fundamental group of $\Bbb R^3\setminus S^1$ with explicit deformation retract

Question

İ am studying Tamma tom Dieck algebraic topology book and i see question 2.8.1

Question : Let $ D = \{(0,0,t) \mbox{ } | -2 \leq t \leq 2 \} $ and $S^2(2) = \{ x \in \mathbb{R}^3 \mbox{ } | \mbox{ } \|x\| = 2 \} $ . Then $S^2(2) \cup D $ is deformation retract of $\mathbb{R}^3\setminus S^1 $

I have seen the question about finding fundamental group of $\mathbb{R}^3\setminus S^1 $ on the net but i have never seen answer with the explicit formula of deformation retract and i can't see it , even i can't get intuition . I want both explicit formula of deformation retract and intuition .

Answer 1

Here is a way to define a retraction $r:\mathbb R^3\backslash S^1\rightarrow S^2(2)\cup D$.

Let $x\in \mathbb R^3\backslash S^1$,

• If $\|x\|\geq 2$, then define the image of $x$ to be $r(x)=2\dfrac{x}{\|x\|}$,
• If $x\in D$, then $r(x)=x$,
• Otherwise, if $x\notin D$ and $0<\|x\|\leq 2$, consider the plane $\mathcal P$ containing the $z$-axis $(Oz)$ and $x$. Let $p$ be the point of $S^1$ in $\mathcal P$ close to $x$. Now consider the half-line $L$ starting at $p$ and passing through $x$. We define $r(x)$ to be the first intersection point of $L$ and $S^2(2)\cup D$.

You can check that, by construction, $r$ is well define on each part and continuous. Moreover, $r$ sends $S^2(2)\cup D$ identically to it self.

The map $$\begin{array}{rccl}F:&\mathbb R^3\backslash S^1\times [0,1]&\longrightarrow &\mathbb R^3\backslash S^1\\ & (x,t)& \longmapsto &t\cdot i\circ r(x)+(1-t)\cdot x\end{array}$$ is well defined, continuous, $F(\cdot,0)=id$, $F(\cdot,1)=i\circ r$ and $F(x,t)=x$ for any $x\in S^2(2)\cup D$ and any $t\in [0,1]$. Hence, $r$ is actually a strong retract deformation.