Computing this fundamental group What is the fundamental group of 
$$X = \left\{\left(\sqrt{x^2+y^2}-2\right)^2 + z^2 = 1\right\}\cup \left\{(x,y,0)\;\; :\;\; x^2 + y^2 \leq 9\right\}\subset\mathbb R^3\,?$$ 
I would say that it is $\,\mathbb Z\,$ cause you can deform one of "the class of paths" that usually would make the fundamental group of $\,S^1\times S^1\,$ be $\,\mathbb Z\oplus \mathbb Z\,$ in the constant path. 
I used the Seifert van Kampen Theorem but I'm not sure if I used it correctly. 
Thanks a lot!

Consider $C_+=$ the intersection of $X$ with the plane $z=+1$ and $C_-=$ the intersection of $X$ with the plane $z=-1$, and define $X_+=X\setminus C_-$ and $X_-=X\setminus C_+$. Their intersection deformation retracts to a point. Also, $X_+\simeq X_-$ are obviously homeomorphic, so we get 
$$\pi_1X=\pi_1X_+\ast \pi_1X_-=G\ast G$$
where $G$ is the fundamental group of $X_+$. $X_+$ is homotopy equivalent to a torus lying flatly on a plane, or the shape you get if you put a circle on a stick and rotate it around a vertical axis, or $Y$ where $Y$ is constructed as $X$ was constructed except that the disk you use only has radius one now. 
We'll work with $Y$. Remove the outmost circle of radius $3$ form $Y$, the result $(Y_0)$ is homotopy equivalent to a point. Remove the origin from the radius one disk, the result $(Y_1)$ is homotopy equivalent to a torus. Their intersection is homotopy equivalent to a circle, so 
$$G=\pi_1Y\simeq 0\ast_{\mathbb Z}\big(\mathbb Z\times \mathbb Z\big)$$
where we should have the following pushout diagram
$$\begin{array}{ccc}
 & x\mapsto (x,0) & \\
\mathbb Z & \longrightarrow & \mathbb Z\times \mathbb Z \\
\downarrow & &\downarrow \\
0 & \longrightarrow & G
\end{array}$$
so we should have $G\simeq\mathbb Z$ and $$\pi_1 X\simeq \mathbb Z\ast\mathbb Z.$$
