Stabilizing a framed point suspends its collapse map

Using [A1] for the Pontryagin–Thom and regular-value degree suppliers, for a finite framed set in $S^n$, the centre of the collapse target has precisely that set as its regular preimage, with differential signs equal to its framing signs. The regular-value degree formula therefore gives degree equal to the signed count. Degree classifies based self-maps of $S^n$, and the fixed-codimension Pontryagin--Thom bijection transfers this classification to framed cobordism. A single positive point has degree $+1$ and its reversal degree $-1$.

(The framed point generates ${\pi }_1(S^1)$.) By [F1] the Pontryagin-Thom map of the positively framed point $(p,\phi )$ is a smooth self-map of $S^1$ whose differential at $p$ has sign $+1$, so its degree is $+1$; by [F2] degree is an isomorphism ${\pi }_1(S^1)\to \mathbb{Z}$, hence the class of the framed point is the generator. A single point realizes $+1$ and its orientation reversal $-1$ in the classification of [F1].

(Its stabilization is a positively framed point.) Under equatorial stabilization the point becomes $i(p)\in S^2$ with the framing ${\nu }_{\mathrm{e}\mathrm{q}}\oplus \phi$, where ${\nu }_{\mathrm{e}\mathrm{q}}$ is the equatorial normal, by [F3]; the normal bundle of the point $i(p)$ in $S^2$ is all of $T_{i(p)}{S}^2$, so this framing is a basis of $T_{i(p)}{S}^2$. By [F4] the pair $({\nu }_{\mathrm{e}\mathrm{q}},\phi )$ is positive for the standard orientation of $S^2$ because $\phi$ is a positive basis of the equator, so the stabilization is again a single positively framed point. By [F1] its Pontryagin-Thom map has degree $+1$ and, by [F2], it generates ${\pi }_2(S^2)\cong \mathbb{Z}$.

(The suspension identity on the generator.) By [F3] the Pontryagin-Thom class of the stabilization is $E$ of the class of the framed point, which is the generator of ${\pi }_1(S^1)$ by step 1.2. By [F4] the suspension sends the class of the identity of $S^1$ to the class of the identity of $S^2$, and the class of the identity is the generator of ${\pi }_1(S^1)$ by [F2]; hence $E(\mathrm{P}\mathrm{T}(p,\phi ))$ is the identity class of ${\pi }_2(S^2)$, i.e. the degree-$+1$ self-map of $S^2$. This agrees with step 1.3, where the Pontryagin-Thom class of the stabilization was computed directly as the generator of ${\pi }_2(S^2)$: the identity $\mathrm{P}\mathrm{T}(\sigma (p,\phi ))=E(\mathrm{P}\mathrm{T}(p,\phi ))$ holds on this nonzero class.

(Iteration and conclusion.) Repeating the two computations one dimension higher: the stabilization of the positively framed point of $S^k$ is the positively framed point of $S^{k+1}$ (the same boundary-orientation computation as step 1.3, where the equatorial normal is prepended to a positive basis), and its Pontryagin-Thom class is the generator of ${\pi }_{k+1}(S^{k+1})\cong \mathbb{Z}$; by [F3] and [F4] these classes are related by the suspension isomorphisms $E:{\pi }_k(S^k)\to {\pi }_{k+1}(S^{k+1})$, which send generator to generator. The stabilized class is therefore a nonzero element of the zeroth stable stem ${\pi }_0^s$ (Stable stems of the sphere), and the example verifies the stabilization-to-suspension compatibility on a nonzero class.