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Im almost ashamed for asking such an elementary question.

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 $\,\mathbb Z+\mathbb Z\,$ in the constant path.

I used SvK Theorem but im not sure if i used it correctly. Sorry if its really basic, but im not having Algebraic Topology classes. Thanks a lot !

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I edited your question with LaTeX symbols. Please check I wrote what you meant. –  DonAntonio Jul 1 '12 at 15:17
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You shouldn't be ashamed of asking simple questions! Personally I think the very idea of a simple algebraic topology question is a bit strange :P but if you included more detail on what you did with SVK, it would be much easier to check your work than to reproduce it. –  Ben Millwood Jul 1 '12 at 15:28
    
A more words-y description of $X$ is a torus of thickness radius 1 and inner-circle radius 1 centred at the origin union a closed disc of radius 3 also centred at the origin. I guess you can't use SVK on the obvious components because the intersection is not path-connected. –  Ben Millwood Jul 1 '12 at 15:33
    
If my description I just gave is right, I'm pretty sure $\mathbb Z$ is wrong. My intuitive guess would be free group on two generators, but don't hold me to that. –  Ben Millwood Jul 1 '12 at 15:38
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It looks like the disc "slices the torus in half." Maybe try Seifert-Van Kampen on the "top" and "bottom", with intersection the disc? –  Neal Jul 1 '12 at 16:50

2 Answers 2

up vote 2 down vote accepted

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 $$\pi_1 X\simeq \mathbb Z\ast\mathbb Z.$$

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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.

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