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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.

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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
@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

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.

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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 $\mathbb Z*\mathbb Z* \mathbb Z* \mathbb Z* \mathbb Z$

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