Stable normal bundle is independent of the embedding

If $N_0+N_1=0$, all tangent and normal fibres are zero and the unique zero-bundle isomorphism proves the assertion; assume the ambient dimension is positive below. In ${\mathbb{R}}^{N_0}\oplus {\mathbb{R}}^{N_1}$ set $j_t(x)=(\cos (\pi t\mathrm{/}2)i_0(x),\sin (\pi t\mathrm{/}2)i_1(x))$. At least one coefficient is nonzero; hence $j_t$ is injective and $d{j}_t$ is injective on each tangent space, because the coefficient that does not vanish already forces equality of $x$, respectively vanishing of the tangent vector. Let $Q_t(x)$ be the orthogonal projection onto $d{j}_t(T_{x}M)$ and $P_t=I-Q_t$. In a smooth local frame $e_1,\dots ,e_n$ of $TM$ from [F3], the matrix $A(t,x)$ with columns $d{j}_t(e_i(x))$ has full column rank, and $Q_t(x)=A(A^{T}A{)}^{-1}{A}^T$ is smooth in $(t,x)$; the same formula applies in boundary charts. For a zero-dimensional $M$, the frame is empty and the formula reads $Q_t=0$, $P_t=I$. Thus $P_t$ is a smooth family of orthogonal projections with $\ker P_t=d{j}_t(TM)=d{j}_t(T_{x}M)$ fibrewise. At $t=0$ one has $\mathrm{im}{P}_0=d{i}_0(T_{x}M{)}^{\perp }\oplus {\mathbb{R}}^{N_1}$ and at $t=1$ one has $\mathrm{im}{P}_1={\mathbb{R}}^{N_0}\oplus d{i}_1(T_{x}M{)}^{\perp }$, and [F1] identifies the orthogonal summands with ${\nu }_{i_0}$ and ${\nu }_{i_1}$ smoothly over $M$, in the half-space form at boundary points.

Put $K_t={\dot{P}}_t{P}_t-P_t{\dot{P}}_t$ and solve ${\dot{U}}_t=K_t{U}_t$, $U_0=I$ in the finite-dimensional ambient space, with $x\in M$ as a parameter. For each fixed $x$ the coefficients $t\mapsto K_t(x)$ are smooth, so [F2] gives a unique solution on all of $[0,1]$; For an interior parameter chart, the coefficients are defined on an open time-state-parameter domain, so [F2] gives local smooth dependence. At a boundary parameter point use [F4] to extend the coordinate functions of $i_0,i_1$ and the local frame to an open Euclidean parameter neighbourhood; shrink it so both embedding differentials and the frame remain full rank. The same formula for $A(t,x)$ then stays full rank on an open time interval containing $[0,1]$, since its sine and cosine coefficients never vanish together. Hence $P_t$ and $K_t$ have smooth extensions on an open time-parameter domain, and the vector field $(t,U,x)\mapsto K_t(x)U$ is smooth on an open time-state-parameter domain as required by [F2]. Apply that theorem on the extension and restrict back to the half-space. On the compact time interval, finitely many local continuations and uniqueness glue these solution maps to a smooth $U_t(x)$; the restricted extensions give smoothness up to the boundary. Since $K_t^T=-K_t$, differentiating $U^{T}U$ gives zero, so every $U_t$ is orthogonal. Differentiating $P_t^2=P_t$ gives $P_t{\dot{P}}_t{P}_t=0$, whence $KP-PK=(\dot{P}P-P\dot{P})P-P(\dot{P}P-P\dot{P})=\dot{P}P+P\dot{P}=\dot{P},$ so $\frac{d}{dt}(U_t^{-1}{P}_t{U}_t)=U_t^{-1}(-K_t{P}_t+{\dot{P}}_t+P_t{K}_t)U_t=0$ and $U_t^{-1}{P}_t{U}_t=P_0$. Passing to $t=1$, the smooth family $x\mapsto U_1(x)$ restricts to a smooth bundle isomorphism $\mathrm{im}{P}_0\to \mathrm{im}{P}_1$ over $M$.

By step 1.1 the initial complement bundle is ${\nu }_{i_0}\oplus {\epsilon }^{N_1}$ and the final one is ${\epsilon }^{N_0}\oplus {\nu }_{i_1}$, so with the reordering of the two orthogonal summands the bundle isomorphism of step 2.1 gives ${\nu }_{i_0}\oplus {\epsilon }^{N_1}\cong {\nu }_{i_1}\oplus {\epsilon }^{N_0}$. This proves the intrinsic nature of the stable normal class and applies verbatim to compact manifolds with boundary, where the same fibrewise orthogonal projections are smooth in boundary charts. For empty $M$ the assertion is the unique map of zero bundles. No summand has been cancelled: both sides retain their unstable ranks. The countable choice of [F1] is the only choice used; the ODE transport selects the solution of a linear equation with Lipschitz coefficients and adds none.