Is the first Stiefel-Whitney class an isomorphism if there is a unique orientable class? Suppose that $X$ is a nice compact manifold such that its reduced real $K$-group $\tilde{K}\mathcal{O}(X)$ has a unique stably equivalent class of orientable bundles, i.e, $\ker(\omega_{1})$ is the identity element in $\tilde{K}\mathcal{O}(X)$ where $\omega_{1}:\tilde{K}\mathcal{O}(X)\to H^{1}(X,\mathbb{Z}_{2})$ is the first Stiefel-Whitney class. Here, we generalize the notion of orientability to stably equivalent classes of vector bundles stating that a stably equivalent class of bundles $[E]$ is orientable if $\omega_{1}([E])=0$. Under this hypothesis, can we deduce that $\omega_{1}$ classifies all stably equivalent vector bundles? In other words, is $\omega_{1}:\tilde{K}\mathcal{O}(X)\to H^{1}(X,\mathbb{Z}_{2})$ an isomorphism?
Indeed, since $\omega_{1}$ restricted to line bundles is an isomorphism, it is onto and its injectivity follows from the hypothesis. Is this argumentation correct? If so, which is the role of higher Stiefel-Whitney classes in this setting?
 A: Yes, under these conditions $w_1 : \widetilde{KO}(X) \to H^1(X; \mathbb{Z}_2)$ is an isomorphism. Here's another way to see it.
Suppose $[E], [F] \in \widetilde{KO}(X)$ satisfy $w_1([E]) = w_1([F]) = \alpha \neq 0$. Let $L$ denote the real line bundle with $w_1(L) = \alpha$. Then $w_1([E\oplus L]) = w_1([F\oplus L]) = 0$ so by assumption $[E\oplus L] = [F\oplus L]$. Therefore, there are non-negative integers $m, n$ with $E\oplus L\oplus\varepsilon^m \cong F\oplus L\oplus\varepsilon^n$. As $X$ is compact and Hausdorff, there is a bundle $L'$ with $L\oplus L' \cong \varepsilon^k$ for some $k \geq 2$. Therefore $E\oplus L\oplus\varepsilon^m\oplus L' \cong E\oplus\varepsilon^{n+k}$ and $F\oplus L\oplus\varepsilon^n\oplus L' \cong F\oplus\varepsilon^{m+k}$, so $[E] = [F]$.
By your assumption, every orientable bundle is stably trivial and hence has trivial Steifel-Whitney classes. So if $w_1([E]) = \alpha \neq 0$ and $L$ is as above, then $w_1([E\oplus L]) = 0$ and hence $w_i([E\oplus L]) = 0$. On the other hand $w_i([E\oplus L]) = w_i([E]) + w_{i-1}([E])w_1([L])$, so $w_i([E]) = w_{i-1}([E])w_1([L]) = w_{i-1}([E])\alpha$ and therefore $w_i([E]) = \alpha^i$. That is, the Stiefel-Whitney classes of $[E]$ are completely determined by $w_1([E])$.
