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This is problem 10-5 from John Lee's ITM.

Compute the fundamental group of the complement of the three coordinate axes in $\mathbb{R}^3$, giving explicit generators.

I figured out that the space $X$ is homotopic to the sphere minus six points. And further since the sphere minus six points is homeomorphic to the plane minus five points, which is homotopic to the bouquet of $5$ circles, $X$ has fundamental group isomorphic to $\mathbb{Z}^{*5}$. Hence it is generated by $5$ loops, each corresponding to a circle around a distinct half-axis. However, I cannot see or imagine how the $6th$ loop, that is the loop around the one left half-axis can be generated by a free product of $5$ other generators. I would greatly appreciate any help with understanding the explicit generator for this group.

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Draw a disk with 5 punctures (this is a 6 punctured sphere). Do you see how to draw the bouquet that this disk deformation retracts onto? The 6th loop, the boundary, should just be the product of the loops around the 5 interior punctures.

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  • $\begingroup$ I am actually not sure about the picture in the five puncture case. I can draw it in the 2 punctured disk by deformation retracting onto a theta space first. $\endgroup$ – nomadicmathematician Nov 4 '19 at 10:26
  • $\begingroup$ math.stackexchange.com/questions/1593137/… Excellent illustration here. Can't you do the same thing, but with the points fixed at the five roots of unity? $\endgroup$ – Prototank Nov 4 '19 at 11:16

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