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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.

Aleph-null
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1 Answers1

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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.

ziggurism
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