I'm trying to find the Fundamental group of the $\mathbb R^3$ without the axis $z$. Intuitively, It's easy to realized that this fundamental group is the same of the circle, i.e., $\mathbb Z$. I don't know yet how to find fundamental groups with coverings and lifts, just with homotopy and Seifert-Van Kampen Theorem. I need help.
Thanks
Best Answer
You can prove that $$\mathbb R^3\setminus\{(0,0,z)\mid z\in\mathbb R\}\simeq \mathbb R^2\setminus\{(0,0)\}\simeq S^1,$$ where $\simeq$ means the spaces are homotopically equivalent. Since equivalent spaces have isomorphic fundamental groups, you're done.
Addition: Proving $A:=\Bbb R^2\setminus\{(0,0)\}\simeq S^1$: First we can define $i:S^1\to A$ to be just the inclusion map. Then take $r:A\to S^1$ to be $$r(x)=\frac x{|x|}.$$ We want to show that $i\circ r\simeq \operatorname{id}_A$ and $r\circ i\simeq \operatorname{id}_{S^1}$. But $r\circ i$ is actually already equal to the identity function, so only the first one is left.
So, define $H:A\times I\to A$ by $$H(x,t)=(1-t)\frac x{|x|}+tx$$ (for a fixed $x$, this is a straight line between $x/|x|$ and $x$). Note that $H$ is well defined and continuous. It is also easy to see that $H_0=i\circ r$ and $H_1=\operatorname{id}_A$.
Showing that $\Bbb R^3\setminus\{(0,0,z)\mid z\in\Bbb R\}\simeq\Bbb R^2\setminus\{(0,0)\}$ can be done in a similar way. Use the maps $r'(x,y,z)=(x,y)$ and $i'(x,y)=(x,y,0)$.