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Let $E\rightarrow B$ be a bundle with an inner product. Using $D(E):=\left\{v\in E \ : \ u(v,v)\leq1\right\}$ and $S(E):=\left\{v\in E \ : \ u(v,v)=1\right\}$, we define a Thom space: $$\text{Th} \ E= D(E)/S(E).$$

In my book there is a fact which I cannot imagine, namely, if $B$ is compact then $\text{Th} \ E$ is homeomorphic to one-point compactification of the space $E$. Probably this is trivial but can anyone show any construction?

In the book there is a hint but still not enough:

We want to compactify all the fibers separately, hence form the pushout of $$B\leftarrow S(E) \rightarrow D(E).$$

The left arrow should be a surjection (because $S(E)\rightarrow B$ is a fiber bundle) and clearly the right one is an inclusion. The inclusion $B\rightarrow P$ (why do we know it's inclusion indeed?) is the 'section at infinity (also don't imagine this). Now one should see that the quotien $P/B$ is the one-point compactification of the space E and the final result follows since $P/B$ is homeomorphic with $D(E)/S(E)$.

If it is possible, can anyone in more basic terms?

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I think the book means the 'fibrewise one-point compactification'.

By assumption The fibre $E_b$ over each point $b\in B$ has the structure of a $\mathbb{K}$-vector space for some field $\mathbb{K}$. Then we can write $E=\bigcup_{b\in B}E_b$, suitably toplogised using the bundle atlas which defines the bundle stucture. The fibrewise one point compactification is then defined by $E^+_B=\bigcup_{b\in B}E_b^+$, where $E_b^*$ denotes the 1-point compactification of the vector space $E_b$, obtained by adding to it a single point at infinity $\infty_b$ . There is an obvious projection $E_B^+\rightarrow B$ extending that of $E\rightarrow B$, defined by $\infty_b\mapsto b$, as well as an obvious section $s_{\infty}:B\rightarrow E^+_B$. The space $E^+_B$ is topologised ion the unique way so that the map $E\rightarrow E_B^+$ over $B$ restricts to each fibre to give a one-point compactification $E_b\hookrightarrow E_b^+$, and so that the section $s_\infty:B\rightarrow E^+_B$ embeds $B$ as a closed subset.

Now define the Thom space $Th'(E)$ as the quotient of $E^+_B$ by the section at infinity,

$Th'(E)=E^+_B/s_\infty(B)$.

Observe that as a set $E^+_B=E\sqcup B$, so as a set $Th'(E)=E\sqcup \ast$, and we do indeed recover that the Thom space is the one-point compactification of $E$. You might like to check that when $B$ is compact, the Alexandroff topology agrees with that given by the quotient $E^+_B/s_\infty(B)$.

Now there is a map $D(E)\rightarrow E^+_B$ defined by $e\mapsto e/(1-|e|)$, which sends points $e_b$ in the fibre of the sphere bundle $S(E)_b$ to the point at infinity $\infty_b$. This map is not continuous, but fibrewise it gives the standard isomorphism $D(E_b)/S(E_b)\cong E_b^+$. Now observe that the composition $D(E)\rightarrow E^+_B\rightarrow Th'(B)$ factors over the quotient $D(E)\rightarrow D(E)/S(E)$ to produce a map

$Th(E)=D(E)/S(E)\xrightarrow{\theta} Th'(E)=E^+_B/s_{\infty}(B)$

that is indeed continuous. This map $\theta$ is the homeomorphism we are looking for.

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  • $\begingroup$ Thank you, almost everything is clear. How do we know that $E^+_B$ is the disjoint union of $E$ and $B$? The same question for $Th'(E)$ and what the $*$ means? And the second question: we have an isomorphism in the last paragraph on the fibres level, so why $\theta$ is a homeo? $\endgroup$ – Yelon Mar 26 '18 at 11:39
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    $\begingroup$ It's not. Only as a set it is, the topology is different. To form $E^+_B$ we have added to each fibre a single point at infinity $\infty_b$, and used these points to define the section $s_\infty:B\rightarrow E^+_B$. The collection of all these points at infinity is the copy of $B$ in $E^+_B$. Any other point in $E^+_B$ not in the image of $s_\infty$ is contained in $E$. $\endgroup$ – Tyrone Mar 26 '18 at 11:44
  • $\begingroup$ Okay, so what about $\theta$ (second part of the question) $\endgroup$ – Yelon Mar 26 '18 at 12:13
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    $\begingroup$ For me $\ast$ denotes an arbitrary one-point space. You can check that the map $D(E)\rightarrow E^+_B\rightarrow Th'(E)$ that induces $\theta$ is bijective away from the (now unique) point at infinity, and the inverse image of this point is $S(E)$. Hence $\theta$ is bijective. Since we are assuming that $B$ is compact, we now have a bijective map between compact spaces, which is thus a homeomorphism for general reasons. $\endgroup$ – Tyrone Mar 26 '18 at 12:31

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