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Calculate the fundamental group of the complement in $\mathbb{R}^3$ of

$$\{ (x,y,z) \ | \ y = 0 , \ x^{2} + z^{2} = 1\} \cup \{ (x,y,z) \ | \ z = 0 , \ (x-1)^{2} + y^{2} = 1\}.$$

Note: this space is $\mathbb{R}^{3}\setminus \{ \mbox{2 linked circles }\}$.

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    $\begingroup$ I see no question above, only an order. You should never copy/paste a question from a textbook, or phrase your question as such. Moreover, you should explain what you have tried, and where you are stuck. $\endgroup$ – M Turgeon May 16 '12 at 14:50
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    $\begingroup$ @MTurgeon While I agree with your sentiments, it does seem like your tone is no different than OP's. We should lead by example. $\endgroup$ – Sasha May 16 '12 at 14:52
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    $\begingroup$ @Sasha Ok. I see no question above, only an order. I encourage you to never copy/paste a question from a textbook, or phrase your question as such. Moreover, you should try to explain what you have tried, and where you are stuck. $\endgroup$ – M Turgeon May 16 '12 at 14:55
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    $\begingroup$ One possible approach: $S^3\setminus \{\text{2 linked circles}\}$ is homotopy equivalent to a torus. Note that adding a point at $\infty$ to turn $\mathbb R^3$ into $S^3$ does not change $\pi_1$. $\endgroup$ – Cheerful Parsnip May 16 '12 at 14:56
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    $\begingroup$ As @JimConant says, you can add a point to make this $S^3-K$, where $K$ is your two linked circles. This does not change the homotopy type. Choosing another point of $S^3$ to be infinity won't change it either, so you can remove a point from $K$ to get a new (homeomorphic) space, $\mathbb{R}^3-L$, where $L$ is the union of a circle in the xy-plane, and the z-axis. Now the deformation retraction to the torus is obvious. $\endgroup$ – user641 May 16 '12 at 20:17

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