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let $p:S\rightarrow B$ be a fibration wiht fiber a rational $r$-homology sphere $\Sigma^r$, i.e., $H_*(\Sigma^r;\mathbb Q)=H_*(S^r;\mathbb Q)$. to such a fibration we associate its Thom space $$MS=(S\times I \cup_p B)/S\times\{1\}$$ where we consider $p$ as a map $S\times\{0\}\rightarrow B$.

1) isn't $MS $ the same as the mapping cone of $p$?

2) how does this relate to the special case of the thom space of a vector bundle?

3) to each $b\in B$ we define $$M\Sigma^r_b:=\Sigma^r_b\times I /(\Sigma^r_b\times \{0\}\cup \Sigma^r_b\times \{1\})$$ what is $M\Sigma^r_b$ and how can we interpret it?

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up vote 1 down vote accepted

1) Yes, you can define Thom spaces as mapping cones.

2) Certainly if your sphere bundle comes from a vector bundle then you can construct the Thom space in either way. I don't know much about rational homology spheres, but of course there are sphere bundles (and hence rational sphere bundles) that do not come from vector bundles. This is easy to state in the language of classifying spaces (or equivalently in the language of transition functions, if you happen to be working with actual fiber bundles): the obvious inclusion $O(n)\rightarrow Homeo(S^{n-1})$ is a homotopy equivalence iff $n\leq 3$ (I think, maybe Homeo should be Diffeo; in any case, the point still stands).

3) It looks like this is supposed to be the "restriction" of the Thom space over the point $b$. I put "restriction" in quotes since the Thom space no longer has a projection to the base space, since there's nowhere to put the new base point. Except when you write the gluing relation as you have, this is usually interpreted to mean that all the points in $\Sigma^r_b \times \partial I$ are identified, whereas I think you just want to crunch $\Sigma^r_b \times \{0\}$ to a point and $\Sigma^r_b \times \{1\}$ to a point. You could write this more suggestively (w/r/t your question 1) as $(\Sigma^r_b \times I\cup_{p|_{p^{-1}(b)}}\{b\})/(\Sigma^r_b \times \{1\})$.

Just as a little pointer, from what I've seen, perhaps the most interesting thing about Thom spaces is the Thom isomorphism theorem. Note that when you take the Thom space of a trivial vector bundle and apply the isomorphism theorem, you get back the usual suspension isomorphism! So you can think of the Thom isomorphism as being a generalization: all Thom spaces have cohomology groups isomorphic to those of the base but shifted by the dimension of the fiber. Of course, different bundles have different Thom classes (not that they could necessarily be compared anyways, since they live in different places), so it's not like the Thom isomorphism theorem is saying that singular cohomology is blind to twistedness...

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