Give the definitions of deformation retract and strong deformation retract for topological spaces. Compute the fundamental group of \(\mathbb R^3-C\) where \(C\) denotes the circle \(x^2+y^2=1\), \(z=0\).
Solution
1. A subspace \(A\subseteq X\) is a deformation retract of \(X\) if there is a retraction\[ r:X\to A, \qquad r|_A=\operatorname{id}_A, \]such that\[ i\circ r\simeq\operatorname{id}_X, \]where \(i:A\hookrightarrow X\) is the inclusion.It is a strong deformation retract if the homotopy can be chosen to fix \(A\) pointwise at every time.
Equivalently, a deformation retract is given by a homotopy \[ H:X\times I\to X \] with \[ H(x,0)=x, \qquad H(x,1)=r(x)\in A. \] For a strong deformation retract one additionally requires \[ H(a,t)=a \] for every \(a\in A\) and \(t\in I\).
Let \[ C=\{(x,y,z)\in\mathbb R^3:x^2+y^2=1,\ z=0\}. \]
2. Under inverse stereographic projection\[ \Phi:\mathbb R^3\longrightarrow S^3\setminus\{N\}\subseteq\mathbb R^4, \]the circle \(C\) becomes the great circle\[ C_0=\{(x_1,x_2,0,0):x_1^2+x_2^2=1\}. \]
Take \[ S^3=\{(x_1,x_2,x_3,x_4)\in\mathbb R^4:x_1^2+x_2^2+x_3^2+x_4^2=1\} \] and \[ N=(0,0,0,1). \] The inverse stereographic projection is \[ \Phi(x,y,z) =\frac{1}{x^2+y^2+z^2+1} \bigl(2x,2y,2z,x^2+y^2+z^2-1\bigr). \] If \((x,y,z)\in C\), then \[ x^2+y^2+z^2=1 \] and \(z=0\), so \[ \Phi(x,y,0)=(x,y,0,0). \] Thus \(\Phi(C)=C_0\). Consequently \[ \mathbb R^3\setminus C \cong S^3\setminus(C_0\cup\{N\}). \]
3. The complement\[ M=S^3\setminus C_0 \]strongly deformation retracts onto the complementary great circle\[ D=\{(0,0,x_3,x_4):x_3^2+x_4^2=1\}\cong S^1. \]
Write a point of \(S^3\) as \[ (u,v)\in\mathbb R^2\times\mathbb R^2, \qquad \|u\|^2+\|v\|^2=1. \] Then \[ C_0=\{(u,0):\|u\|=1\}, \] so on \(M\) one has \(v\ne0\).
Define \[ H:M\times I\to M \] by \[ H((u,v),t) = \frac{((1-t)u,v)}{\sqrt{(1-t)^2\|u\|^2+\|v\|^2}}. \] The denominator is nonzero because \(v\ne0\). The second coordinate remains a nonzero scalar multiple of \(v\), so the image stays in \(M\). At \(t=0\), \[ H((u,v),0)=(u,v). \] At \(t=1\), \[ H((u,v),1)=\left(0,\frac{v}{\|v\|}\right)\in D. \] If \((u,v)\in D\), then \(u=0\) and \(\|v\|=1\), so \[ H((0,v),t)=(0,v) \] for all \(t\). Thus \(H\) is a strong deformation retraction of \(M\) onto \(D\).
4. Therefore\[ \pi_1(M)\cong\mathbb Z. \]
By 3, the inclusion \(D\hookrightarrow M\) is a homotopy equivalence. Hence \[ \pi_1(M)\cong\pi_1(D)\cong\pi_1(S^1)\cong\mathbb Z. \]
5. The inclusion\[ M\setminus\{N\}\hookrightarrow M \]induces an isomorphism on fundamental groups.
The point \(N\) lies in \(D\) and is disjoint from \(C_0\). Choose a sufficiently small open \(3\)-ball \[ U\subseteq M \] around \(N\). Set \[ V=M\setminus\{N\}. \] Then \[ M=U\cup V, \qquad U\cap V=U\setminus\{N\}. \] The set \(U\) is contractible, while \[ U\setminus\{N\}\cong B^3\setminus\{0\} \] deformation retracts onto \(S^2\) and is therefore path-connected and simply connected.
The space \(V\) is path-connected. Indeed, by 2 it is homeomorphic to \(\mathbb R^3\setminus C\). Write a point there as \((x,y,z)\) and put \[ \rho=\sqrt{x^2+y^2}. \] If \(z\ne0\), the radial path \[ t\longmapsto((1-t)x,(1-t)y,z) \] joins the point to the \(z\)-axis and never meets \(C\), because its third coordinate is nonzero. If \(z=0\) and \(\rho<1\), the same radial path stays strictly inside the unit circle and joins the point to the origin. If \(z=0\) and \(\rho>1\), first move vertically to \((x,y,1)\), then radially at height \(1\) to \((0,0,1)\), and finally down the \(z\)-axis to the origin. None of these three path segments meets \(C\). Thus every point can be joined to the origin in \(\mathbb R^3\setminus C\), so \(V\) is path-connected.
Choose a basepoint in \(U\cap V\). The Seifert–van Kampen theorem gives \[ \pi_1(M) \cong \pi_1(U)*_{\pi_1(U\cap V)}\pi_1(V). \] Since \[ \pi_1(U)=0 \qquad\text{and}\qquad \pi_1(U\cap V)=0, \] the canonical homomorphism \[ \pi_1(V)\to\pi_1(M) \] is an isomorphism.
6. Hence\[ \boxed{\pi_1(\mathbb R^3\setminus C)\cong\mathbb Z.} \]
By 2, \[ \mathbb R^3\setminus C \cong M\setminus\{N\}. \] By 5, \[ \pi_1(M\setminus\{N\})\cong\pi_1(M), \] and 4 computes the latter group as \(\mathbb Z\).