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Feb 22, 2015 at 14:27 comment added ThiKu I guess one has to invoke the tubular neighborhood theorem to see that all meridians are conjugate to each other.
Feb 22, 2015 at 12:50 comment added Alex Degtyarev Oops! Sorry. Still, this does not seem necessary. One can probably derive this from Zariski--van Kampen, but an easier way is to take a loop $\gamma$ in $V\setminus M$, assume that it bounds a disk, and then make this disk transverse to $M$. Then it's clear that $\gamma$ is the product of a bunch of loops conjugate to the meridian.
Feb 22, 2015 at 12:45 history edited user283635 CC BY-SA 3.0
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Feb 22, 2015 at 12:31 comment added user283635 The condition is $f_*\colon \pi_1(M)\to \pi_1(V)$ is surjective not $\operatorname{inc}_*\colon \pi_1(V-f(M))\to \pi_1(V)$ is surjective.
Feb 22, 2015 at 12:22 comment added Alex Degtyarev This is not a condition, as it always holds. E.g., because any loop in $V$ can be made disjoint from $M$.
Feb 22, 2015 at 11:58 review First posts
Feb 22, 2015 at 12:22
Feb 22, 2015 at 11:43 history asked user283635 CC BY-SA 3.0