Let $X:=\{(x,y,z)\in \mathbb{R}^3: (x^2+y^2-1)(x^2+z^2-9)=0\}$. I would like to compute the fundamental group of $X$. The plot of $X$ is this.
My attempt:
The space $X$ is homotopy equivalent to the sphere $S^2$ with four holes and a tube stuffed inside. In turn, this new space is homotopy equivalent to a cylinder with two handles.
Now, let $Y$ be a path connected space and $Y'$ be the space obtained by attaching a $1$-cell to $Y$. As it is shown in many basic textbooks in Algebraic Topology, we have $$\pi_1(Y')\simeq \pi_1(Y)\ast \mathbb Z.$$ Thus, as $\pi_1(\textit{cylinder})\simeq \mathbb Z$, we get $$\pi_1(X)\simeq \mathbb Z\ast\mathbb Z\ast\mathbb Z.$$
I don't know if my solution is right, but if it were, we would have $H_1(Z)=\mathbb Z\times\mathbb Z\times \mathbb Z$. In particular, this answer would be incorrect.
