# Clarification about the Thom-Pontrjagin construction as explained in Bredon's book

In Bredon's book, at page 118-119, there is a little chapter about the Thom-Pontrjagin construction, and I'm trying to follow the reason depicted there.

He starts with a map $f \colon R^{n+k}\to R^n_+$ (constant outside a compact) coming from a map $f \colon S^{n+k}\to S^n$ where we identify the spheres with the one-point compactification of the respective real spaces (denoted with $R^n_+$), and he builds a fattened manifold $M^k\times E^n \xrightarrow{embedded} R^{n+k}$ using such map $f$. Then he introduce this map $\theta$, homotopic to the identity of $R^n_+$ which spreads $E^n$ on all $R^n$ and everything else to $\infty$. and now it comes the part which is confusing me:

Then the composition $\theta\circ f\simeq f$ is the same as the map taking $N\cong M^k\times E^n \to E^n$ by the projection followed by a diffeomorphism $E^n\to R^n$ and taking everything else to $\infty$. Therefore every fattened $k-$manifold $g\colon M^k\times E^n \to R^{n+k}$ gives rise to a map $\phi_g \colon R^{n+k} \to R^n_+$ of this form, and every map $R^{n+k} \to R^n_+$ as above, is homotopic to a map arising in this way. I don't get how a fattened manifold induces the map $\phi_g$, because the only idea which came to my mind is to include the fat. manifold in $R^{n+k}$, project to the second factor $E^n$, and then use the map $\theta$ (as suggested by the picture above?) but in this way the induced map is always null-homotopic, because $E^n \neq S^n$ and therefore by composing with a stereographic projection we can "shrink" it to a point.

So how does this "assignment" works precisely?

## 1 Answer

I think, "fattern manifold" is $k$-submanifold $M\subset\mathbb R^{n+k}$ with trivialized normal bundle, so you can identify neighborhood of $M$ with $M\times E^n$, $g$ is the inclusion $M\times E^n\subset \mathbb R^{n+k}$. And the map $\theta_g$ is constructed as follows: you should project $M\times E^n$ on the $E^n$, then cover $S^n$ without $\infty$ by this $E^n$, and send all the $\mathbb R^{n+k}\setminus (M\times E^n)$ to $\infty$.

If you continue $\theta_g$ on the $S^{n+k}=\mathbb R^{n+k}\cup\{\infty\}$ (let $\infty\mapsto\infty$), this mapping is not always null-homotopic. For example, if $n=2$, $k=1$ and $M$ is $S^1$ with standart embedding into $\mathbb R^3$, the map $\theta_g:S^3\to S^2$ will be homotopic to Hopf fibration (you can prove is as a beautiful geometric-topology exercise).

EDIT: But fixed trivialization of normal bundle (or neighborhood) of $S^1$ with $S^1\times \mathbb R^2$ must be not "trivial", it must be twisted so that any two constant sections form the Hopf link.

• (sorry for late answering your nice answer) so there must be a fault in my reasoning about the null-homotopicness of the map. What can possibly go wrong so? – Riccardo Jul 31 '15 at 8:19
• of course, the map $M\times E^n\to S^n$ can be null-homotopic, but when the source becomes $S^{n+k}$ it already cannot. the easiest example is when as the $M$ you take one point (so $k=0$), this brings us the identity map $S^{n+0}\to S^n$ . – Andrey Ryabichev Jul 31 '15 at 9:50
• i was wondering.. do you have some hints on how to do the "exercise" of the Hopf Fibration? I tried writing down the maps but I don't see where the Hopf map comes out let's say – Riccardo Jul 31 '15 at 14:32
• sorry, there is a mistake: the trivialization of normal bundle with $S^1\times \mathbb R^2$, that you fix, must be not "trivial", it must be twisted so that any two constans sections form the Hopf link (otherwise preimages of different points of $S^2$ will be not linking and the map really will be null-homotopic). if you, conversely, take the Hopf map, just look on the preimages of different points, identifying $S^3\setminus\{\infty\}$ with $\mathbb R^3$. – Andrey Ryabichev Jul 31 '15 at 15:04
• Dear Andrey, I posted another question about this topic, do you have time to give it a look? because I really like your answer here math.stackexchange.com/questions/1386435/… – Riccardo Aug 6 '15 at 11:30