Pontryagin Whitney product away from two

On each fiber define ${\mathrm{\Phi }}_b:((E_b\oplus F_b){\otimes }_{\mathbb{R}}\mathbb{C})\to (E_b{\otimes }_{\mathbb{R}}\mathbb{C})\oplus (F_b{\otimes }_{\mathbb{R}}\mathbb{C})$ by ${\mathrm{\Phi }}_b((e,f)\otimes z)=(e\otimes z,f\otimes z)$ and extend additively. The tensor balancing relations make this well defined, and the inclusions of the two direct summands give its inverse. In simultaneous real bundle charts, a Whitney-sum transition is $\mathrm{diag}(g_{ji},h_{ji})$ by [F4]; complexification reads the same real matrix over $\mathbb{C}$, which is exactly the block-diagonal transition for $E_{\mathbb{C}}\oplus F_{\mathbb{C}}$. Thus the ${\mathrm{\Phi }}_b$ are locally the same fixed coordinate isomorphism, hence continuous and compatible with all transitions, and [F4] glues them to $(E\oplus F{)}_{\mathbb{C}}\cong E_{\mathbb{C}}\oplus F_{\mathbb{C}}$. Multiplicativity [F3] now gives $c((E\oplus F{)}_{\mathbb{C}})=c(E_{\mathbb{C}})c(F_{\mathbb{C}})$; expanding in degree $2i$ gives the sum of the even-even and odd-odd terms.

In the even-even terms write $a=2r$, $b=2s$ with $r+s=i$: then $(-1{)}^i{c}_{2r}(E_{\mathbb{C}})c_{2s}(F_{\mathbb{C}})=(-1{)}^{r+s}{c}_{2r}(E_{\mathbb{C}})c_{2s}(F_{\mathbb{C}})=p_r(E)p_s(F)$ by [F1], since $(-1{)}^{r+s}=(-1{)}^i$.

Every odd-odd term has the form $c_{2r+1}(E_{\mathbb{C}})c_{2s+1}(F_{\mathbb{C}})$ with $r+s=i-1$, and each factor is two-torsion by [F2]; after inverting two these terms vanish, and the same holds in any coefficient ring in which $2$ is invertible.

Multiplying the degree-$2i$ expansion of step 1.1 by $(-1{)}^i$ and using steps 2.1 and 2.2, $p_i(E\oplus F)={\sum }_{r+s=i}{p}_r(E)p_s(F)$ over $\mathbb{Z}[1\mathrm{/}2]$, which is the degree-$i$ component of $p(E\oplus F)=p(E)p(F)$.

The statement asserts no integral identity: the discarded odd-odd terms need not vanish integrally, and the companion examples page exhibits an integral counterexample for the universal real line.

Boundary cases. For $i=0$ both sides are $1$; if one summand has rank zero the formula reduces to $p(F)=p(F)$ by stability. The empty base is excluded by the path-connected hypothesis; the coefficient ring $\mathbb{Z}[1\mathrm{/}2]$ is nonzero and $2$ is invertible there by construction, so no division by zero occurs. AC is used only through [A1].