- Main Idea: Using a funky deformation retract. See Hatcher, PDF page 55, Example 1.23. Add picture!!
Deformation retract \(\RR^3 - S^1\) onto \(S^2 - U\), where \(U\) is a diameter inside \(S^2\) also passing through the middle of \(S^1\) in the interior. This can be done by moving points outside of \(S^2\) towards the surface, and points inside \(S^2\) just move away from the \(S^1\) inside (either towards \(U\) or towards the surface of \(S^2\), so they don’t hit \(S^1\)).
Then take a geodesic between the endpoints of the diameter on \(S^2\), pick any point \(p\) on the geodesic, and move both diameter points towards it. This yields \(S^2 \vee S^1\) at the point \(p\).
Solution
1. Let \(C\subset\mathbb R^3\) be the standard unknotted circle and choose a closed tubular neighborhood \(N\cong S^1\times D^2\) of \(C\).
The normal bundle of the standard circle in \(\mathbb R^3\) is trivial.
2. The complement \(\mathbb R^3\setminus C\) deformation retracts onto \(\mathbb R^3\setminus\operatorname{int}N\).
Inside each normal disk of \(N\), radially push every noncentral point away from the core circle to the boundary \(S^1\times S^1\), while fixing the exterior of \(N\). This gives a strong deformation retraction.
3. After one-point compactifying \(\mathbb R^3\) to \(S^3\), the closure of \(\mathbb R^3\setminus\operatorname{int}N\) is the complementary solid torus \(V\) of the unknot; the point at infinity lies in the interior of \(V\). Thus\[\mathbb R^3\setminus\operatorname{int}N\cong V\setminus\{\infty\}.\]
The complement in \(S^3\) of a tubular neighborhood of an unknot is again a solid torus.
4. A solid torus with one interior point removed deformation retracts onto the wedge of its core circle with a small \(2\)-sphere surrounding the deleted point.
Remove a small open \(3\)-ball around the missing point. The resulting punctured solid torus has a handle decomposition with one \(0\)-handle, one \(1\)-handle, and the additional spherical boundary component. Collapse the \(0\)- and \(1\)-handle part to the core circle while retaining that boundary sphere, joining them by a single arc and then collapsing the arc to the wedge point.
5. Consequently\[\boxed{\mathbb R^3\setminus S^1\simeq S^1\vee S^2.}\]
Combine <1>2–<1>4.