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I want to calculate the fundamental group of this space: $$X=\{(x,y,z)\in\mathbb{R}^3:(x^2+y^2+z^2-100)[(x-10)^2+y^2+z^2-1]=0\}.$$ I’m not good with this kind of exercises: I know the Seifert-van Kampen theorem and the basic results about covering spaces.

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  • $\begingroup$ Your decomposition of $X$ as a product $Y\times S^1$ is not correct. Why do you think this is the case? $\endgroup$ – Christoph Dec 16 '18 at 18:18
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$X$ is equal to the union of two spheres which intersect in a circle. You can apply Van Kampen's Theorem and compute $\pi_1(X)$ using the amalgamated product. An easier way to get the result is observing that $X \simeq S^2 \vee S^2 \vee S^2$. This is because the two halves of the smaller sphere can be homotoped into two spheres by contracting the circle of intersection with the larger sphere to a point.

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    $\begingroup$ I'm not quite sure this is correct. $\endgroup$ – Aweygan Dec 16 '18 at 18:41
  • $\begingroup$ You were right. I think I've corrected it now. $\endgroup$ – Lukas Kofler Dec 16 '18 at 18:49
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We have $$X=\{(x,y,z):x^2+y^2+z^2=100\}\cup\{(x,y,z):(x-10)^2+y^2+z^2=1\}$$ These two spheres intersect in the circle, so after a little visualization we can see that $X\cong S^2\vee S^2\vee S^2$. Applying Seifert-van Kampen twice to obtain the result.

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