One-point compactification of $S^3\setminus S^1$

Question

Let $S^1$ be a circle embedded in $S^3$. Is the one-point compactification of $S^3\setminus S^1$ homeomorphic to $S^3$?

Answer 1

As noted in the comments, the one point compactification of $S^3-S^1$ is homeomorphic to $S^3/S^1$. We will not care about how $S^1$ is embedded. We will use the tools of homology from section 2.1 of Hatcher. Then the long exact sequence of the pair $(S^3, S^1)$ yields

$$0\rightarrow H_2(S^3,S^1)\rightarrow \mathbb{Z}\rightarrow 0$$

Hence the middle map is an isomorphism. However, $H_2(S^3/S^1)\cong H_2(S^3, S^1)\cong \mathbb{Z}$, but $H_2(S^3)=0$, so $S^3/S^1$ is not homeomorphic to $S^3$.

Answer 2

I think I may have found another rather elementary solution.

If $Y = (S^3\setminus S^2)\cup \{\infty\}$ is the one-point compactification of $(S^3\setminus S^2)$, then there exists a point in $Y$ (namely, $\infty$) whose removal results in the space $S^3\setminus S^2$ which is not simply connected. However the removal of any point from $S^3$ results in $\mathbb{R}^3$, which is simply connected. Hence $Y\not\cong S^3$.