[proofplan] The proof translates the statement into Chern classes of complexified bundles. Complexification commutes with direct sums, and the Whitney product formula for total Chern classes gives a product formula for $c((E\oplus F)_{\mathbb C})$. Since the odd Chern classes of complexified real bundles vanish in de Rham cohomology, only even Chern classes contribute, and the identity $p_k(V)=(-1)^k c_{2k}(V_{\mathbb C})$ converts the resulting even-degree formula into the Pontryagin formula. [/proofplan]

[step:Identify the complexification of the direct sum] For a smooth real vector bundle $V\to M$, let \begin{align*} V_{\mathbb C}:=V\otimes_{\mathbb R}\mathbb C \end{align*} denote its complexification. Define the complex vector bundle map \begin{align*} \Phi:(E\oplus F)_{\mathbb C} &\to E_{\mathbb C}\oplus F_{\mathbb C} \end{align*} fiberwise by \begin{align*} \Phi_m((e,f)\otimes z):=(e\otimes z,f\otimes z) \end{align*} for $m\in M$, $e\in E_m$, $f\in F_m$, and $z\in\mathbb C$, extended by complex linearity. This is a smooth complex vector bundle isomorphism, with inverse induced by \begin{align*} (e\otimes z,f\otimes w)\mapsto (e,0)\otimes z+(0,f)\otimes w. \end{align*} Therefore \begin{align*} (E\oplus F)_{\mathbb C}\cong E_{\mathbb C}\oplus F_{\mathbb C}. \end{align*} [/step]

[step:Apply the Whitney formula for Chern classes]For a smooth complex vector bundle $A\to M$, write \begin{align*} c(A):=\sum_{q\ge 0}c_q(A)\in H_{\mathrm{dR}}^{2*}(M;\mathbb C) \end{align*} for its total Chern class, where $c_q(A)\in H_{\mathrm{dR}}^{2q}(M;\mathbb C)$, $c_0(A)=1$, and $H_{\mathrm{dR}}^{2*}(M;\mathbb C):=\bigoplus_{q\ge 0}H_{\mathrm{dR}}^{2q}(M;\mathbb C)$ is equipped with the graded cup product. Real de Rham cohomology classes are viewed in complex de Rham cohomology through the natural coefficient extension map $\iota:H_{\mathrm{dR}}^*(M)\to H_{\mathrm{dR}}^*(M;\mathbb C)$. Since $M$ is paracompact, the Chern and Pontryagin classes cited below are available as the standard de Rham characteristic classes on smooth vector bundles over $M$. By the Whitney product formula for total Chern classes of complex vector bundles, equivalently the multiplicativity of the Chern class multiplicative sequence [citetheorem:9797], applied to the complex vector bundles $E_{\mathbb C}\to M$ and $F_{\mathbb C}\to M$, the isomorphism from the previous step gives \begin{align*} c((E\oplus F)_{\mathbb C})=c(E_{\mathbb C})c(F_{\mathbb C}) \end{align*} in $H_{\mathrm{dR}}^{2*}(M;\mathbb C)$.[/step]

[guided]The reason we pass to complex bundles is that Pontryagin classes are defined through the Chern classes of the complexification. The [direct sum](/page/Direct%20Sum) causes no ambiguity: the previous step constructed a complex vector bundle isomorphism \begin{align*} (E\oplus F)_{\mathbb C}\cong E_{\mathbb C}\oplus F_{\mathbb C}. \end{align*} Chern classes are invariant under complex vector bundle isomorphism, so \begin{align*} c((E\oplus F)_{\mathbb C})=c(E_{\mathbb C}\oplus F_{\mathbb C}). \end{align*} For any smooth complex vector bundle $A\to M$, the notation $c(A)=\sum_{q\ge 0}c_q(A)$ denotes the total Chern class in $H_{\mathrm{dR}}^{2*}(M;\mathbb C)$. The coefficient ring is complex here because Chern classes of complex bundles naturally live in complex de Rham cohomology in this Chern-Weil formulation; later, Pontryagin classes are real classes and are compared after applying the natural coefficient extension map $\iota:H_{\mathrm{dR}}^*(M)\to H_{\mathrm{dR}}^*(M;\mathbb C)$. Now we use the [Whitney product formula for Chern classes](/theorems/7052), equivalently the multiplicativity of the Chern class multiplicative sequence [citetheorem:9797], which states that for complex vector bundles $A\to M$ and $B\to M$, \begin{align*} c(A\oplus B)=c(A)c(B). \end{align*} The hypotheses are satisfied with $A=E_{\mathbb C}$ and $B=F_{\mathbb C}$, since both are smooth complex vector bundles over the same paracompact smooth manifold $M$. Hence \begin{align*} c((E\oplus F)_{\mathbb C})=c(E_{\mathbb C})c(F_{\mathbb C}) \end{align*} in the graded [cohomology ring](/theorems/2271).[/guided]

[step:Extract the even Chern class component] Write \begin{align*} c(E_{\mathbb C})=\sum_{a\ge 0}c_a(E_{\mathbb C}) \end{align*} and \begin{align*} c(F_{\mathbb C})=\sum_{b\ge 0}c_b(F_{\mathbb C}). \end{align*} The degree-$4k$ component of the Whitney identity is \begin{align*} c_{2k}((E\oplus F)_{\mathbb C})=\sum_{a+b=2k}c_a(E_{\mathbb C})c_b(F_{\mathbb C}). \end{align*} By [[Vanishing of Odd Chern Classes of Complexified Real Vector Bundles](/theorems/9779)][citetheorem:9779], every odd Chern class of $E_{\mathbb C}$ and $F_{\mathbb C}$ vanishes in de Rham cohomology. Thus all summands with $a$ odd or $b$ odd vanish, and only $a=2i$ and $b=2j$ remain. Therefore \begin{align*} c_{2k}((E\oplus F)_{\mathbb C})=\sum_{i+j=k}c_{2i}(E_{\mathbb C})c_{2j}(F_{\mathbb C}). \end{align*} [/step]

[step:Translate even Chern classes into Pontryagin classes] By [[Pontryagin Classes from Chern Classes of the Complexification](/theorems/9780)][citetheorem:9780], for every smooth real vector bundle $V\to M$ and every integer $\ell\ge 0$, \begin{align*} p_\ell(V)=(-1)^\ell c_{2\ell}(V_{\mathbb C}). \end{align*} Equivalently, \begin{align*} c_{2\ell}(V_{\mathbb C})=(-1)^\ell p_\ell(V). \end{align*} Applying this to $V=E\oplus F$, $V=E$, and $V=F$, the formula from the previous step becomes \begin{align*} (-1)^k p_k(E\oplus F)=\sum_{i+j=k}(-1)^i p_i(E)(-1)^j p_j(F). \end{align*} Since $i+j=k$ in every summand, $(-1)^i(-1)^j=(-1)^k$. Multiplying both sides by $(-1)^k$ gives the displayed equality after applying $\iota$: \begin{align*} \iota\left(p_k(E\oplus F)\right)=\iota\left(\sum_{i+j=k}p_i(E)p_j(F)\right). \end{align*} The map $\iota$ is injective because complex de Rham cohomology is obtained from real de Rham cohomology by extension of scalars, $H_{\mathrm{dR}}^*(M;\mathbb C)\cong H_{\mathrm{dR}}^*(M)\otimes_{\mathbb R}\mathbb C$, and the map $u\mapsto u\otimes 1$ is injective over the field $\mathbb R$. Hence \begin{align*} p_k(E\oplus F)=\sum_{i+j=k}p_i(E)p_j(F) \end{align*} in $H_{\mathrm{dR}}^{4k}(M)$. This holds for every integer $k\ge 0$, so the equivalent total-class identity follows: \begin{align*} p(E\oplus F)=p(E)p(F). \end{align*} [/step]