# 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?

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 $$w_1([E])$$.