Metric independence of the Thom space

Define $r_{h,k}(0_b)=0_b$ and $r_{h,k}(v)=(\mathrm{\parallel }v{\mathrm{\parallel }}_h\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_k)v$ for $v\mathrm{\ne }0$, as in [F2]. Homogeneity gives $\mathrm{\parallel }{r}_{h,k}v{\mathrm{\parallel }}_k=\mathrm{\parallel }v{\mathrm{\parallel }}_h$, so $r_{h,k}$ maps the $h$-disk to the $k$-disk and the $h$-sphere to the $k$-sphere. The formula is continuous on $E\setminus \{0\}$, where both norms are continuous and nonzero, and it is continuous at $0$ because $\mathrm{\parallel }{r}_{h,k}v{\mathrm{\parallel }}_k=\mathrm{\parallel }v{\mathrm{\parallel }}_h\to 0$; its inverse is $r_{k,h}$.

The pair homeomorphism descends to a based quotient homeomorphism ${\mathrm{Th}}_h(E)\to {\mathrm{Th}}_k(E)$. For $v\mathrm{\ne }0$, substituting the formulas gives $r_{k,l}(r_{h,k}(v))=\frac{\mathrm{\parallel }{r}_{h,k}(v){\mathrm{\parallel }}_k}{\mathrm{\parallel }{r}_{h,k}(v){\mathrm{\parallel }}_l}{r}_{h,k}(v)=\frac{\mathrm{\parallel }v{\mathrm{\parallel }}_k}{\mathrm{\parallel }v{\mathrm{\parallel }}_l}\cdot \frac{\mathrm{\parallel }v{\mathrm{\parallel }}_h}{\mathrm{\parallel }v{\mathrm{\parallel }}_k}v=r_{h,l}(v),$ and all three maps fix $0$, so the composition law holds exactly. The positive-definite family $h_t=(1-t)h+tk$ gives the radial isotopy $r_{h,h_t}$ from the identity to $r_{h,k}$. Empty bases and rank zero have identity maps with the conventions of [F1].