The collapse of a regular preimage is homotopic to the original map

First suppose $y=y_0$, $b=b_0$. Shrink a framed tube so both maps lie in the centre coordinate chart there; use $u={\phi }_s(v)$ as the fibre coordinate. In these coordinates $F(s,0)=G(s,0)=0$ and $D_{u}F(s,0)=D_{u}G(s,0)=I$, with $G(s,u)=u$ on a sufficiently small tube by [F1]. A finite cover of compact $N$ bounds the second fibre derivatives of $F$; integrating the derivative along each fibre segment gives $\mathrm{\mid }F(s,u)-u\mathrm{\mid }\le C\mathrm{\mid }u{\mathrm{\mid }}^2$ uniformly. Choose $c>0$ small enough that $Cc<1\mathrm{/}2$, the estimates hold for $\mathrm{\mid }u\mathrm{\mid }\le c$, and $G(s,u)=u$ there. Thus for $0<\mathrm{\mid }u\mathrm{\mid }\le c$, $F(s,u)\cdot u\ge \mathrm{\mid }u{\mathrm{\mid }}^2-C\mathrm{\mid }u{\mathrm{\mid }}^3>0$ and $G(s,u)\cdot u=\mathrm{\mid }u{\mathrm{\mid }}^2>0$. This uses framing coordinates, not a false positivity inference about an arbitrary invertible matrix.

Let $\lambda (\mathrm{\mid }u{\mathrm{\mid }}^2)$ be a smooth cutoff equal to one for $\mathrm{\mid }u\mathrm{\mid }\le c\mathrm{/}2$ and zero for $\mathrm{\mid }u\mathrm{\mid }\ge c$. In the target centre chart put $F_t=(1-t\lambda )F+t\lambda G$ and use $f$ elsewhere. On the support both vectors have positive dot product with $u$ when $u\mathrm{\ne }0$, so no extra centre preimage is created. The family equals $f$ on a neighbourhood of the tube boundary, hence pastes smoothly. Its final map $f_1$ agrees with $g$ on $V=\{\mathrm{\mid }u\mathrm{\mid }<c\mathrm{/}2\}$ and with $f$ outside a compact tube $U$.

On $X\setminus N$, both $f_1$ and $g$ avoid $y_0$. In stereographic coordinates $h:S^k\setminus \{y_0\}\to {\mathbb{R}}^k$ interpolate $h(f_1)$ linearly to $h(g)$. On $V$ use the constant family $f_1=g$. These formulas agree on $V\setminus N$, so they paste to a smooth homotopy constant near $N$. Endpoint-flat reparametrization makes its concatenation with step 2.1 smooth. If $N=\mathrm{\varnothing }$, take $U=V=\mathrm{\varnothing }$ and simply interpolate the two maps in the chart avoiding $y_0$.

For general $y,b$, choose a rotation $R$ joined smoothly to the identity with $R(y)=y_0$; the usual plane rotation handles nonantipodal points, the identity handles equality, and a $\pi$ rotation handles antipodes. Put $f^{\mathrm{\prime }}=Rf$. Its fibre at $y_0$ is $N$, and the pushed basis $d{R}_{y}b$ induces exactly $\phi$. By [F4], changing that positive basis to $b_0$ gives a framed cobordism from $(N,\phi )$ to the framed preimage $(N,{\phi }^{\mathrm{\prime }})$ of $(f^{\mathrm{\prime }},y_0,b_0)$. Their collapses are homotopic. Steps 1.1–3.1 compare $f^{\mathrm{\prime }}$ to the collapse of $(N,{\phi }^{\mathrm{\prime }})$, while the rotation path compares $f$ to $f^{\mathrm{\prime }}$. The concatenation proves the claimed homotopy; [F4] makes it smooth.

If $f$ is only continuous away from its regular fibre, the local deformation of steps 1.1–2.1 is still smooth on a small tube and continuous elsewhere, and the chart interpolation of step 3.1 is continuous. For step 4.1, basis independence is the explicit framed cylinder using a path of positive bases, which requires only the local differential on the fibre. The same constructions therefore give a continuous homotopy to the collapse. The regular fibre is compact because it is closed in $X$. All choice is inherited from the stated suppliers.