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181:. The usual thing is then to say, take both boundary circles of the annulus and identify them to one point P. We can get this in stages: identify (b,0) first with (b,1), and that's a 2-torus T. Then we collapse a circle on T to the point P. No problem about projecting T to B; but when we collapse the circle? There would have to be some symmetry breaking going on.
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OK, I see what my problem was. If we do as
Charles has suggested above and first take the bundle of one-point compactifications, we do get a sphere bundle. But the Thom complex is the quotient of the total space of the sphere bundle obtained by gluing all the points at infinity together. I edited the
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I think the answer to your question is yes. If E is a k-vector bundle then T(E) is a k-sphere bundle. So we are taking the one-point compactification of each fiber, remembering that we've only added one new point. About T(p): maybe we need more hypotheses for this. Here is my reasoning. Since E is
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Forgive me if I'm being stupid, but I don't see how T(p) : T(E) → B is a sphere bundle. To which point in the base does the "new point" project? In order to get a sphere bundle it seems like it would have to project everywhere (it lies in the "fiber" above every point). --
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We can loosely interpret the theorem in the following geometric sense. Since E is a vector bundle it retracts onto the base B. So we might suppose that E would be cohomologically equivalent to B. In a way, the theorem bears out this expectation
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Not such a great idea to use the same notation for the Thom space — T(M) — that is used for the tangent bundle of the manifold M. Especially in an article like this one where there is constant danger of confusion. Why not Th(M) or Thom(M) or
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I am the one who is stupid; you're absolutely right. So can we agree that T(E) is not so much a bundle over anything (not sure why I ever thought so, at this point...)? And that all references to such should be removed?
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Funnily enough, I was just reading it now, came across that passage, and wondered, what on earth? Please include your comments into the article itself. There is no need to be timid. --
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Bear with me a moment. Take B a circle and E the trivial bundle BxR. Then what we do is take Bx inside E; think of this as an
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Is it not firstly the bundle of one-point compactifications, though? I don't see how T(p) is defined.
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Knowledge. If you would like to participate, please visit the project page, where you can join
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In the name of precision, I just want to point out that the Thom space of the trival bundle over
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B is continuous there is a unique map T(p) extending p to all of T(E). Am I missing something?
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definition to reflect this. Thanks both of you for your helpful and timely corrections!
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The article attempts to explain the meaning of the Thom isomorphism as follows:
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I agree and will go ahead and change the notation. —-
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407:{\displaystyle \Sigma ^{n}B}
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