The framed preimage class is independent of regular value and positive basis

(Basis independence (i).) Let $N=f^{-1}(y)$ and let $b,b^{\mathrm{\prime }}$ be positive bases of $T_y{S}^k$. The change-of-basis matrix carries $b$ to $b^{\mathrm{\prime }}$ and has positive determinant, so [F1] gives a smooth path $b_t$, $t\in [0,1]$, of positive bases with $b_0=b$, $b_1=b^{\mathrm{\prime }}$; by the reparametrisation recorded in [F2] we may take $b_t$ smooth and constant near $t=0$ and $t=1$. On $N\times I\subseteq X\times I$ the normal bundle is canonically ${\mathrm{pr}}_N^{\ast }\nu (N\subseteq X)$, and the formula $\mathrm{\Psi }(x,t):=f_{\ast }{b}_t(x)$ defines a smooth bundle isomorphism $\nu (N\times I\subseteq X\times I)\to N\times I\times {\mathbb{R}}^k$ which over the end collar $t\in [0,\epsilon )$ equals the pullback of $f_{\ast }b$ and over $t\in (1-\epsilon ,1]$ the pullback of $f_{\ast }{b}^{\mathrm{\prime }}$. Hence $(N\times I,\epsilon ,\mathrm{\Psi })$ is a framed cobordism from $(N,f_{\ast }b)$ to $(N,f_{\ast }{b}^{\mathrm{\prime }})$ by [F2].

(A rotation family and a homotopy.) Let $y,y^{\mathrm{\prime }}\in S^k$ be regular values of $f$ and choose an orthonormal pair $u,v$ in ${\mathbb{R}}^{k+1}$ spanning a plane containing $y$ when $y^{\mathrm{\prime }}=-y$; when $y^{\mathrm{\prime }}=y$ take the constant identity family; otherwise there is an explicit rotation $R_1$, identity on the orthogonal complement of a two-plane and equal to a plane rotation there, with $R_1(y)=y^{\mathrm{\prime }}$: rotate in the plane $\mathrm{span}\{y,y^{\mathrm{\prime }}\}$ if $y^{\mathrm{\prime }}\mathrm{\ne }\pm y$, and by angle $\pi$ in a plane containing $y$ if $y^{\mathrm{\prime }}=-y$. Let $R_t$ be the corresponding family of rotations through angle $t\theta$, so that $t\mapsto R_t$ is smooth, $R_0=\mathrm{i}\mathrm{d}$, $R_1(y)=y^{\mathrm{\prime }}$, and each $R_t$ has determinant one by [F5]. Put $F(x,t):=R_t(f(x))$; this is a smooth homotopy from $f_0:=f$ to $f_1:=R_1\circ f$. The value $y^{\mathrm{\prime }}$ is a regular value of $f_0=f$ by hypothesis, and of $f_1$ because $f_1^{-1}(y^{\mathrm{\prime }})=f^{-1}(R_1^{-1}{y}^{\mathrm{\prime }})=f^{-1}(y)$ and $d{f}_1=d{R}_1\circ df$ is surjective there.

(The cobordism from the homotopy.) Apply [F3] to the smooth homotopy $F$ from $f$ to $R_1\circ f$ and the common regular value $y^{\mathrm{\prime }}$ with the positive basis $b^{\mathrm{\prime }}$: the framed preimages $(f^{-1}(y^{\mathrm{\prime }}),f_{\ast }{b}^{\mathrm{\prime }})$ and $((R_{1}f{)}^{-1}(y^{\mathrm{\prime }}),(R_{1}f{)}_{\ast }{b}^{\mathrm{\prime }})$ are framed cobordant. The second preimage is $f^{-1}(y)$; and since $d{R}_1$ is invertible and orientation-preserving, the basis $b^{\mathrm{\prime }\mathrm{\prime }}:=d{R}_1^{-1}(b^{\mathrm{\prime }})$ is positive at $y$ and $(R_{1}f{)}_{\ast }{b}^{\mathrm{\prime }}=f_{\ast }{b}^{\mathrm{\prime }\mathrm{\prime }}$ by the chain rule: the framing of the preimage of a composition is the framing of the inner map read through the invertible differential. Hence $(f^{-1}(y^{\mathrm{\prime }}),f_{\ast }{b}^{\mathrm{\prime }})$ is framed cobordant to $(f^{-1}(y),f_{\ast }{b}^{\mathrm{\prime }\mathrm{\prime }})$.

(Independence of the regular value (ii).) By step 1.1 applied to the regular value $y$ of $f$ and the two positive bases $b^{\mathrm{\prime }\mathrm{\prime }}$ and $b$, the framed preimages $(f^{-1}(y),f_{\ast }{b}^{\mathrm{\prime }\mathrm{\prime }})$ and $(f^{-1}(y),f_{\ast }b)$ are framed cobordant. Concatenating this cobordism with the one of step 2.1 by [F4] gives a framed cobordism from $(f^{-1}(y^{\mathrm{\prime }}),f_{\ast }{b}^{\mathrm{\prime }})$ to $(f^{-1}(y),f_{\ast }b)$, which is (ii).

(Consequence for smooth homotopies.) Let $f_0,f_1:X\to S^k$ be smoothly homotopic via $F$, and let $y_i$ be regular values of $f_i$ with positive bases $b_i$, $i=0,1$. By [F6] choose a common regular value $p$ of $f_0$ and $f_1$; choose any positive basis $c$ of $T_p{S}^k$. Applying [F3] to the homotopy $F$ and the common regular value $p$ gives a framed cobordism between $(f_0^{-1}(p),f_{0\ast }c)$ and $(f_1^{-1}(p),f_{1\ast }c)$; applying (ii) of step 3.1 to $f_0$ gives one between $(f_0^{-1}(y_0),f_{0\ast }{b}_0)$ and $(f_0^{-1}(p),f_{0\ast }c)$, and applying it to $f_1$ gives one between $(f_1^{-1}(y_1),f_{1\ast }{b}_1)$ and $(f_1^{-1}(p),f_{1\ast }c)$. Concatenating the three framed cobordisms by [F4] gives a framed cobordism between $(f_0^{-1}(y_0),f_{0\ast }{b}_0)$ and $(f_1^{-1}(y_1),f_{1\ast }{b}_1)$, as asserted.

(Conclusion.) Steps 1.1 and 3.1 prove (i) and (ii); step 4.1 proves that smoothly homotopic maps with any choices of regular values and positive bases have framed-cobordant preimages, so the framed cobordism class of a regular preimage depends only on the smooth homotopy class of the map. Empty preimages are included. The hypothesis $k\ge 1$ is essential: for a constant map $X\to S^0$, the two regular-value preimages are $X$ and $\mathrm{\varnothing }$, which need not be framed cobordant. Only ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited from the cited suppliers, is used.