The oriented normal bundle of a knot is trivial

Use an orientation-preserving identification of the standard three-sphere with the unit sphere in ${\mathbb{R}}^4$. Write the knot as $\gamma :S^1\to S^3\subset {\mathbb{R}}^4$ and set $\tau (s)={\gamma }^{\mathrm{\prime }}(s)\mathrm{/}\parallel {\gamma }^{\mathrm{\prime }}(s)\parallel$. Since $\gamma (s)$ is unit and $\gamma$ is an immersion, $\gamma (s)$ and $\tau (s)$ are orthonormal. Put $P_s=\{v\in {\mathbb{R}}^4:⟨v,\gamma (s)⟩=⟨v,\tau (s)⟩=0\}$, oriented by the rule in the statement. Near any $s_0$, project two fixed vectors spanning $P_{s_0}$ onto $P_s$ by $Q_s(v)=v-⟨v,\gamma (s)⟩\gamma (s)-⟨v,\tau (s)⟩\tau (s)$. They remain independent on a small arc; Gram–Schmidt and, if necessary, a sign change give a smooth oriented orthonormal frame $e_1(s),e_2(s)$. Equip the quotient set in [F2] with the charts $[v]\mapsto (s,⟨v,e_1(s)⟩,⟨v,e_2(s)⟩)$ for $v\in T_{\gamma (s)}{S}^3$. These are well-defined bijections on each fibre, with inverse $(s,a,b)\mapsto [a{e}_1(s)+b{e}_2(s)]$; their transition matrices consist of smooth inner products of the local frames. They give the standard quotient smooth structure: in any slice chart, changing the tangent frame to $(\tau ,e_1,e_2)$ is a smooth invertible matrix, and the induced change on the two transverse quotient coordinates is smooth with smooth inverse. Thus $P$ and ${\nu }_K$ are smooth rank-two bundles here, and $v\mapsto v-⟨v,\tau (s)⟩\tau (s)$ on $T_{\gamma (s)}{S}^3$ induces a smooth orientation-preserving isomorphism ${\nu }_K\to P$. This constructs the needed smooth structure directly, without an additional choice assumption.

Cut $S^1$ at one point and fix an oriented local frame on an arc about the cut. Compactness of the remaining interval gives a finite subdivision whose closed subintervals each lie in a bundle-trivializing chart. Inductively extend a frame from one subinterval to the next: on a small overlap, the incoming and chart frames differ by a smooth map into ${\mathrm{G}\mathrm{L}}^{+}(2,\mathbb{R})$. Shrink the overlap so that this map lies in a neighbourhood of its value at the junction where straight interpolation remains in ${\mathrm{G}\mathrm{L}}^{+}$. Then join that constant value to the identity by a path in ${\mathrm{G}\mathrm{L}}^{+}$ and use a smooth cutoff to extend the transition across the new subinterval, making the adjusted frame agree on a collar of the junction and be a fixed chart frame away from it. This can be done at each of the finitely many junctions, with the first and last collars expressed in the fixed cut-arc frame by constant matrices. The required paths exist because Gram–Schmidt writes every matrix in ${\mathrm{G}\mathrm{L}}^{+}(2,\mathbb{R})$ as a rotation times an upper-triangular matrix with positive diagonal; rotations are path-connected and the positive-diagonal triangular matrices contract linearly to the identity. We obtain a smooth oriented interval frame with constant transition matrices at its endpoint collars.

Under the endpoint identification, the two endpoint frame germs differ by a constant matrix $A\in {\mathrm{G}\mathrm{L}}^{+}(2,\mathbb{R})$ by step 1.2. Choose a smooth path $B(t)$ from the identity to $A^{-1}$ that is constant on collars of $0$ and $1$, and multiply the interval frame by $B(t)$. The endpoint values and germs now agree under the identification, because the original germs differ by the constant $A$ and $B$ is constant on their collars. The adjusted frame descends smoothly to an oriented frame on $S^1$, giving the required bundle isomorphism.