# How to use the Prontrjagin-Thom construction to obtain the Gysin map?

I need help to understand the diagram in Miller's script Vector Fields on Spheres, etc. Chapter 23, p.82 on the bottom of the page.

Before, Miller introduces the Prontrjagin-Thom construction: It is for a locally compact Hausdorff space $X$ and an open subspace $U\subseteq X$ the collaps map $P:\bar X\to \bar X/(\bar X\setminus U)\simeq \bar U$ where the upper bar denotes the one-point compactification.

For the diagram in question, he takes a smooth fibration of compact mainfolds $F\to E\to B$, an embedding $E\subset \mathbb{R}^n$ and the induced map $$E\xrightarrow{i} B\times \mathbb{R}^n$$ over $B$. Moreover, $\nu(i)$ is the normal bundle of $i$ and $N\subseteq B\times \mathbb{R}^n$ a tubular neighbourhood.

The Prontrjagin-Thom construction provides a map $P:\overline{B\times \mathbb{R}^n}\to \overline{N}$.

There is a nice picture on top of page 83: $B=[0,1]$ is an interval, $n=1$ and $E=B+B+B$ is just a trivial $3$-sheeted covering. Then, $\overline{B\times \mathbb{R}^n}\simeq S^2$ and $\overline{N}\simeq S^2\vee S^2\vee S^2$, so far I understand.

Now in the diagram on the bottom of page 82, it is claimed that $B^{n\epsilon}\simeq \overline{B\times \mathbb{R}^n}$ and that $E^{\nu(i)}\simeq \overline{N}$ where the exponent refers to the Thom space construction and $n\epsilon$ is the $n$-dimensional trivial bundle over $B$. I don't understand that. In the example above, the Thom space of $1\epsilon:\mathbb{R}^1\times [0,1]\to [0,1]$ is homotopy equivalent to $S^1$ (and not to $S^2$!): One-point compactification in every fiber gives a cylinder and collapsing $\infty\times [0,1]$ to a point is homotopy equivalent to $S^1$.

Where is my misunderstanding?

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