I have the following problem that I am stuck on.
Let $S^{3}$ be the one-point compactification of $\mathbb{R}^{3}$. Compute the fundamental group of $S^{3}-X$, where:
(a) $X=S^{1}=\{(x,y,0)\in\mathbb{R}^{3}\mid x^{2}+y^{2}=1\}$.
(b) $X$ is the union of two unlinked circles.
(c) $X$ is the union of two linked circles.
(d) $X$ is the union of three linked circles.
I think that I should be using the Seifert-van Kampen Theorem, but I am not sure how to proceed. I'm really confused as to how to determine which sets are homeomorphic to more common spaces that we already know the fundamental groups of. I think that $S^{3}-S^{1}$ would be homeomorphic to a more familiar space, but I am having a hard time coming up with the connection. Thanks in advance for any help or recommendations on any part of this problem!
(a) You're right in thinking $S^3 - S^1$ is homeomorphic to a more familiar space. Intuitively, you can think about $S^3$ as $\mathbb{R}^3$ with a point at infinity, but then you can 'translate' $S^1$ inside $S^3$ so that it goes through this point at infinity, making the missing part just a straight line through infinity. So, $S^3 - S^1 \cong \mathbb{R}^3 - \{(0,0,z) \mid z \in \mathbb{R}\}$, i.e. $\mathbb{R}^3$ without the $z$-axis. From here, it should be easier to think about the fundamental group!
(b) Split the new space in two (using SvK) and think about the fundamental group of each. You might like to use a similar idea as for (a), but be careful because you've just cut out half of $S^3$.
(c) Translate as above, but you can't cut it up so nicely. You might need a cleverer idea to finish off, but thinking about how the space looks will help you here.