Show that \(\mathbb{R}^3 - S^1 \simeq S^1 \vee S^2\).
Solution
1. Let \(C\subset\mathbb R^3\) be the standard embedded circle and let \(N(C)\) be a closed tubular neighborhood. Then\[ \mathbb R^3\setminus C\simeq \mathbb R^3\setminus\operatorname{int}N(C). \]
Inside the tubular neighborhood \(N(C)\cong S^1\times D^2\), radially push each nonzero normal vector in \(D^2\setminus\{0\}\) to the boundary circle. Extend this deformation by the identity outside \(N(C)\). This gives a deformation retraction of the complement of the core \(C=S^1\times\{0\}\) onto the exterior of the open tubular neighborhood.
2. After one-point compactifying \(\mathbb R^3\) to \(S^3\), the exterior \(S^3\setminus\operatorname{int}N(C)\) of the unknot is a solid torus \(S^1\times D^2\), and the point at infinity lies in its interior. Hence\[ \mathbb R^3\setminus\operatorname{int}N(C)\cong (S^1\times D^2)\setminus\{p\} \]for an interior point \(p\).
The complement in \(S^3\) of a tubular neighborhood of the unknot is again a solid torus. Passing from \(S^3\) back to \(\mathbb R^3\) removes the point at infinity, which is disjoint from \(N(C)\) and therefore lies in the complementary solid torus.
3. A solid torus with one interior point removed has homotopy type \(S^1\vee S^2\).
Remove a small open \(3\)-ball about \(p\). The resulting compact manifold is a solid torus with an additional spherical boundary component and deformation-retracts onto a spine obtained from the usual core circle of the solid torus together with that boundary sphere, joined at one point by an arc. Contracting the joining arc gives the wedge \(S^1\vee S^2\).
4. Therefore\[ \boxed{\mathbb R^3\setminus S^1\simeq S^1\vee S^2}. \]
Combine <1>1–<1>3.