4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
  • 1
    $\begingroup$ See also Fundamental group of $S^2\cup\{xyz = 0\}$. $\endgroup$ – Martin Sleziak Jan 18 '17 at 16:52
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.