega^*)$. Note that $A^*|_{U_{\alpha_0\cdots\alpha_p}}$ commutes with $d$ locally (properties of pullback operation), so $A^*$ commutes with differentiation $d$. Also, $A^*$ commutes with $\delta$ since $A^*$ is linear.
Hence, $A^*$ is a double complex morphism. By $A\circ A = \textrm{id}_E$, we get $(A^*)^{-1} = (A^{-1})^* = A^*$ is a double complex isomorphism.
Consider the "Tic-Tac-Toe" digram which defines the Euler class:
\begin{array}{|c|c|c|c|c|c}
  \beta_0   & & & & & \\
     & \beta_1& & &  &\\
     & & \beta_2 & & &\\
     & & & \ddots & &\\
     & & & & \beta_n & - \pi^*\epsilon\quad\quad\quad\\
     \hline
\end{array}
where $\beta_0 = (\sigma_\alpha)$ is the orientation of $E$ and $\epsilon$ is a representative of euler class. By applying $A^*$ to every term involved, we get $A^*(-\pi^*\epsilon) = -(\pi\circ A)^* \epsilon = -\pi^*\epsilon $ is the Euler class corresponding to the orientation $A^*\beta_0 = -\beta_0$. Hence, for opposition orientation, we get the same Euler class, i.e. $e(E) = -e(E)$. This shows $e(E) = 0$.
