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