2
$\begingroup$

I would be happy if one can help me with how to approach this problem.

Suppose that $K$ is a knot in $\mathbb{R}^3$, show that $\pi_1(\mathbb{R}^3 \setminus K)$ is isomorphic to $\pi_1(\mathbb{S}^3 \setminus K)$. I know I need to use Van-Kampen's Theorem to show the desired isomorphism, but I don't know how to choose $U$ and $V$. Thanks.

$\endgroup$
2
$\begingroup$

Let $N$ be a point of $S^3$ not contained in $K$. By stereographic projection we have a homeomorphism $S^3\setminus \{N\}\cong\mathbb{R}^3.$ Let $U = \mathbb{R}^3\setminus K$ and $V$ be a neighborhood of $N$ disjoint from $K$.

Then we have $\pi_1(S^3\setminus K) = \pi_1(\mathbb{R}^3\setminus K)\underset{\pi_1(D^3\setminus N)}{*}\pi_1(D^3)$ which is isomorphic to $\pi_1(\mathbb{R}^3\setminus K)$ since $\pi_1(D^3)$ and $\pi_1(D^3\setminus N)$ are trivial.

$\endgroup$
  • $\begingroup$ A neighborhood of N disjoint from K. $\endgroup$ – Mariano Suárez-Álvarez Dec 23 '16 at 1:55
  • $\begingroup$ @MarianoSuárez-Álvarez Thanks $\endgroup$ – ziggurism Dec 23 '16 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.