7
$\begingroup$

I'm a student taking my first course in algebraic topology. I've stumbled across this exercise: calculate the fundamental group of $S^3-\gamma$, where $\gamma$ is a circumference in $\mathbb{R}^3$ (i.e. $\gamma=S^1$ in $\mathbb{R}^3$) and $S^3=\mathbb{R}^3\cup\lbrace\infty\rbrace$ is the one-point compactification of $\mathbb{R}^3$. I thought of $\mathbb{R^3}$ as $\mathbb{R}^1\times\mathbb{R}^2$, then I can think of $\gamma$ as an $S^1$ in $\mathbb{R}^2$ which has the same homotopy type as $\mathbb{R}^2\setminus\lbrace (0,0)\rbrace$. So $\mathbb{R^3}\cup\lbrace\infty\rbrace$ without $\gamma$ has the same homotopy type as $\mathbb{R}\cup\lbrace\infty\rbrace$, and its one-point compactification gives $S^1$.

In the end, I would get $\pi_1(S^3\setminus\gamma)=\pi_1(S^1)=\mathbb{Z}$, but I don't know if it's correct.

Maybe I could think of $S^3$ as $S^1\ast S^1$, but I don't know if that could help.

Thank you in advance

  • 0
    @user32240 I didn't know that. But in that case, I would assume that $\gamma$ is the boundary of an embedded disc. That also fits with OP's description of $\gamma$ as a "circumference".2012-12-27

1 Answers 1

2

If you "rotate" your $S^3 = \mathbb{R}^3\cup\lbrace\infty\rbrace$ so that one of the points of $\gamma$ is at $\infty$, you get $S^3 \setminus \gamma$ homeomorphic to $\Bbb R^3 \setminus \text{a line}$, which can be deformation retracted to a circle, so $\Bbb Z$ it is!

Also, your reasoning seems sound enough, and you get the same answer, so I would assume it is correct.

  • 0
    Oh ok. I thought it was more like removing $S^1$ from $\mathbb{R}^3$ and then compactify with the point at $\infty$, but I guess your point of view is more correct.2012-12-26