6

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 $\mathbb{R}^{3}\setminus \{ \mbox{2 linked circles }\}$.

M Turgeon
  • 10,419
WLOG
  • 11,436
  • 5
    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. – M Turgeon May 16 '12 at 14:50
  • 2
    @MTurgeon While I agree with your sentiments, it does seem like your tone is no different than OP's. We should lead by example. – Sasha May 16 '12 at 14:52
  • 4
    @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. – M Turgeon May 16 '12 at 14:55
  • 4
    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 $\pi_1$. – Cheerful Parsnip May 16 '12 at 14:56
  • Look up the Wirtinger presentation. – Neal May 16 '12 at 16:41
  • 3
    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. –  May 16 '12 at 20:17

0 Answers0