The Pontryagin-Thom correspondence in fixed codimension

(The map $\mathrm{\Phi }$ on classes.) First take the free homotopy class of $f_{(N,\phi )}{\mathrm{\mid }}_{S^n}$, and use the inverse of the forgetful bijection [F7] to define $\mathrm{\Phi }(N,\phi )\in {\pi }_n(S^k)$. The disjoint basepoint of $(S^n{)}_{+}$ does not by itself make that restriction based at a preselected point of $S^n$. By [F2], framed-cobordant framed submanifolds have based homotopic Pontryagin-Thom maps, so $\mathrm{\Phi }$ is constant on framed cobordism classes and induces a map $\mathrm{\Phi }$ from framed cobordism classes to ${\pi }_n(S^k)$.

(The map $\mathrm{\Psi }$ on classes.) For a continuous map $f:S^n\to S^k$, choose a smooth map $f^{\mathrm{\prime }}$ homotopic to it by [F3], a regular value $y$ of $f^{\mathrm{\prime }}$ and a positive basis $b$ of $T_y{S}^k$ (a positive basis exists since the orientation of $T_y{S}^k$ has two classes and one flips sign by negating a vector), and set $\mathrm{\Psi }(f):=[(f^{\mathrm{\prime }-1}(y),f_{\ast }^{\mathrm{\prime }}b)]$, the framed cobordism class of the framed regular preimage. This is well defined: any two smooth maps homotopic to $f$ are smoothly homotopic by [F3], and [F5] gives framed cobordism of the resulting preimages for different smooth representatives, different regular values and different positive bases. Hence $\mathrm{\Psi }$ induces a map $\mathrm{\Psi }:{\pi }_n(S^k)\to \{\text{framed cobordism classes}\}$, defined without choosing a representative of the class: for every representative and every admissible choice the value is the same.

($\mathrm{\Psi }\circ \mathrm{\Phi }$ is the identity.) Let $(N,\phi )$ be a closed framed $(n-k)$-submanifold and $f=f_{(N,\phi )}$ its Pontryagin-Thom map, which is smooth by [F1]. Its centre $y_0$ is a regular value and the framed preimage of $(f,y_0,b)$ is $(N,\phi )$ on the nose by [F6], where $b$ is the positive basis corresponding to the structure identification. Therefore one admissible choice in the definition of $\mathrm{\Psi }$ gives the class of $(N,\phi )$, and by the well-definedness proved in step 1.2 every admissible choice gives it: $\mathrm{\Psi }(\mathrm{\Phi }(N,\phi ))=[(N,\phi )]$.

($\mathrm{\Phi }\circ \mathrm{\Psi }$ is the identity.) Let $[f]\in {\pi }_n(S^k)$ and let $(N,\phi )$ be the framed preimage produced by $\mathrm{\Psi }$ from a smooth map $f^{\mathrm{\prime }}$ homotopic to $f$, a regular value $y$ and a positive basis $b$. Then $\mathrm{\Phi }(\mathrm{\Psi }[f])=[f_{(N,\phi )}]$, and $f_{(N,\phi )}$ is smoothly homotopic to $f^{\mathrm{\prime }}$ by [F6]; since $f^{\mathrm{\prime }}$ is homotopic to $f$, the classes agree: $\mathrm{\Phi }(\mathrm{\Psi }[f])=[f]$.

(Conclusion.) Steps 1.1-1.2 define the two maps, and steps 2.1-2.2 show that their composites are the identities on the two sets; hence they are mutually inverse bijections. Composing $\mathrm{\Psi }$ with the identification of based and free classes [F7] gives the corresponding bijection with $[S^n,S^k]$. The case $n=k$ is included: the preimages are zero-dimensional, and the empty manifold is allowed as a framed submanifold and as a preimage. Only ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited through the transversality, approximation and normal-bundle suppliers, is used.