Fundamental group of $R^3$ minus two disjoint circles I know the fundamental group of $X=\mathbb{R}^3-(A\cup B)$ should be $\mathbb{Z}*\mathbb{Z}$, where $A=\{(x,y,1)\in\mathbb{R}^3:x^2+y^2=1\}$ and $B=\{(x,y,2)\in\mathbb{R}^3:x^2+y^2=1\}$.
Could someone show me how to use Van Kampen theorem to do the calculation? This is totally new to me, so I really have no idea.
Let $a=(0,0,0)$ be the base point. I know we should find two path-connected open sets $U$ and $V$ s.t. $X=U\cup V$, and $U\cap V$ is also path-connected and contains $a$. But what is a proper choice of $U$ and $V$? I guess $\pi_1(U,a)$ and $\pi_1(V,a)$ should both equal to $\mathbb{Z}$. Not sure if this is true.
 A: Let us first compute the fundamental group of $\mathbb{R}^3 \setminus S^1$. For this we can use the following homotopy equivalences:

An easy application of the van Kampen theorem (or some other result you can use to compute fundamental groups of wedges) thus yields $$\pi_1(\mathbb{R}^3 \setminus S^1) \cong \pi(S^2 \vee S^1) = \mathbb{Z}.$$ Here I was assuming that you know the fundamental groups of spheres. Now consider the case of two missing circles. I will denote that space by $X$ as you did. We choose open sets $U$ and $V$ as follows:

Note that both $U$ and $V$ and their intersection are path-connected. Moreover, the intersection has trivial fundamental group and $U$ and $V$ are both homotopy equivalent to $\mathbb{R}^3 \setminus S^1$. Thus by the van Kampen theorem (and our computation before) we get $$\pi_1(X) \cong \pi_1 (U) \ast_{\pi_1(U \cap V)} \pi_1(V) = \pi_1 (\mathbb{R}^3 \setminus S^1) \ast_{1} \pi_1(\mathbb{R}^3 \setminus S^1) = \mathbb{Z} \ast \mathbb{Z},$$ as wanted.
Edit: I updated the drawings and the text, so that one can actually read the text.
