Timeline for Is the following 3-manifold always a trivial I-bundle over a surface?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 12, 2015 at 10:04 | answer | added | H1ghfiv3 | timeline score: 0 | |
| Dec 11, 2015 at 20:11 | vote | accept | H1ghfiv3 | ||
| Dec 11, 2015 at 20:11 | comment | added | H1ghfiv3 | It is a remarkable skill, being able to even write coherent setences while fast asleep. A small mistake does not matter | |
| Dec 11, 2015 at 19:45 | comment | added | Danny Ruberman | Yes, $H_3$. Serves me right for answering questions before waking up. | |
| Dec 11, 2015 at 19:32 | history | edited | Sam Nead | CC BY-SA 3.0 |
last change, I hope.
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| Dec 11, 2015 at 19:27 | answer | added | Sam Nead | timeline score: 2 | |
| Dec 11, 2015 at 13:02 | comment | added | H1ghfiv3 | You probably mean $H_3(M,\partial M)$, don't you ? I was aware of the equality of homological degree and covering degree for closed manifolds, I guess the generalization to compact manifolds is straightforward. | |
| Dec 11, 2015 at 12:50 | comment | added | Danny Ruberman | $f$ has degree 1 as a map of manifolds with boundary, ie the degree of the map on $H_2(M,\partial M) \cong Z$. This is because the degree of the map on the boundary is 1. That the degree is the same as the covering degree is a basic fact that you can find in textbooks. | |
| Dec 11, 2015 at 12:00 | comment | added | H1ghfiv3 | What exactly do you mean by degree here ? I assume you are talking about the induced map in second homology, since we have manifolds with boundary here. Necessarily, $H_2(M)$ should always be $\mathbb Z$, but why is this given ? And why does this degree again coincide with the covering degree of a covering map ? | |
| Dec 11, 2015 at 11:50 | comment | added | user83633 | Show that $f$ is a degree 1 map. Degree 1 maps must be surjective on the fundamental group (since they can't lift to a non-trivial cover). | |
| Dec 11, 2015 at 11:13 | comment | added | H1ghfiv3 | I have changed the notation. Everything should be clear now | |
| Dec 11, 2015 at 11:11 | history | edited | H1ghfiv3 | CC BY-SA 3.0 |
added 108 characters in body
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| Dec 11, 2015 at 11:05 | review | Close votes | |||
| Dec 11, 2015 at 15:41 | |||||
| Dec 11, 2015 at 10:49 | comment | added | Ryan Budney | Could you explain your notation? Your question makes no sense to me. I assume $S_g$ is a surface, but $\cong$ is what? | |
| Dec 11, 2015 at 10:04 | comment | added | ThiKu | Perhaps a possible approach: using Morse theory your bordism can be decomposed into handles. If $M$ is not a product, then it should be obtained by adding handles to $S_g\times I$. And then your homotopy can be made transversal to the handles of that decomposition. | |
| Dec 10, 2015 at 22:35 | history | asked | H1ghfiv3 | CC BY-SA 3.0 |