Let \(X \subset \mathbb{R}^3\) be the union of \(n\) lines through the origin. Compute \(\pi_1(\mathbb{R}^3 - X)\).
Solution
Assume the \(n\) lines are distinct. Let \[ L_1,\dots,L_n \] be these lines and set \[ Y=\mathbb R^3\setminus X. \]
1. Radial projection gives a deformation retraction\[ Y\simeq S^2\setminus\bigl(X\cap S^2\bigr). \]
Since every line \(L_i\) passes through the origin, membership in \(X\) is invariant under multiplication by a positive scalar. For \(y\in Y\), define \[ H(y,t)=\left((1-t)+\frac{t}{\|y\|}\right)y. \] The scalar factor is positive for all \(t\in I\), so \(H(y,t)\) lies on the same ray as \(y\) and therefore never enters \(X\). At \(t=0\) one has \(H(y,0)=y\), and at \(t=1\) one has \[ H(y,1)=\frac{y}{\|y\|}\in S^2. \] Points already on \(S^2\) are fixed. Thus \(H\) is a deformation retraction onto \[ S^2\setminus(X\cap S^2). \]
2. The set \(X\cap S^2\) consists of exactly \(2n\) points.
Each line through the origin meets \(S^2\) in a pair of antipodal points. Distinct lines give disjoint antipodal pairs. Hence \[ X\cap S^2=\{p_1,-p_1,\dots,p_n,-p_n\} \] has cardinality \(2n\).
3. A sphere with \(k\ge1\) punctures has fundamental group free of rank \(k-1\).
Choose one puncture \(q\). Stereographic projection from \(q\) gives a homeomorphism \[ S^2\setminus\{q\}\cong\mathbb R^2. \] Removing the remaining \(k-1\) punctures therefore gives \[ S^2\setminus\{q,q_2,\dots,q_k\} \cong \mathbb R^2\setminus\{a_2,\dots,a_k\}. \]
Choose pairwise disjoint small closed disks around the points \(a_2,\dots,a_k\) and join their boundary circles to a common basepoint by pairwise disjoint arcs arranged as a tree. The complement deformation retracts onto the union of these \(k-1\) circles and the joining tree. Collapsing the tree gives a wedge of \(k-1\) circles. Hence \[ \pi_1\bigl(S^2\setminus\{q_1,\dots,q_k\}\bigr) \cong F_{k-1}, \] the free group on \(k-1\) generators.
4. Therefore\[ \pi_1(\mathbb R^3\setminus X)\cong F_{2n-1}. \]
By <1>1–<1>2, \[ Y\simeq S^2\setminus\{2n\text{ points}\}. \] Apply 3 with \(k=2n\). Thus \[ \boxed{\pi_1(\mathbb R^3\setminus X)\cong F_{2n-1}}. \]