Let \(L\) be the union of the \(z\)-axis and the unit circle in the \(xy\dash\)plane. Compute \(\pi_1 (\RR^3 \backslash L, \ast)\).
Solution
Write \[ A=\{(0,0,z):z\in\RR\} \] for the \(z\)-axis and \[ C=\{(x,y,0):x^2+y^2=1\} \] for the unit circle, so \(L=A\cup C\).
1. There is a homeomorphism\[ \RR^3\setminus A \cong S^1\times(0,\infty)\times\RR \]under which \(C\) corresponds to\[ S^1\times\{(1,0)\}. \]
For \((x,y,z)\notin A\), put \[ r=\sqrt{x^2+y^2}>0, \qquad u=\frac{x+iy}{r}\in S^1. \] Then \[ \Phi(x,y,z)=(u,r,z) \] defines a continuous map \[ \Phi:\RR^3\setminus A\longrightarrow S^1\times(0,\infty)\times\RR. \] Its inverse is \[ \Phi^{-1}(u,r,z) = (r\operatorname{Re}u,r\operatorname{Im}u,z), \] which is continuous, so \(\Phi\) is a homeomorphism.
A point of \(C\) has \(r=1\) and \(z=0\), while \(u\) ranges over all of \(S^1\). Hence \[ \Phi(C)=S^1\times\{(1,0)\}. \]
2. Consequently,\[ \RR^3\setminus L \cong S^1\times \left(((0,\infty)\times\RR)\setminus\{(1,0)\}\right). \]
Since \(L=A\cup C\), \[ \RR^3\setminus L = (\RR^3\setminus A)\setminus C. \] Apply the homeomorphism from 1 and remove its image of \(C\).
3. The space\[ ((0,\infty)\times\RR)\setminus\{(1,0)\} \]deformation retracts onto a circle.
The map \[ h:(0,\infty)\times\RR\longrightarrow\RR^2, \qquad h(r,z)=(\log r,z), \] is a homeomorphism with inverse \[ h^{-1}(s,z)=(e^s,z). \] It sends \((1,0)\) to \((0,0)\). Hence it restricts to a homeomorphism \[ ((0,\infty)\times\RR)\setminus\{(1,0)\} \cong \RR^2\setminus\{0\}. \]
The punctured plane strongly deformation retracts onto the unit circle by \[ H(v,t) = \left((1-t)+\frac{t}{\lVert v\rVert}\right)v, \qquad v\in\RR^2\setminus\{0\},\quad 0\le t\le1. \] Therefore the stated space deformation retracts onto \(S^1\).
4. The complement \(\RR^3\setminus L\) deformation retracts onto the torus \(S^1\times S^1\).
3. Taking the product of the identity map of \(S^1\) with the deformation retraction from 3 gives a deformation retraction\[ \RR^3\setminus L\simeq S^1\times S^1. \]
5. Therefore\[ \pi_1(\RR^3\setminus L,*)\cong\ZZ^2. \]
Fundamental groups are invariant under deformation retraction, and for path-connected spaces \[ \pi_1(X\times Y)\cong\pi_1(X)\times\pi_1(Y). \] Thus 4 gives \[ \pi_1(\RR^3\setminus L,*) \cong \pi_1(S^1\times S^1) \cong \pi_1(S^1)\times\pi_1(S^1) \cong \ZZ\times\ZZ. \] Hence \[ \boxed{\pi_1(\RR^3\setminus L,*)\cong\ZZ^2.} \]