# How does the one-point compactification affect the fundamental group of the complement of a knot?

I've just begun to study knot theory and I'm wondering how does the one-point compactification of $\mathbb{R}^3$ change the knot group of a knot $K$, e.g. if $\pi(\mathbb{R}^3 \backslash K)$ is isomorphic to $\pi(\mathbb{S}^3 \backslash K)$. My intuition is that yes, they're isomorphic, but how could I proceed the formal proof?

Let $\infty$ denote the point added to $\mathbb{R}^3$ to make $S^3$, and let $U$ be a contractible open neighbourhood of $\infty$ in $S^3\setminus K$. Then $S^3\setminus K = (\mathbb{R}^3\setminus K)\cup U$. Now apply the Seifert-van Kampen Theorem.
• And am I right thinking that the intersection of them would be the interior of three-dimensional ball $D^3$ without a point and hence it's fundamental group is trivial? – Maria WJ Mar 13 '17 at 23:37
• Exactly. The intersection deformation retracts onto $S^2$ and hence has trivial fundamental group. – Michael Albanese Mar 13 '17 at 23:40