Timeline for Can the fundamental group of an intersection of a homeomorphic image of a ball with a complement of a ball in $R^3$ be perfect?
Current License: CC BY-SA 3.0
13 events
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| Aug 28, 2012 at 9:20 | comment | added | Pawel Goldstein | I still wonder, however, if one can prove that no component has nontrivial perfect $\pi_1$... By the way, Misha, could you give a reference to the result you mention, on non-triviality of $H_1$ for compact 3-manifolds? It seems something one should know; thanks in advance. | |
| Aug 28, 2012 at 9:04 | comment | added | Pawel Goldstein | No, one is enough, thanks a lot for your answer. The source of the problem is metric topology, in particular questions of local connectivity of certain sets. The standard question is: taking a homotopy trivial arc in A, that is contained in a small ball, can you glue a disk into it that is contained in a ball twice the size? And the same for the complement of A and arcs outside a ball - how much space contracting the arc might take. This is phrased in homotopy terms, and I wondered if there is an obstruction for rephrasing it homologically. | |
| Aug 28, 2012 at 8:53 | vote | accept | Pawel Goldstein | ||
| Aug 28, 2012 at 2:18 | answer | added | Ian Agol | timeline score: 10 | |
| Aug 28, 2012 at 1:48 | comment | added | Ian Agol | To clarify, do you want to know if each component of $A\backslash B$ has perfect fundamental group? | |
| Aug 27, 2012 at 23:20 | answer | added | Roberto Frigerio | timeline score: 1 | |
| Aug 27, 2012 at 21:49 | comment | added | Misha | @Igor: If you use, say, PL balls, then perfection is equivalent to triviality of the fundamental group: A compact 3-manifold with boundary different from a union of spheres, always has nontrivial $H_1$. On the other hand, if you remove a doubly wild arc from an open ball, the complement has perfect nontrivial $\pi_1$: However, this is an example for $B-A$, not $A-B$ as requested. | |
| Aug 27, 2012 at 14:34 | comment | added | Igor Rivin | Where does this question come from? WHy should perfection be hard to attain? | |
| Aug 27, 2012 at 11:56 | comment | added | Pawel Goldstein | Of course, that is what I had in mind | |
| Aug 27, 2012 at 11:23 | comment | added | Mark Grant | In the generic case the fundamental group is trivial, hence perfect. But I guess you are asking if the fundamental group of (some component of) $A\setminus B$ can be a non-trivial perfect group. | |
| Aug 27, 2012 at 11:00 | comment | added | Pawel Goldstein | As I wrote, I am curious about all 4 possibilities, but the case of $A$ closed, $B$ open (to have $B\setminus A$ compact) would probably do. | |
| Aug 27, 2012 at 10:15 | comment | added | YCor | homeomorphic to closed or open ball? | |
| Aug 27, 2012 at 10:13 | history | asked | Pawel Goldstein | CC BY-SA 3.0 |