Lawson and Michelsohn's proof of the splitting principle for oriented vector bundles I'm trying to understand the proof of the following result in Lawson & Michelsohn's Spin Geometry:

Proposition 11.2. Let $E$ be an oriented real vector bundle of dimension $2n$ over a manifold $X$. Then there is a smooth proper fibration $\pi\colon \mathscr S_E\to X$ such that $\pi^*\colon H^*(X)\to H^*(\mathscr S_E)$ is injective and the bundle $\pi^*(E\otimes \mathbb C)$ splits into complex line bundles:
$$
\pi^*(E\otimes \mathbb C) \cong \ell_1\oplus \overline {\ell_1} \oplus \dots \oplus \ell_n \oplus \overline {\ell_n}
$$
where $\overline{\ell_j}$ denotes the "inverse" or "complex conjugate" bundle to $\ell_j$. In fact, there is a splitting
$$
\pi^*(E) \cong E_1\oplus \dots \oplus E_n
$$
into oriented real $2$-plane bundles such that $E_k\otimes \mathbb C = \ell_k \oplus \overline{\ell_k}$ for each $k$.

The proof is based on constructing the bundle $p:G(E)\to X$ whose fiber at a point $x\in X$ is the set of all oriented $2$-dimensional subspaces of $E_x$, so the pullback bundle $p^*E$ splits into the tautological oriented $2$-plane bundle $E_1\to G(E)$ plus its orthogonal complement, and then proceeding inductively. This part I understand.
What is perplexing me is their argument that $p$ induces an injection on cohomology. It goes like this: "The argument given for Theorem C.14 adapts immediately to prove that the homomorphism $p^*: H^*(X;\mathbb Z) \to H^*(G(E);\mathbb Z)$ is injective." The theorem referred to is this:

Theorem C.14. Suppose $E\to X$ is a complex vector bundle of rank $k$. Then $H^*(\mathbb P(E);\mathbb Z)$ is a free $H^*(X;\mathbb Z)$-module with basis $1,u,u^2,\dots,u^{k-1}$ [where $\mathbb P(E)$ is the projectivization of $E$, $\ell$ is the tautological complex line bundle over $\mathbb P(E)$,  and $u = c_1(\ell)$].

The proof of Theorem C.14 is a fairly standard application of the Leray-Hirsch theorem, using the fact that the restriction of $u$ to each fiber generates the integral cohomology ring of the fiber, which is a copy of $\mathbb C\mathbb P^{k-1}$.
But in the case at hand, the fiber of $G(E)\to X$ is a copy of the oriented Grassmannian $\widetilde G_2(\mathbb R^{2n})$, and I'm not aware of any argument that there exist global cohomology classes on $G(E)$ whose restrictions to each fiber freely generate the integral cohomology groups of the fiber (as an abelian group), which is what's needed to apply Leray-Hirsch. What am I missing?
 A: For odd vector bundles with non-trivial euler class $p^*$ cannot be injective, and it seems, one can transport this problem to the even case. Please let know if I made any mistakes...
Consider the long exact Bockstein sequence
$$ \ldots \to H^*(-;\mathbb{Z}) \overset{2\cdot}{\to} H^*(-;\mathbb{Z}) \overset{\rho_2}{\to} H^*(-;\mathbb{Z}/2) \overset{\beta_2}{\to} H^{*+1}(-;\mathbb{Z}) \to \ldots$$
Let $V \to X$ be an oriented vector bundle of rank $2n-1=3$ with

*

*$w_{3}(V) \neq 0 \in H^{3}(X;\mathbb{Z}/2)$

*$\beta_2 w_2(V) = e(V) \in H^{3}(X;\mathbb{Z})$
(I believe 2. is a general fact, but I couldn't find a reliable source..)
Add a trivial line bundle $E = V \oplus 1$. We have $w_{3}(E) = w_{3}(V) \neq 0$.
Now
For $p: G(E) \to X$ as described above, we have $p^*(E) \overset{\sim}{=} P_1 \oplus P_2$ where $P_1$ and $P_2$ are oriented 2-plane bundles.
It is a general fact that the $\rho_2$ maps the euler class of an oriented vector bundle to its top Stiefel Whitney class, which means that the top SW class is in the kernel of $\beta_2$, for example $\beta_2 w_2(P_i) = 0$.
Since $w_1(P_i) = 0$, we have, that
$$ p^*w_{2}(E) = w_{2}(P_1) + w_2(P_2) \in \ker\beta_2.$$
Using that the Bockstein is natural we have
$$p^*e(V) = p^*\beta_2w_2(V) = \beta_2p^*w_2(E) = 0.$$
Since $e(V) \neq 0$ that means $p^*$ is not injective.
In this answer Qiaochu Yuan constructed such a vector bundle $V \to \mathbb{RP}^2\times \mathbb{RP}^2$ with the $w_3(V) \neq 0$ property. Using $$ Sq^1w_2(V) = \rho_2 \beta_2(w_2(V)) = w_3(V) = \rho_2(e(V))$$
one gets $\beta_2 w_2(V) = e(V)$, since $H^3(\mathbb{RP}^2\times \mathbb{RP}^2;\mathbb{Z}) $ only consists of $2$- torsion.
