Adding a trivial normal line suspends the Thom space

Write points of $E\oplus {\epsilon }^1$ as pairs $(v,t)$ with $v\in E$ and $t\in \mathbb{R}$, let $\mathrm{\parallel }\cdot \mathrm{\parallel }$ denote the metric of $E$ and $s(v,t)=\sqrt{\mathrm{\parallel }v{\mathrm{\parallel }}^2+t^2}$, $m(v,t)=\max (\mathrm{\parallel }v\mathrm{\parallel },\mathrm{\mid }t\mathrm{\mid })$ the sum and max norms. The fiberwise radial map $\chi (v,t)=\frac{s(v,t)}{m(v,t)}(v,t)$ for $(v,t)\mathrm{\ne }(0,0)$ and $\chi (0,0)=(0,0)$ is continuous, has continuous inverse $y\mapsto \frac{m(y)}{s(y)}y$ on the punctured set, and is continuous at zero along every direction because $1\le s\mathrm{/}m\le \sqrt{2}$ is bounded there. It carries the sum-norm disk onto the max-norm disk and, since $\mathrm{\parallel }\chi (v,t){\mathrm{\parallel }}_{\max }=s(v,t)$, carries the sum-norm sphere onto the max-norm sphere $S(E)\times [-1,1]\cup D(E)\times \{-1,1\}$. Thus $\chi$ is a homeomorphism of the two disk/sphere pairs.

When $S(E)\mathrm{\ne }\mathrm{\varnothing }$, by [F1] and step 1.1 the Thom quotient of $E\oplus {\epsilon }^1$ is the quotient of $D(E)\times [-1,1]$ by the union $A=S(E)\times [-1,1]\cup D(E)\times \{-1,1\}$. By [F2] the product of the two disk-to-quotient maps is a quotient map onto $(D(E)\mathrm{/}S(E)){\times }_k([-1,1]\mathrm{/}\{-1,1\})$. Compose it with the smash quotient, which collapses the two basepoint axes. This composite is a quotient map with one fibre $A$ and singleton fibres elsewhere, so it induces a homeomorphism $\mathrm{Th}(E\oplus {\epsilon }^1)\cong (D(E)\mathrm{/}S(E))\wedge ([-1,1]\mathrm{/}\{-1,1\})=\mathrm{Th}(E)\wedge S^1=\mathrm{\Sigma }\mathrm{Th}(E)\mathrm{.}$ When $B$ is compact the same identification is the ordinary quotient map of compact Hausdorff spaces.

In rank zero $S(E)=\mathrm{\varnothing }$, the convention $X\mathrm{/}\mathrm{\varnothing }=X_{+}$ supplies $B_{+}$ with its disjoint basepoint. The smash $B_{+}\wedge S^1$ is presented as $(B_{+}\times [-1,1])\mathrm{/}(B_{+}\times \{-1,1\}\cup \{\ast \}\times [-1,1])$. Collapsing the added basepoint component and the two endpoint copies gives exactly $(B\times [-1,1])\mathrm{/}(B\times \{-1,1\})$ with the based convention, which is the disk/sphere quotient of the trivial line bundle. Thus the same homeomorphism holds without treating $B\to B_{+}$ as an onto quotient map; for empty $B$ every space involved is a point. Every bundle map preserving the metrics and trivializations acts by the identity product formula and commutes with $\chi$ and with the quotient maps, so the homeomorphism is natural. Applying the result to the successive sums $E\oplus {\epsilon }^1\oplus \cdots \oplus {\epsilon }^1$ adds one suspension per specified trivial normal direction.