 Solution

The radial gradient $F=D\mathbf{y}$ is symmetric positive definite. If it belongs to a well $SO(3)U_i$, write $F=Q{U}_i$. Then $F^{T}F=U_i^2$, while positive symmetry makes $F$ itself the unique positive square root of an operator of $F^{T}F$. Hence $F=U_i$. Likewise a symmetric positive-definite matrix in $SO(3)$ must be $I$. ThereforeFor each radius, the tangential stretch $r(R)/R$ must belong to the finite set of eigenvalues of these matrices. The ratio is continuous on the connected radial interval, so it is one constant $v>0$. Thus $r(R)=vR$ and differentiation gives $r^{\prime }=v$ as well. HenceThis proof uses finiteness and positive definiteness of the wells, not isotropy or the Baker-Ericksen inequalities. It also covers $R_1=0$: the punctured three-dimensional ball remains connected, and no condition at the excluded center was used.