[proofplan] The proof is a formal consequence of the splitting principle and the Whitney product formula. First we spell out that the expressions $\prod_i Q(x_i)$ and $\prod_j P(y_j)$ are interpreted degree by degree as symmetric polynomials in the corresponding characteristic classes, so they define elements of completed even cohomology. For the complex case, pull back to a splitting space where $E$ and $F$ split into line bundles; there the Chern roots of $E\oplus F$ are the combined roots of $E$ and $F$, so the product factors. The injectivity part of the splitting principle descends the equality to $M$, and the Pontryagin case is identical with Pontryagin roots and the Whitney formula for total Pontryagin classes. [/proofplan]

[step:Interpret the power-series root expressions degree by degree]Let $E\to M$ be a complex vector bundle of rank $r$. Write \begin{align*} Q(z)=\sum_{m=0}^{\infty}a_m z^m \end{align*} with $a_m\in R$ and $a_0=1$. For every integer $n\ge 0$, define $\Phi_{Q,r,n}$ to be the homogeneous degree-$n$ component of the symmetric formal [power series](/page/Power%20Series) \begin{align*} \prod_{i=1}^{r}Q(x_i)\in R[[x_1,\dots,x_r]], \end{align*} where each $x_i$ has formal degree $1$. Since $\Phi_{Q,r,n}$ is a symmetric polynomial in $x_1,\dots,x_r$, the [fundamental theorem of symmetric polynomials](/theorems/5179) gives a unique polynomial \begin{align*} \Psi_{Q,r,n}\in R[e_1,\dots,e_r] \end{align*} such that \begin{align*} \Phi_{Q,r,n}(x_1,\dots,x_r)=\Psi_{Q,r,n}(e_1(x),\dots,e_r(x)). \end{align*} Thus the degree-$2n$ component of $K_Q(E)$ is \begin{align*} K_Q(E)_n:=\Psi_{Q,r,n}(\rho_R(c_1(E)),\dots,\rho_R(c_r(E)))\in H^{2n}(M;R). \end{align*} This defines \begin{align*} K_Q(E):=(K_Q(E)_n)_{n\ge 0}\in \prod_{n\ge 0}H^{2n}(M;R)=\widehat H^{\mathrm{even}}(M;R). \end{align*} The same construction applies to a real vector bundle $V\to M$ of rank $d$. Define $q:=\lfloor d/2\rfloor$. If \begin{align*} P(t)=\sum_{m=0}^{\infty}b_m t^m \end{align*} with $b_0=1$, and if the formal Pontryagin roots $y_1,\dots,y_q$ are assigned cohomological degree $4$, then the degree-$4n$ component of $\prod_{j=1}^{q} P(y_j)$ is a symmetric polynomial in the Pontryagin roots, hence a polynomial in the Pontryagin classes $p_1(V),\dots,p_q(V)$. Components in cohomological degrees not divisible by $4$ are zero. This gives a well-defined element \begin{align*} K_P^{\mathbb R}(V)\in \widehat H^{\mathrm{even}}(M;R). \end{align*} The completed product is well-defined because the degree-$2n$ component of a product in $\widehat H^{\mathrm{even}}(M;R)$ is a finite sum: \begin{align*} (\alpha\beta)_n=\sum_{i+j=n}\alpha_i\smile\beta_j. \end{align*}[/step]

[guided]The only subtle point in the statement is that the products $\prod_i Q(x_i)$ and $\prod_j P(y_j)$ are not finite ordinary products in cohomology when $Q$ and $P$ are arbitrary power series. They must be read one cohomological degree at a time. For the complex case, suppose $E\to M$ has rank $r$. We introduce formal variables $x_1,\dots,x_r$, each of formal degree $1$, and write \begin{align*} Q(z)=\sum_{m=0}^{\infty}a_m z^m \end{align*} with $a_m\in R$ and $a_0=1$. In the formal power series ring $R[[x_1,\dots,x_r]]$, the product \begin{align*} \prod_{i=1}^{r}Q(x_i) \end{align*} has a homogeneous part of degree $n$ for every $n\ge 0$. Denote that homogeneous part by $\Phi_{Q,r,n}$. Because the product is unchanged when the variables $x_1,\dots,x_r$ are permuted, $\Phi_{Q,r,n}$ is a symmetric polynomial in these variables. Therefore the fundamental theorem of symmetric polynomials gives a unique polynomial \begin{align*} \Psi_{Q,r,n}\in R[e_1,\dots,e_r] \end{align*} such that \begin{align*} \Phi_{Q,r,n}(x_1,\dots,x_r)=\Psi_{Q,r,n}(e_1(x),\dots,e_r(x)). \end{align*} Now the formal Chern-root notation means precisely that the elementary symmetric functions in the roots are replaced by the Chern classes: \begin{align*} e_i(x_1,\dots,x_r)=c_i(E). \end{align*} Thus the degree-$2n$ component of the class defined by $\prod_i Q(x_i)$ is \begin{align*} K_Q(E)_n:=\Psi_{Q,r,n}(\rho_R(c_1(E)),\dots,\rho_R(c_r(E)))\in H^{2n}(M;R). \end{align*} The degree doubles because each Chern root has cohomological degree $2$, while $\Phi_{Q,r,n}$ has formal degree $n$. The real case is the same construction with Pontryagin roots. If $V\to M$ is a real vector bundle of rank $d$ and $q:=\lfloor d/2\rfloor$, then the formal Pontryagin roots are $y_1,\dots,y_q$, each with cohomological degree $4$. The homogeneous degree-$n$ part of $\prod_{j=1}^{q}P(y_j)$ therefore lands in $H^{4n}(M;R)$, and the odd even-degree components, namely degrees $4n+2$, are zero. Since the expression is symmetric in the Pontryagin roots, it is a polynomial in the integral Pontryagin classes $p_1(V),\dots,p_q(V)$, and the actual $R$-coefficient class is obtained by substituting $\rho_R(p_1(V)),\dots,\rho_R(p_q(V))$. Hence it defines an element \begin{align*} K_P^{\mathbb R}(V)\in \widehat H^{\mathrm{even}}(M;R). \end{align*} Finally, the completed [cohomology ring](/theorems/2271) is needed because $Q$ and $P$ may have infinitely many nonzero coefficients and $M$ may have nonzero cohomology in infinitely many even degrees. Multiplication is still well-defined: to compute the degree-$2n$ component of $\alpha\beta$, only finitely many terms contribute, namely \begin{align*} (\alpha\beta)_n=\sum_{i+j=n}\alpha_i\smile\beta_j. \end{align*}[/guided]

[step:Prove multiplicativity after splitting the complex bundles into line bundles]Let $E\to M$ and $F\to M$ be complex vector bundles of ranks $r$ and $s$. By the splitting principle for complex vector bundles, applied to the bundle $E\oplus F\to M$, there is a CW complex $S$, a continuous map \begin{align*} \sigma:S\to M, \end{align*} and complex line bundles $L_1,\dots,L_r\to S$ and $N_1,\dots,N_s\to S$ such that \begin{align*} \sigma^*E\cong \bigoplus_{i=1}^{r}L_i \end{align*} and \begin{align*} \sigma^*F\cong \bigoplus_{j=1}^{s}N_j. \end{align*} The splitting space may be constructed as an iterated flag bundle. At each flag-bundle stage the projective-bundle theorem gives cohomology as a free module over the previous stage with $1$ among the basis elements; iterating, $H^*(S;R)$ is a free $H^*(M;R)$-module through $\sigma^*$ with $1$ as one basis element. Hence if $\alpha\in H^k(M;R)$ and $\sigma^*\alpha=0$, the coefficient of the basis vector $1$ in $\sigma^*\alpha$ is $\alpha$, so $\alpha=0$. Consequently \begin{align*} \sigma^*:\widehat H^{\mathrm{even}}(M;R)\to \widehat H^{\mathrm{even}}(S;R) \end{align*} is injective degree by degree. Here $\sigma^*$ denotes the degreewise pullback homomorphism on completed even cohomology. Let $\rho_{S,R}:H^*(S;\mathbb Z)\to H^*(S;R)$ denote the coefficient homomorphism induced by $\mathbb Z\to R$. For each $i\in\{1,\dots,r\}$, define \begin{align*} u_i:=\rho_{S,R}(c_1(L_i))\in H^2(S;R), \end{align*} and for each $j\in\{1,\dots,s\}$, define \begin{align*} v_j:=\rho_{S,R}(c_1(N_j))\in H^2(S;R). \end{align*} By [citetheorem:9814] for Chern classes, compatibility of coefficient change with cup products, and the [Whitney product formula for Chern classes](/theorems/7052), the $R$-coefficient total Chern classes satisfy \begin{align*} \rho_{S,R}(c(\sigma^*E))=\prod_{i=1}^{r}(1+u_i) \end{align*} and \begin{align*} \rho_{S,R}(c(\sigma^*F))=\prod_{j=1}^{s}(1+v_j). \end{align*} Moreover, \begin{align*} \sigma^*(E\oplus F)\cong \sigma^*E\oplus\sigma^*F\cong \left(\bigoplus_{i=1}^{r}L_i\right)\oplus\left(\bigoplus_{j=1}^{s}N_j\right). \end{align*} Applying the Whitney product formula again and then applying $\rho_{S,R}$ gives \begin{align*} \rho_{S,R}(c(\sigma^*(E\oplus F)))=\prod_{i=1}^{r}(1+u_i)\prod_{j=1}^{s}(1+v_j). \end{align*} Therefore the formal Chern roots of $\sigma^*(E\oplus F)$ are the combined list $u_1,\dots,u_r,v_1,\dots,v_s$. Hence \begin{align*} K_Q(\sigma^*(E\oplus F))=\prod_{i=1}^{r}Q(u_i)\prod_{j=1}^{s}Q(v_j). \end{align*} The completed cup product separates this finite product as \begin{align*} K_Q(\sigma^*(E\oplus F))=K_Q(\sigma^*E)K_Q(\sigma^*F). \end{align*}[/step]

[guided]The purpose of the splitting space is to replace the formal Chern-root notation by actual first Chern classes of line bundles. The splitting principle, applied to $E\oplus F$, supplies a CW complex $S$, a continuous map $\sigma:S\to M$, and line bundles $L_1,\dots,L_r\to S$ and $N_1,\dots,N_s\to S$ such that $\sigma^*E\cong \bigoplus_{i=1}^{r}L_i$ and $\sigma^*F\cong \bigoplus_{j=1}^{s}N_j$. We use the standard iterated flag-bundle splitting space. At each stage, the projective-bundle theorem identifies the new cohomology as a free module over the previous cohomology ring with $1$ among the basis elements. Iterating this description, $H^*(S;R)$ is a free $H^*(M;R)$-module through $\sigma^*$ with $1$ as one basis element. Thus, if $\alpha\in H^k(M;R)$ and $\sigma^*\alpha=0$, the coefficient of the basis vector $1$ is $\alpha$, so $\alpha=0$. Therefore the degreewise pullback $\sigma^*:\widehat H^{\mathrm{even}}(M;R)\to\widehat H^{\mathrm{even}}(S;R)$ is injective. Let $\rho_{S,R}:H^*(S;\mathbb Z)\to H^*(S;R)$ denote the coefficient homomorphism. For each $i\in\{1,\dots,r\}$, define $u_i:=\rho_{S,R}(c_1(L_i))\in H^2(S;R)$, and for each $j\in\{1,\dots,s\}$, define $v_j:=\rho_{S,R}(c_1(N_j))\in H^2(S;R)$. By [citetheorem:9814] for Chern classes, compatibility of coefficient change with cup products, and the Whitney product formula for Chern classes, the $R$-coefficient total Chern classes satisfy \begin{align*} \rho_{S,R}(c(\sigma^*E))=\prod_{i=1}^{r}(1+u_i) \end{align*} and \begin{align*} \rho_{S,R}(c(\sigma^*F))=\prod_{j=1}^{s}(1+v_j). \end{align*} Pullback commutes with direct sums of vector bundles, so \begin{align*} \sigma^*(E\oplus F)\cong \sigma^*E\oplus\sigma^*F\cong \left(\bigoplus_{i=1}^{r}L_i\right)\oplus\left(\bigoplus_{j=1}^{s}N_j\right). \end{align*} Applying the Whitney product formula once more and then applying $\rho_{S,R}$ gives \begin{align*} \rho_{S,R}(c(\sigma^*(E\oplus F)))=\prod_{i=1}^{r}(1+u_i)\prod_{j=1}^{s}(1+v_j). \end{align*} Thus the Chern roots of $\sigma^*(E\oplus F)$ on the splitting space are exactly the combined list $u_1,\dots,u_r,v_1,\dots,v_s$. Substituting this combined list into the degreewise definition of $K_Q$ gives \begin{align*} K_Q(\sigma^*(E\oplus F))=\prod_{i=1}^{r}Q(u_i)\prod_{j=1}^{s}Q(v_j). \end{align*} The first finite product is $K_Q(\sigma^*E)$ and the second finite product is $K_Q(\sigma^*F)$, with multiplication interpreted by the completed cup product. Therefore \begin{align*} K_Q(\sigma^*(E\oplus F))=K_Q(\sigma^*E)K_Q(\sigma^*F). \end{align*}[/guided]

[step:Descend the complex identity from the splitting space] Naturality of the Chern classes implies naturality of every polynomial expression in the Chern classes. By the degreewise definition of $K_Q$, this gives \begin{align*} K_Q(\sigma^*E)=\sigma^*K_Q(E), \end{align*} \begin{align*} K_Q(\sigma^*F)=\sigma^*K_Q(F), \end{align*} and \begin{align*} K_Q(\sigma^*(E\oplus F))=\sigma^*K_Q(E\oplus F). \end{align*} Using that $\sigma^*$ is a ring homomorphism for the completed cup product, the equality from the splitting space becomes \begin{align*} \sigma^*K_Q(E\oplus F)=\sigma^*(K_Q(E)K_Q(F)). \end{align*} Since $\sigma^*$ is injective, it follows that \begin{align*} K_Q(E\oplus F)=K_Q(E)K_Q(F) \end{align*} in $\widehat H^{\mathrm{even}}(M;R)$. [/step]

[step:Prove the Pontryagin multiplicativity by the same symmetric-polynomial calculation]Let $V\to M$ and $W\to M$ be real vector bundles of ranks $d$ and $e$. Define $m:=\lfloor d/2\rfloor$, $\ell:=\lfloor e/2\rfloor$, and $q:=\lfloor(d+e)/2\rfloor$. Introduce formal variables $y_1,\dots,y_m$ for $V$ and $z_1,\dots,z_\ell$ for $W$, each assigned cohomological degree $4$. By the [Whitney product formula for Pontryagin classes](/theorems/9781) [citetheorem:9781], \begin{align*} p(V\oplus W)=p(V)p(W). \end{align*} Thus, after applying the coefficient map $\rho_R$, the elementary symmetric functions in the formal Pontryagin roots of $V\oplus W$ are the elementary symmetric functions in the combined list $y_1,\dots,y_m,z_1,\dots,z_\ell$, with one additional zero root when $q=m+\ell+1$. The latter case occurs exactly when both $d$ and $e$ are odd, and the extra root contributes the factor $P(0)=1$. Therefore the degreewise symmetric-polynomial definition gives \begin{align*} K_P^{\mathbb R}(V\oplus W)=\prod_{i=1}^{m}P(y_i)\prod_{j=1}^{\ell}P(z_j). \end{align*} The first factor is $K_P^{\mathbb R}(V)$, the second factor is $K_P^{\mathbb R}(W)$, and their product is interpreted by the completed cup product. Hence \begin{align*} K_P^{\mathbb R}(V\oplus W)=K_P^{\mathbb R}(V)K_P^{\mathbb R}(W) \end{align*} in $\widehat H^{\mathrm{even}}(M;R)$.[/step]

[guided]We do not need a separate geometric splitting space for real bundles. The definition of $K_P^{\mathbb R}$ is already algebraic: in each degree it is the symmetric polynomial in the formal Pontryagin roots obtained after replacing elementary symmetric functions by the $R$-coefficient Pontryagin classes $\rho_R(p_i(-))$. Let $m:=\lfloor d/2\rfloor$, $\ell:=\lfloor e/2\rfloor$, and $q:=\lfloor(d+e)/2\rfloor$. Use formal Pontryagin roots $y_1,\dots,y_m$ for $V$ and $z_1,\dots,z_\ell$ for $W$, each of cohomological degree $4$. The Whitney product formula for Pontryagin classes [citetheorem:9781] gives \begin{align*} p(V\oplus W)=p(V)p(W). \end{align*} This identity says exactly that the elementary symmetric functions of the Pontryagin roots of $V\oplus W$ are obtained from the combined list $y_1,\dots,y_m,z_1,\dots,z_\ell$. If both ranks $d$ and $e$ are odd, then $q=m+\ell+1$, so the formal definition for $V\oplus W$ has one more possible Pontryagin root than the combined lists for $V$ and $W$. The Whitney product formula forces that additional elementary-root contribution to be zero, so we append a zero root. Since $P(0)=1$, appending this root does not change the product. Substituting the combined list into the degreewise definition of $K_P^{\mathbb R}$ gives \begin{align*} K_P^{\mathbb R}(V\oplus W)=\prod_{i=1}^{m}P(y_i)\prod_{j=1}^{\ell}P(z_j). \end{align*} By the same degreewise definition, the first product is $K_P^{\mathbb R}(V)$ and the second product is $K_P^{\mathbb R}(W)$. Multiplication in $\widehat H^{\mathrm{even}}(M;R)$ is the completed cup product, so this equality is precisely \begin{align*} K_P^{\mathbb R}(V\oplus W)=K_P^{\mathbb R}(V)K_P^{\mathbb R}(W). \end{align*}[/guided]