thom space associated to a fibration

allowed $p:S\rightarrow B$ be a fibration wiht fiber a sensible $r$ - homology round $\Sigma^r$, i.e., $H_*(\Sigma^r;\mathbb Q)=H_*(S^r;\mathbb Q)$. to such a fibration we associate its Thom room $$MS=(S\times I \cup_p B)/S\times\{1\}$$ where we take into consideration $p$ as a map $S\times\{0\}\rightarrow B$.

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

2) just how does this connect to the grandfather clause of the thom room of a vector package?

3) per $b\in B$ we specify $$M\Sigma^r_b:=\Sigma^r_b\times I /(\Sigma^r_b\times \{0\}\cup \Sigma^r_b\times \{1\})$$ what is $M\Sigma^r_b$ and also just how can we analyze it?

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2022-06-07 14:31:23
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Answers: 1

1) Yes, you can specify Thom rooms as mapping cones.

2) Certainly if your round package originates from a vector package after that you can construct the Thom room in either means. I do not recognize much concerning sensible homology rounds, yet certainly there are round packages (and also therefore sensible round packages) that do not originate from vector packages. This is very easy to state in the language of identifying rooms (or equivalently in the language of change features, if you take place to be collaborating with real fiber packages): the noticeable incorporation $O(n)\rightarrow Homeo(S^{n-1})$ is a homotopy equivalence iff $n\leq 3$ (I assume, possibly Homeo needs to be Diffeo ; regardless, the factor still stands).

3) It resembles this is intended to be the "restriction" of the Thom room over the factor $b$. I placed "restriction" in quotes given that the Thom room no more has an estimate to the base room, given that there is no place to place the new base factor. Other than when you write the gluing relationship as you have, this is generally analyzed to suggest that all the factors in $\Sigma^r_b \times \partial I$ are recognized, whereas I assume you simply intend to crisis $\Sigma^r_b \times \{0\}$ to a factor and also $\Sigma^r_b \times \{1\}$ to a factor. You can write this even more suggestively (w/r/t your inquiry 1) as $(\Sigma^r_b \times I\cup_{p|_{p^{-1}(b)}}\{b\})/(\Sigma^r_b \times \{1\})$.

Equally as a little reminder, from what I've seen, probably one of the most intriguing feature of Thom rooms is the Thom isomorphism theory. Keep in mind that when you take the Thom room of a unimportant vector package and also use the isomorphism theory, you come back the common suspension isomorphism! So you can consider the Thom isomorphism as being a generalization: all Thom rooms have cohomology teams isomorphic to those of the base yet changed by the measurement of the fiber. Certainly, various packages have various Thom courses (not that they can always be contrasted anyways, given that they stay in various areas), so it is not such as the Thom isomorphism theory is claiming that single cohomology is callous twistedness

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2022-06-07 14:57:04
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