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Nov 28, 2011 at 14:31 comment added Autumn Kent I am saying that if being interior to a ball were the same as being trivial in the fundamental group, then you would have a new proof of the Poincare conjecture, which suggests that the assumption is false. The fact that they are not the same is easy to see: one component of the Whitehead link is null homotopic in the exterior of the other, but is not interior to a ball there.
Nov 28, 2011 at 5:19 comment added yanqing @binyu : Are you from tongji university, Shanghai?
Nov 27, 2011 at 21:47 comment added Autumn Kent Also, you need more reputation to comment. This is to prevent you from commenting until you've been around a while.
Nov 27, 2011 at 21:45 comment added Autumn Kent Being interior to a ball is not the same as being trivial in the fundamental group. In fact, it is a theorem of Bing that M is the 3-sphere if and only if every knot is interior to a ball. So, the Poincare conjecture would follow from your "visual meaning."
Nov 27, 2011 at 21:24 comment added Bin Yu Where do you use $[K]=0$ in $\pi_1{M}$? I feel that your answer is an explain of Dehn's lemma. I guess that "$[K]=0$" visual meaning is "$K\subset D^3 \subset M$" where $D^3$ is a 3 ball(similar to knot in $S^3$).
Nov 26, 2011 at 12:05 history edited Sam Nead CC BY-SA 3.0
Its now fairly detailed!
Nov 26, 2011 at 12:01 comment added Sam Nead @Yanqing - My answer gives a necessary and sufficient condition for $K$ to bound an embedded disk in $M$. I'll restate this at the end of the answer.
Nov 26, 2011 at 1:17 comment added yanqing To Sam: So it is not clear whether $K$ bounds an embedded or not.
Nov 26, 2011 at 1:12 vote accept yanqing
Nov 25, 2011 at 21:58 history edited Sam Nead CC BY-SA 3.0
many typos!
Nov 25, 2011 at 17:35 history answered Sam Nead CC BY-SA 3.0