1
$\begingroup$

Suppose that $M^n$, $V^{n+2}$ are connected, compact smooth manifolds.

Let $f\colon M^n\to V^{n+2}$ is a smooth embedding.

Let $K_f$ be the kernel of the inclusion induced homomorphism $\pi_1(V-f(M))\to \pi_1(V)$.

Choose a meridian $\mu\in \pi_1(V-f(M))$.

Proposition 1.3 of Smith's paper, Complements of codimension two sub-manifolds I: The Fundamental group, Illinois J. Math. 1978 p. 232-239 says that if $f_*\colon \pi_1(M)\to \pi_1(V)$ is a surjection, then $K_f$ is normally generated by $\mu$.

Question: Is the condition that $f_*\colon \pi_1(M)\to \pi_1(V)$ is surjective necessary? That is, is there an embedding $f\colon M\to V$ such that $K_f$ is not normally generated by just one meridian?

Improve this question
$\endgroup$
4
  • $\begingroup$ This is not a condition, as it always holds. E.g., because any loop in $V$ can be made disjoint from $M$. $\endgroup$
    – Alex Degtyarev
    Commented Feb 22, 2015 at 12:22
  • $\begingroup$ 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. $\endgroup$
    – user283635
    Commented Feb 22, 2015 at 12:31
  • 2
    $\begingroup$ 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. $\endgroup$
    – Alex Degtyarev
    Commented Feb 22, 2015 at 12:50
  • $\begingroup$ I guess one has to invoke the tubular neighborhood theorem to see that all meridians are conjugate to each other. $\endgroup$
    – ThiKu
    Commented Feb 22, 2015 at 14:27

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .