Opposite local signs give the compatible Whitney-circle framing

Choose a metric near the bigon that is product in the corner charts and along the sheet collars. The complementary-sheet product charts identify the inward disk tangent at each end of the $B$ arc with the $A$-arc velocity at its first end and its negative at the other end. Along the $A$ arc choose an oriented orthonormal $(a-1)$-frame $E$ tangent to $A$ and orthogonal to the disk. This bundle over an interval can be framed explicitly: embed the ambient manifold in Euclidean space, take its smooth orthogonal projection onto the relevant tangent subbundle, and transport an initial basis along the interval by $U^{\mathrm{\prime }}=[P^{\mathrm{\prime }},P]U$. For $K=[P\mathrm{\prime },P]$, $K^T=-K$ and $[K,P]=P\mathrm{\prime }$, so uniqueness gives $U^{T}U=I$ and $UP(0)U^T=P$. This yields a smooth basis, which [F7] orthonormalizes in the chosen product metric. The same procedure gives a frame for the normal-to-$B$, disk-orthogonal bundle over its arc.

Compare $E$ at the two corners in the oriented normal bundle of $B$. The full $a$-frame consisting of the inward disk tangent followed by $E$ has sign $\epsilon (p)$ at the first corner and $-\epsilon (q)$ at the other: the inward tangent is the $A$-arc velocity at the first corner and its negative at the second. The assumed opposite signs therefore place both prescribed endpoint $(a-1)$-frames in the same oriented frame component. In a trivialization of this interval bundle the comparison matrices lie in $SO(a-1)$. This group is path connected: finitely many plane rotations align its columns, with the final one-dimensional determinant forced to be one. For $a-1=1$ both comparison matrices are the identity. Interpolate by such a smooth rotation path, constant in the fixed endpoint collars. This extends $E$ over the $B$ arc as a frame normal to $B$ and to the disk, giving an admissible partial frame all around the boundary.

On the boundary take the orthogonal complement of $E$ inside the disk normal bundle. Its rank is $b-1$, and on the $B$ arc it is precisely the tangent-to-$B$, disk-orthogonal bundle. It is oriented by the tube and the chosen frame $E$. Its frames can be interpolated along the two arcs with matched corner values by the same interval transport and connected-$SO(b-1)$ argument. This produces the required admissible boundary frame. Adjustments by based loops within a summand fix the subspace and corner values but can change the total frame class; orientation and interval path connectivity do not remove this freedom. Hence there is no preferred class.

To justify the local orientation assertion, embed the ambient manifold in Euclidean space and along $W$ take the smooth projection $P$ onto the orthogonal complement of $TW$ in $TX{\mathrm{\mid }}_W$. Use its pullback to the convex bigon coordinates, centered at an interior point. Radial transport $U^{\mathrm{\prime }}=[P^{\mathrm{\prime }},P]U$ from the centre gives a smooth full normal frame over the disk, by the matrix ODE and parameter-dependence suppliers. Orient the disk and this normal frame. Their wedge is a nowhere-zero section of the determinant line of $TX$ along $W$. Extend this section locally in ambient charts, using smooth extensions of the cornered embedding, and patch the extensions by [F6]; on $W$ they all restrict to the same section. The patched section stays nonzero on an open neighbourhood of the compact disk and hence orients that neighbourhood. Rounding the cornered boundary adds the inward tangent-to-disk line to this rank-$(m-2)$ normal bundle, giving boundary-circle rank $m-1$. Thus all sign comparisons needed above use local orientations alone.