Fundamental group of $X:=\Bbb{R}^3\setminus{l}.$ Let $l$ be a line in $\Bbb{R^3}.$
I would like to compute the fundamental group of $X:=\Bbb{R}^3\setminus{l}.$
In fact if I can identified $l$ to be $(0,0,z)$ then the punctured plane $\Bbb{R^2}\setminus{(0,0)}$ is a strong retract so that $\pi_1(X,x_0)=\Bbb{Z.}$
So I am tempted to say WLOG suppose $l$ is $(0,0,z)$ but it's a bit dishonest because I am not sure how can I prove this.
So perhaps it's incorrect. 
Otherwise, let $(x(t),y(t),z(t))$ be a parametrization of $l$, I mean that let $A(x_A,y_A,z_A)$ be a point of $L$ and $u(a,b,c)$ the director vector we have a parametrization $$x(t)=x_A+ta$$$$y(t)=y_A+tb$$$$z(t)=z_A+tc.$$
So we have $X=\{\{(x,y,t):x\ne x(t),y\ne y(t),z\ne z(t)\}:t\in\Bbb{R}\}$ but not sure it help my problem.

A translation is a homeomorphism of $\mathbb{R}^3$, so you may suppose that $l$ goes through the origin. Now take any non-zero vector spanning $l$ and extend it to a basis of $\mathbb{R^3}$ (ordered however you want). There is a unique linear map that sends this ordered basis to the standard basis; in particular, $l$ is mapped to a coordinate axis. It remains to note that any linear automorphism of $\mathbb{R}^3$ is a homeomorphism, as is evident by the coordinate description of such a map. The rest of your argument goes through.
