Let $\quad r_1 = \{ (x,y,z) \in \mathbb{R}^3 \mid x=0,z=0 \}$ ( $y$ axis), $\quad r_2 = \{(x,y,z) \in \mathbb{R}^3 \mid x=0,y=0 \}$( $z$ axis) and $\quad \mathbb{S}^1 =\{(x,y,z) \in \mathbb{R}^3 \mid z=0,x^2+y^2=1\}$. Call $H=r_1 \cup r_2 \cup \mathbb{S}^1$ and define $X = \mathbb{R}^3 \setminus H$.
I'm trying to compute $\pi_1(X)$.
I think I could find a deformation retract of $X$ which is similar to a Toro, in particular i suppose this deformation retract is $\mathbb{T} \setminus \{\text{4 points} \}$. If this works then I'm able to say that $\pi_1(X)=\mathbb{Z} *\mathbb{Z} *\mathbb{Z} *\mathbb{Z} *\mathbb{Z} $.
So, can anyone confirm that $\mathbb{T} \setminus \{\text{4 points} \}$ is actually a deformation retract of $X$?
$\mathbf{Edit:}$ Idea:
Ok, maybe I can see $\mathbb{R}^3 \setminus ( \mathbb{S}^1 \cup \text{$z$ axis} )$ admits as deformation retract a empty cilinder (only side surface) which is homeomorphic to $\mathbb{T}$. Now, if I stick this Toro with $x$ axis then I take away $4$ points from it. Is this idea correct?
$\mathbb{R}^3- x$axis is $S^1\times [0,1]$
$\mathbb{R}^3- x$axis$ -\{ (x,y,z)|x^2+y^2=1,\ z=0,\ x\geq 0\}$ is union of $S^1\times [0,1] $ and interval so that it is $ S^1\vee S^1$
$\mathbb{R}^3- x$axis$ -\{ (x,y,z)|x^2+y^2=1,\ z=0,\ x\geq 0\}$
$ -\{ (x,y,z)|(x-3)^2+y^2=1,\ z=0,\ x\geq 0\}$ is union of $S^1\times [0,1] $ and two intervals
$\mathbb{R}^3-x$axis $-y$axis is $S^1\times [0,1]-$two points.
$\mathbb{R}^3- x$axis$ -\{ (x,y,z)|(x-2)^2+y^2=1,\ z=0,\ x\geq 0\} $
$-\{ (x,y,z)|(x-5)^2+y^2=1,\ z=0,\ x\geq 0\} -y$axis is a union of $S^1\times [0,1] -$two points and two intervals.