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I am trying to calculate the fundamental group of $\mathbb{R}^3- \{ x\text{-axis}\cup y\text{-axis}\cup z\text{-axis}\}$.

Idea: I think we can show it deformation retracts on 2-sphere minus 4 points.

Jared
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hina
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    I think it's the $2$-sphere minus $6$ points...? – Qiaochu Yuan Nov 29 '12 at 11:19
  • @ Yuan please more explanation ! – hina Nov 29 '12 at 11:24
  • @QiaochuYuan: I don't think so. isn't the fundamental group of what you're saying isomorphic to Z? I don't think hina's group is... – akkkk Nov 29 '12 at 11:25
  • @ all ,I am not all comfortable with this subject any comment requires some explanation for me. – hina Nov 29 '12 at 11:26
  • @akkkk: No, I agree with Qiaochu. Consider the homotopy $F(t,x)=(1-t)x+t\frac{x}{||x||}$ between the identity and the deformation retraction. – Espen Nielsen Nov 29 '12 at 11:28
  • @hina: what is unclear. – akkkk Nov 29 '12 at 11:29
  • I mean writing down a precise homotopy requires first an intuition.So could you please tell how one will proceed for such questions? – hina Nov 29 '12 at 11:31
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    @hine: It's six points, because you have the negative and positive ray on each axis. Since the sphere minus one point is homeomorphic to the plane, your space is homotopy-equivalent to the plane minus five points. – Rhys Nov 29 '12 at 11:35
  • @ Rhys: Then it's fundamental group should be isomorphic to the fundamental group of wedge of 5 circles,which is free group on 5 symbols. – hina Nov 29 '12 at 11:39
  • @Yuan: It will be really great for me if you can explain how the my space is homotopy equivalent to sphere minus 6 points. – hina Nov 29 '12 at 11:41
  • @all: thanks a lot ,finally got it completely !! :) – hina Nov 29 '12 at 13:40

2 Answers2

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As Qiaochu Yuan mentioned, $\mathbb{R}^3\setminus(\text{union of the $x$- $y$- and $z$-axes})$ deformation retracts onto the sphere $S^2$ with 6 missing points. To see this, note that $\mathbb{R}^3\setminus\{0\}$ deformation retracts onto $S^2$ by $$ (\mathbb{R}^3\setminus\{0\})\times[0,1] \to \mathbb{R}^3\setminus\{0\} : (x,y,z)\mapsto t \frac{(x,y,z)}{||(x,y,z)||} + (1-t) (x,y,z) .$$ The same formula will retract $\mathbb{R}^3$ without the axes onto $S^2$ without the points $(\pm1,0,0)$, $(0,\pm1,0)$ and $(0,0,\pm1)$.

Now as pointed out by user61223, the sphere with 6 missing points is homeomorphic to the plane with 5 missing points, which has as fundamental group the free group on five generators. This is the fundamental group you're looking for.

Daan Michiels
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Since sphere delete one point is homeomorphic to $R^2$, so $S^2$\ $\{ X-axis, Y-axis,Z-axis\}$ is homotopy equivalent to unit disk $D^2$\ $\{ 5 points\}$, so fundamental group will be free product $\mathbb Z*\mathbb Z* \mathbb Z* \mathbb Z* \mathbb Z$

Ram
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Hoang Nguyen
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