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Jul 31, 2014 at 5:48 vote accept Oliver Straser
Jul 31, 2014 at 1:06 answer added S. Carnahan timeline score: 7
Jul 30, 2014 at 14:42 comment added Daniel Valenzuela I guess you can freely homotop an element in $[S^1,\mathbb R^3 -(Z-F)]$ parallel to $Z$ into an orthogonal plane of $Z$ based at a point of $F$. You can easily contract it in this plane to a point without problems. Hence the statement.
Jul 30, 2014 at 14:18 comment added ThiKu The same should hold true for any loop in $R^3-Z$, just that in this case the disks won't be embedded anymore.
Jul 30, 2014 at 14:17 comment added ThiKu I think so. Look at a loop representing the generator of $\pi_1(\R^3-Z)$. As soon as you remove at least one point from Z, then this loop will bound some embedded disk.
Jul 30, 2014 at 13:53 history asked Oliver Straser CC BY-SA 3.0