Top Chern class equals Euler class of the underlying real bundle

If $n=0$, both sides are $1$ by the rank-zero conventions. Assume $n\ge 1$. Pulling back to the flag bundle, $q^{\ast }E=L_1\oplus \cdots \oplus L_n$ by [F1].

On the flag bundle the top Chern class of the split bundle is the product of the line classes: $q^{\ast }{c}_n(E)={\prod }_{i=1}^n{c}_1(L_i)$, by multiplicativity in [F2] and the vanishing $c_k(L_i)=0$ for $k\ge 2$.

On the flag bundle the Euler class of the underlying real bundle is the product of the line Euler classes: $q^{\ast }e(E_{\mathbb{R}})=e((q^{\ast }E{)}_{\mathbb{R}})={\prod }_{i=1}^{n}e(L_{i,\mathbb{R}})$, using naturality of the Euler class and [F3] for the ordered sum.

By [F2] each line contributes $c_1(L_i)=e(L_{i,\mathbb{R}})$, so the right sides of steps 2.1 and 2.2 are equal; hence $q^{\ast }(c_n(E))=q^{\ast }(e(E_{\mathbb{R}}))$.

Injectivity of $q^{\ast }$ on $H^{2n}(-;\mathbb{Z})$ from [F1] gives $c_n(E)=e(E_{\mathbb{R}})$, which is the assertion.

Boundary cases. The case $n=0$ was discharged in step 1.1. For $n=1$ the assertion is the line normalization $c_1(L)=e(L_{\mathbb{R}})$ of [F2], and no splitting is needed. The empty base is excluded by the path-connected hypothesis, and the coefficient ring $\mathbb{Z}$ is nonzero. AC enters only through [A1] in the splitting and Thom/Euler suppliers.