A framing, not just the submanifold, determines the Pontryagin-Thom class

The Hopf map $h(z_1,z_2)=[z_1:z_2]$ has fibre $U=\{z_2=0\}$. The two affine charts give smooth local sections $(1,w)\mathrm{/}\sqrt{1+\mathrm{\mid }w{\mathrm{\mid }}^2}$ and $(v,1)\mathrm{/}\sqrt{1+\mathrm{\mid }v{\mathrm{\mid }}^2}$; circle multiplication supplies bundle charts. With $a=\mathrm{\mid }{z}_1{\mathrm{\mid }}^2$ on unit representatives, the functions $\sigma (4a-1)$ and $\sigma (3-4a)$, divided by their positive sum, give a finite support-subordinate partition on these charts. Thus under AC the supplied numerable circle bundle is a Hurewicz fibration. In the $w=z_2\mathrm{/}{z}_1$ chart the normal differential along $U$ is multiplication by $1\mathrm{/}{z}_1$, of real rank two, so $[1:0]$ is regular and a positive basis gives the Hopf framing. The regular-preimage collapse lemma makes its Pontryagin–Thom map homotopic to $h$.

In real coordinates, $U=\{x_3=x_4=0\}\subset S^3$ bounds $D=\{x_4=0,x_3\ge 0\}$. Let $h(t)=2t\sigma (8t-1)$ and put $W=\{(x,t):x\in S^3,x_4=0,x_3=h(t),0\le t\le 1\mathrm{/}2\}$. It has the literal collar $U\times [0,1\mathrm{/}8)$ and no end at time one. At the cap $x_3=1,t=1\mathrm{/}2$, $h^{\mathrm{\prime }}=2$, so it is the smooth graph $t=x_3\mathrm{/}2$ in the sphere $x_4=0$; elsewhere the height gradient there is nonzero. Thus it is a compact neat disk with boundary only $U\times \{0\}$. Its normal quotient is framed by the ordered classes of $e_4$ and $({\mathrm{\nabla }}_{S^2}{x}_3,-h^{\mathrm{\prime }})$, which are smooth, nonzero and independent everywhere. On the product collar this pair is $(e_4,e_3)$: the normal of $D$ and the inward normal of $U$ within $D$. Taking the dual trivialization defines precisely ${\phi }_D$, constant throughout that collar. Hence this framing is framed null-cobordant, and its collapse is nullhomotopic by [F3].

(The two framed submanifolds have the same underlying manifold.) The Hopf framing $(U,h_{\ast }b)$ and the bounding framing $(U,{\phi }_D)$ are framings of the normal bundle of the same standard unknot $U\subseteq S^3$: the underlying closed $1$-submanifold is $U$ in both cases.

The unit circle agrees with the quotient-circle model of the supplied universal cover by $[t]\mapsto (\cos 2\pi t,\sin 2\pi t)$ is a homeomorphism from $\mathbb{R}\mathrm{/}\mathbb{Z}$ to the unit circle. For $j\ge 2$, $S^j$ is path-connected, locally path-connected (small chart balls suffice), and simply connected. The subgroup criterion Lifting criterion for maps from path-connected locally path-connected spaces therefore lifts each based map $S^j\to S^1$ to $\mathbb{R}$. Linear contraction of the based lift gives ${\pi }_j(S^1)=0$. The long exact sequence of the Hopf bundle consequently makes $h_{\ast }:{\pi }_3(S^3)\to {\pi }_3(S^2)$ an isomorphism, since the adjacent groups ${\pi }_3(S^1)$ and ${\pi }_2(S^1)$ vanish. Degree identifies ${\pi }_3(S^3)$ with $\mathbb{Z}$ and its identity with $1$, so $[h]=h_{\ast }[\mathrm{i}\mathrm{d}]$ generates ${\pi }_3(S^2)$ and is nonzero.

(Their Pontryagin-Thom classes differ.) By [F1] the Pontryagin-Thom class of $(U,h_{\ast }b)$ is the generator of ${\pi }_3(S^2)\cong \mathbb{Z}$, computed in steps 1.1 and 2.1. By [F2] the Pontryagin-Thom map of $(U,{\phi }_D)$ is nullhomotopic, so by [F3] the framed submanifold $(U,{\phi }_D)$ is framed null-cobordant and its Pontryagin-Thom class is the zero element of ${\pi }_3(S^2)$. In the isomorphism ${\pi }_3(S^2)\cong \mathbb{Z}$ of [F1] the generator is not zero, so the two classes differ.

(Conclusion.) The two framings of the same underlying submanifold $U$ have different Pontryagin-Thom classes; equivalently, by [F4], the two framed submanifolds are not framed cobordant. Hence the Pontryagin-Thom class is not determined by the underlying submanifold alone, and the framing is load-bearing data; the statement refuted is false.