Stably trivial bundles over spheres below the rank are trivial

If $p=0$, the base consists of two points and one finite basis choice in each fibre trivializes $E$. If $k=0$, the stable trivialization is already a trivialization. Otherwise equip the total trivial bundle with its Euclidean metric and orthonormalize the supplied $k$-frame spanning the added trivial summand. It gives a map $T:S^p\to V_k({\mathbb{R}}^{q+k})$, and its orthogonal complement is isomorphic to $E$ by projection along the given direct sum. Since $p<q=(q+k)-k$, the choice-free part (i) of the preceding Stiefel frame-fields lemma makes $T$ nullhomotopic; choose a continuous extension over $D^{p+1}$.

The projection $P=I-T{T}^T$ has constant rank $q$ over this ball. Its explicit finite-subdivision projection-and-Gram–Schmidt transport from the centre, given in that lemma, supplies a continuous complementary frame throughout. Restricting to the boundary trivializes $E$ continuously. To obtain a smooth frame, approximate its columns on the compact sphere by smooth ambient columns using a finite cover of sufficiently small round balls: choose smooth nonnegative bumps positive on smaller balls covering the sphere, normalize their finite sum, and form weighted averages of the column values at the finitely many centres. Uniform continuity makes these averages uniformly as close as desired to the original columns. Project them into the smooth complementary subbundle on the sphere. For sufficiently close approximations the Gram determinant remains positive by compactness, so Gram–Schmidt yields a smooth global frame. This is a finite construction and needs no countable choice.

A global smooth frame gives a smooth bundle trivialization. If the classifying-space identifications stated in the reformulation are supplied, stable triviality corresponds to a zero stable class and this implication gives injectivity of ${\pi }_p(BO(q))\to {\pi }_p(BO)$ in the stated range. Only injectivity is claimed; the stronger isomorphism at the endpoint would require an additional surjectivity argument. The cases $p=0$ and $k=0$ were handled separately and no choice principle was used.