Timeline for composition of covering map and bundle projection
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2011 at 8:16 | comment | added | Torsten Ekedahl | It depends on your definition of fibration, I have used the one which says that it locally is a product. (You did say fibre bundle though which is more likely to mean one where all the fibres are the same.) It is then a fact that you have the same fibre if the base is connected. Without that condition I am not totally sure the statement is true. In any case that is something I think is most easily shown a posteriori. | |
| Aug 26, 2011 at 6:21 | comment | added | Michael | To Torsten: Thanks for your explanation. I think your argument is really a local one, in view of the words "By shrinking $Z$", where you needn't worry about the fibers being the same. But once you have dealt with local neighborhoods, you need worry about whether the fibers in the various neighborhoods are homeomorphic so that you do get a universal typical fiber, which is required in the definition of fiber bundle. And I think connectedness of the base is a sufficient condition (so that every two neighborhoods are connected by a finite chain of neighborhoods). | |
| Aug 26, 2011 at 5:04 | comment | added | Torsten Ekedahl | Two things. One doesn't need a connected base to have a covering space theory and what I said about a homotopy equivalence inducing a category equivalence is true in general. Also that $Z$ is connected is part of the condition that $Z$ be contractible (or just simply connected). | |
| Aug 26, 2011 at 3:36 | comment | added | Michael | I think I get it now. I also think that one needs to assume the base space Z is connected so that the fibers associated with the various local neighborhoods are homeomorphic. | |
| Aug 26, 2011 at 1:48 | comment | added | Michael | Still don't understand. Torsten seems to be saying if $X \rightarrow Z \times F$ is a covering, then $\exists$ a covering $F' \rightarrow F$ such that $X \cong Z \times F'$. I can prove the existence of $F'$, but am unable to demonstrate a homeomorphism $X \cong Z \times F'$, unless $X$ is connected (so that covering space theory applies). | |
| Aug 25, 2011 at 11:11 | history | edited | Mark Grant | CC BY-SA 3.0 |
fixed typo
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| Aug 25, 2011 at 7:48 | history | edited | Torsten Ekedahl | CC BY-SA 3.0 |
deleted 1 characters in body
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| Aug 25, 2011 at 7:40 | history | edited | Torsten Ekedahl | CC BY-SA 3.0 |
Elaboration
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| Aug 25, 2011 at 6:32 | comment | added | Michael | As I tried to unravel Torsten's concise proof today, I found myself unable to prove the existence of $X'$. The difficulty lies in that one really needs to argue locally and there is no guarantee that the total space of the cover is connected, which needs to be true for the existence of $X'$. Could Torsten elaborate on why $X'$ must exist? Thanks. | |
| Aug 24, 2011 at 7:56 | vote | accept | Michael | ||
| Aug 24, 2011 at 7:55 | comment | added | Michael | Thanks to Torsten's answer. The existence of $X'$ is especially enlightening. | |
| Aug 24, 2011 at 6:51 | comment | added | Chris Gerig | obviously the fiber. | |
| Aug 24, 2011 at 6:33 | comment | added | user5810 | What is $F\hspace{.04 in}$? | |
| Aug 24, 2011 at 6:32 | history | answered | Torsten Ekedahl | CC BY-SA 3.0 |