# Computing fundamental group of $S^3$ minus mutually linked circles

Let $$S^3 \subset \mathbb{C}^2$$ be a $$3$$-sphere and consider the map $$p\colon S^3 \rightarrow \mathbb{C}P^1$$ defined by $$p(z_1, z_2) = [z_1 \colon z_n]$$. Given distinct points $$x_1, \cdots, x_r \in \mathbb{C}P^1$$, I'm asked to compute the fundamental group of $$X_r := S^3 \setminus \cup_{i=1}^{r} p^{-1}(x_i)$$. The points are given arbitrarily.

To solve this problem, I first investigate what is $$p^{-1}(x_i)$$. If we put $$x_i=[a_i : b_i]$$ for $$(a_i, b_i) \in S^3$$, we obtain $$p^{-1}(x_i) = \{ (z_1, z_2) \in S^3 | \exists \lambda \in S^1, (z_1, z_2) = \lambda(a_i, b_i)\}$$. Thus $$p^{-1}(x_i)$$ is a circle.

In particular, it is a "great circle", for which what I mean is that we can obtain $$\{(x,y,z,w)\in \mathbb{R}^4 | x^2+y^2=1, z=w=0 \}$$ by rotating $$p^{-1}(x_i)$$. Here the identification $$\mathbb{C}^2 = \mathbb{R}^4$$ is used.

I've tried to compute first $$\pi_1(X_r)$$ for small values of $$r$$. I'll describe my attempts:

Case 1: $$r=1$$

$$S^3 \setminus p^{-1}(x_1)$$ is homeomorphic to $$\mathbb{R}^3$$ minus $$z$$-axis via stereographic projection, which deformation retracts to $$S^1 \times \mathbb{R}^2$$. Thus $$\pi_1(X_1) = \mathbb{Z}$$.

Case 2: $$r=2$$

Take a stereographic projection $$\psi \colon S^3 \setminus \{\text{pole} \}$$ so that $$\psi(p^{-1}(x_1))$$ is the $$z$$-axis. Then $$\psi(p^{-1}(x_2))$$ is a circle around $$z$$-axis. It follows that $$X_2$$ is homeomrphic to $$R^3$$ minus the union of the $$z$$-axis and a circle around the $$z$$-axis, which deformation retracts to the torus. Thus $$\pi_1(X_2) = \mathbb{Z} \times \mathbb{Z}$$.

I realized that $$X_r$$ is homeomorphic to $$Y_r :=\mathbb{R}^3$$ minus $$\{ z \text{-axis} \} \cup \bigcup \{ \text{circles around the z-axis} \}$$, where circles are mutually linked and the number of circles is $$r-1$$. I'm get stucked here, since I couldn't find a familiar space which is a deformation retract of $$Y_r$$ even the $$r=3$$ case. I think the main obstacle is the linking of circles. If they are not linked, I can find a deformation retraction from $$Y_r$$ to $$S^1 \times ( S^1 \vee \cdots \vee S^1$$). I'm also tried to apply Van Kampen theorem, but there is no progress.

Question:

(1) What is the fundamental group of $$X_r$$ for $$r \geq 3$$?

(2) Is there a familiar space (whose fundamental group can be computed easily) to which $$Y_r$$ deformation retracts for $$r \geq 3$$?

The desired fundamental group (1) is $$\mathbb{Z} \times F_{r-1}$$ and the desired space (2) is $$S^1$$ cross a graph.

Also, the sentence "If they are not linked, I can find a deformation retraction from..." is not correct. If the circles are not linked then the space deformation retracts to a one point union of a disjoint collection of circles and two-spheres (with one more circle than two-sphere).

Here are a few hints coming from geometric topology to get you started on the proofs. (There is a more general, but less "hands-on", approach using tools that are more solidly algebraic topological.)

The map you describe, from $$S^3$$ to $$S^2$$ is called the "Hopf map" after Heinz Hopf. The collection of circles in $$S^3$$ (or $$\mathbb{R}^3$$) you describe are called a "Hopf link" (with $$r = 2$$ being the most commonly used).

When $$r = 1$$, the space $$S^3 - S^1$$ is an open solid torus. Let's call this open solid torus $$V$$. Now, $$V$$ is homeomorphic to $$S^1 \times D$$ where $$D$$ is an open disk. This structure on $$V$$ is called a "fibered solid torus".

We typically think, under the homeomorphism, that all of the circles "going around $$V$$ in parallel". However, we can apply a homeomorphism of $$V$$ (called a Dehn twist) so that, instead of "going around in parallel" the circles "twist" at a constant rate -- the result is that any pair of circles is linked exactly once.