[proofplan]
We compare the total Pontryagin classes of the two stably isomorphic bundles. The Whitney product formula rewrites the Pontryagin class of each [direct sum](/page/Direct%20Sum) as the product of the Pontryagin class of the original bundle and the Pontryagin class of the added summand. The added product-bundle summands have total Pontryagin class $1$, because their complexifications are direct products $M \times \mathbb C^m$ and hence have no positive-degree Chern classes. Comparing homogeneous components then gives equality of every individual Pontryagin class.
[/proofplan]
[step:Pass the stable isomorphism to equality of total Pontryagin classes]
For any smooth real vector bundle $V\to M$, define its total Pontryagin class by $p(V):=\sum_{k\ge 0}p_k(V)\in H_{\mathrm{dR}}^{4*}(M)$, with the convention $p_0(V)=1$. Here $H_{\mathrm{dR}}^{4*}(M):=\bigoplus_{k\ge 0}H_{\mathrm{dR}}^{4k}(M)$ denotes the graded de Rham [cohomology ring](/theorems/2271) in degrees divisible by $4$.
Let $\Phi:E\oplus \varepsilon_M^a\to F\oplus \varepsilon_M^b$ be the given real vector bundle isomorphism over $\operatorname{id}_M$. Pontryagin classes are invariant under real vector bundle isomorphism, since they are characteristic classes of real vector bundles. Therefore $p(E\oplus \varepsilon_M^a)=p(F\oplus \varepsilon_M^b)$ in the graded cohomology ring $H_{\mathrm{dR}}^{4*}(M)$.
[/step]
[step:Use the Whitney product formula to separate the added summands]
Apply the [[Whitney Product Formula for Pontryagin Classes](/theorems/9781)][citetheorem:9781] to the two direct sums. Its hypotheses are satisfied because $E$, $F$, $\varepsilon_M^a$, and $\varepsilon_M^b$ are smooth real vector bundles over the same paracompact smooth manifold $M$. Hence
\begin{align*}
p(E\oplus \varepsilon_M^a)=p(E)p(\varepsilon_M^a)
\end{align*}
and
\begin{align*}
p(F\oplus \varepsilon_M^b)=p(F)p(\varepsilon_M^b).
\end{align*}
Combining these two identities with the equality from the stable isomorphism gives
\begin{align*}
p(E)p(\varepsilon_M^a)=p(F)p(\varepsilon_M^b).
\end{align*}
[guided]
The stable isomorphism relates $E$ and $F$ only after adding product-bundle summands, so the purpose of this step is to remove the direct sums from the notation. The tool that does exactly this is the [Whitney Product Formula for Pontryagin Classes][citetheorem:9781].
We may apply that formula because all four bundles involved,
\begin{align*}
E,\qquad F,\qquad \varepsilon_M^a,\qquad \varepsilon_M^b,
\end{align*}
are smooth real vector bundles over the same paracompact smooth manifold $M$. Applying the formula to $E\oplus \varepsilon_M^a$ gives
\begin{align*}
p(E\oplus \varepsilon_M^a)=p(E)p(\varepsilon_M^a).
\end{align*}
Applying it to $F\oplus \varepsilon_M^b$ gives
\begin{align*}
p(F\oplus \varepsilon_M^b)=p(F)p(\varepsilon_M^b).
\end{align*}
The previous step gave
\begin{align*}
p(E\oplus \varepsilon_M^a)=p(F\oplus \varepsilon_M^b).
\end{align*}
Substituting the two Whitney product identities into this equality yields
\begin{align*}
p(E)p(\varepsilon_M^a)=p(F)p(\varepsilon_M^b).
\end{align*}
This identity is the exact point where the stable isomorphism has been converted into an equation involving the ordinary total Pontryagin classes of $E$ and $F$.
[/guided]
[/step]
[step:Compute the total Pontryagin class of a product real bundle]
Let $m\ge 0$ be an integer. For any complex vector bundle $W\to M$, let $c_j(W)\in H_{\mathrm{dR}}^{2j}(M;\mathbb C)$ denote its $j$-th Chern class. The complexification of the product real bundle $\varepsilon_M^m=M\times \mathbb R^m$ is the product complex bundle $(\varepsilon_M^m)_{\mathbb C}:=\varepsilon_M^m\otimes_{\mathbb R}\mathbb C\cong M\times \mathbb C^m$. The total Chern class of a product complex vector bundle is $1$, so $c_j((\varepsilon_M^m)_{\mathbb C})=0$ for every integer $j\ge 1$. By the defining relation between Pontryagin classes and the Chern classes of the complexification, as recorded in [[Pontryagin Classes from Chern Classes of the Complexification](/theorems/9780)][citetheorem:9780],
\begin{align*}
p_k(\varepsilon_M^m)=(-1)^k c_{2k}((\varepsilon_M^m)_{\mathbb C})
\end{align*}
for every integer $k\ge 0$. Thus $p_0(\varepsilon_M^m)=1$, and for every integer $k\ge 1$,
\begin{align*}
p_k(\varepsilon_M^m)=0.
\end{align*}
Therefore
\begin{align*}
p(\varepsilon_M^m)=1.
\end{align*}
In particular,
\begin{align*}
p(\varepsilon_M^a)=1
\end{align*}
and
\begin{align*}
p(\varepsilon_M^b)=1.
\end{align*}
[/step]
[step:Cancel the added factors and compare homogeneous components]
Substituting $p(\varepsilon_M^a)=1$ and $p(\varepsilon_M^b)=1$ into $p(E)p(\varepsilon_M^a)=p(F)p(\varepsilon_M^b)$ gives $p(E)=p(F)$ in $H_{\mathrm{dR}}^{4*}(M)$. Since equality in a graded cohomology ring is equality in each homogeneous degree, the degree-$4k$ components agree for every integer $k\ge 0$, hence $p_k(E)=p_k(F)$ in $H_{\mathrm{dR}}^{4k}(M)$ for every integer $k\ge 0$. This is the desired stable invariance of Pontryagin classes.
[/step]
