Stabilized upper gauge cancellation by block rotation

Write the upper component as $g(x,v)$ over $D_{+}V$, with $g_0(x)=g(x,0)$. Stabilize $\varphi$ by the pulled-back isomorphism $g_0^{-1}$ on ${\pi }^{\ast }F$. Its upper gauge is $U=\mathrm{diag}(g,g_0^{-1})$. Radial contraction of the disk deforms this to $\mathrm{diag}(g_0,g_0^{-1})$. For any orthogonal endomorphism $a$ of $F$, set $R_t=\left(\begin{array}{cc}\cos tI& -\sin tI\\ \sin tI& \cos tI\end{array}\right)\mathrm{.}$ The path $\mathrm{diag}(a,I)R_t\mathrm{diag}(I,a^{-1})R_t^{-1}$, $0\le t\le \pi \mathrm{/}2$, starts at $\mathrm{diag}(a,a^{-1})$ and ends at $I$. Hence there is an explicit orthogonal homotopy $U_s$ from $U$ to $I$ over the entire upper disk.

Let $\beta$ be the stabilized target clutching map. Use $U^{-1}$ as an automorphism of its upper disk bundle. The required lower automorphism on the equator is $\beta U^{-1}{\beta }^{-1}$. Extend it over the lower disk by $h_{-}(x,rv)=\beta (x,v)U_{1-r}(x,v{)}^{-1}\beta (x,v{)}^{-1}(r>0),h_{-}(x,0)=I\mathrm{.}$ Because $U_1=I$, orthogonality gives uniform continuity at the center, even though $\beta$ need not extend there. The two automorphisms obey the equatorial clutching relation and therefore define a genuine automorphism of the fixed target bundle. Composing it with the stabilized $\varphi$ gives upper gauge $U^{-1}U=I$, as required.