let $X=S^2 \cup \{xyz=0\}\subset\mathbb{R}^3$ be the union of the unit sphere with the 3 coordinate planes. I'd like to find the fundamental group of $X$.

These are my ideas:

I think the first thing I should do is to retract all the points outside the sphere to the sphere (is that possible? how?)

then using spherical coordinates I could make the following deformation: $(1,\phi,\theta)\to ((\sin\phi \sin\theta \cos \phi\cos \theta)^t,\phi,\theta)$. This collapses all the point in $S^2 \cap \{xyz=0\}$ to $0$ obtaining $8$ deformed spheres touching each other in $0$ (how can I prove rigorously that they are simply connected), using Van Kampen theorem we can say that $X$ is simply connected.

  • $\begingroup$ It is possible to retract all points outside the sphere to the sphere. It can be done for all of $\Bbb R^3$, and you can just restrict that deformation retraction to your $X$. More specifically, restricted to each coordinate plane you want to deformation retract the entire plane to the closed unit disc. $\endgroup$ – Arthur Aug 18 '15 at 11:53
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    $\begingroup$ See also Fundamental group of $S^2\cup\{xyz = 0\}$. $\endgroup$ – Martin Sleziak Jan 18 '17 at 16:52

If you want to pursue the argument you started (which seems fine to me), I suggest that you consider the map $$ F: R^3 \times I \to R : (x, s) \mapsto \begin{cases} x & |x| \le 1 \\ (1-s) \frac{x}{\|x\|} + s (x - \frac{x}{\|x\|}) & \text{otherwise} \end{cases} $$ That retracts things onto the sphere plus the three coordinate disks within it.

If you now let $U'$ be the exterior of a ball of radius $1/2$, and $V'$ be the interior of a ball of radius $3/4$ in 3-space, and $U$ and $V$ be the intersections of these with your space, you can apply van Kampen and find that $\pi_1$ is trivial.

  • $\begingroup$ I like you solution, but is there a way to prove that the 8 deformed spheres I obtain through the second deformation are simply connected? $\endgroup$ – miles Aug 18 '15 at 13:39
  • $\begingroup$ Your 8 deformed spheres are still each homeomorphic to a sphere. Each consists of a "filled' spherical triangle with its boundary triangle collapsed to a single point. Consider applying vanKampen to $U =$ interior of triangle and $V = neighborhood of triangle boundary. This will let you see that it's simply connected. $\endgroup$ – John Hughes Aug 18 '15 at 13:48
  • $\begingroup$ @JohnHughes Why $U \cap V$ is arcwise connected? I don't see it. $\endgroup$ – TheWanderer Jan 16 '17 at 18:21
  • $\begingroup$ The intersection of $U' \cap V'$ with any coordinate plane is an annulus in that plane. Annuli from two different planes intersect in a pair of segments. So you can get from any annulus to any other, and since the annuli are connected, the whole set is. Roughly: take the prime meridian + int;k dateline, the longitude 90 and longitude 270 arcs, and the equator. Their union is clearly connected. $U \cap V$ looks like this, extruded one unit in the radial direction. $\endgroup$ – John Hughes Jan 17 '17 at 0:03

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