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$?
 A: 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. 
