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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.

enter image description here

Vincenzo Zaccaro
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  • I think you are correct and the linked post is wrong. – JWL Jul 07 '19 at 02:50
  • More precisely, the answer in the linked post seems to be making some wrong deformation retract. – JWL Jul 07 '19 at 02:52
  • So do I. I don't know why but the author of the answer claims that the space $S$ is hom. eq. to $S^2$ with 2 handles. – Vincenzo Zaccaro Jul 07 '19 at 02:57
  • I think the answer in the linked post is correct. The part of the wider cylinder within the narrower cylinder, ${x^2+z^2=9, x^2+y^2 \leq 1 }$, is part of $X$. In other words the top and bottom holes in your picture should not be there (assuming I'm understanding your picture correctly). – Chi Cheuk Tsang Jul 11 '19 at 06:09

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