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!
 A: Following Neal's suggestion in the comments. The "top" ($z \geq 0$) and "bottom" ($z \leq 0$) are each themselves homeomorphic to a torus with a disk "plugging in the hole," which has fundamental group $\mathbb{Z}$, and the intersection has trivial $\pi_1$, so you should get the free group on two generators.
A: 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 $$\pi_1 X\simeq \mathbb Z\ast\mathbb Z.$$
A: Here's an alternative solution.  We have that $X$ is a torus unioned with a disk which bisects the torus.  Let $X_1 \subset X$ be the space containing the disk, two small strips on the torus where the torus and disk intersect, and a small strip over the top of the torus connecting the inside and outside circles (see my diagram below).  

Let $X_2 \subset X$ be the torus along with small strips near were the torus and disk intersect (again, see below.)

Then, $X_1$ and $X_2$ are open, $X_1 \cup X_2 = X$ and  $X_1 \cap X_2$ is path connected, so w emay apply Seifert van Kampen.  We have that $X_1$ retracts to a circle, so $\pi_1(X_1) = \langle \alpha \rangle$.  We can deformation retract $X_2$ to the torus, so $\pi_1(X_2) = \langle \beta, \gamma\rangle ~\mid~\beta\gamma\overline{\beta}\overline{\gamma}\rangle$. Finally, $X_1 \cap X_2 \sim S^1 \wedge S^1$, so $\pi_1(X_1 \cap X_2) = \langle \delta, \epsilon\rangle$.  All these loops are shown on $X$ below.

From the diagram, we find that $\delta$ and $\epsilon$ are null homotopic in $X_1$ and that $\delta \sim \epsilon \sim \beta$ in $X_2$, so by SvK the fundamental group of $X$ is
$$\pi_1(X) = \langle \alpha, \beta, \gamma ~\mid \beta\gamma\overline{\beta}\overline{\gamma}, \beta\rangle \cong \langle \alpha, \gamma\rangle.$$
