Fundamental group of $S^2\cup\{xyz = 0\}$ [duplicate]

We want to compute the fundamental group of $S^2 \cup \{xyz=0\}$. It is easy to see that it retracts to a sphere joined with three disks. How can I show that its fundamental group is trivial?

• How exactly is that sphere "joined" with three disks? Where are the disks "joined" to the sphere...? Jan 16, 2017 at 17:36
• The three disks are the coordinate disks joined to the sphere in the points $(1,0,0)$, $(-1,0,0)$ and so on... Jan 16, 2017 at 18:23
• "Coordinate disks"? I don't think that even means anything. What you do can do is to shrink each coordinate plane to a unit circle around the unit sphere, so that we'd have the sphere and three disks around and inside it. You though don't need to shrink the coordinate planes... Jan 16, 2017 at 20:41

So the space you are interested in, call it $X$, is the union of the unit sphere and the three coordinate planes. You can use the Seifert-van Kampen Theorem by choosing one set, $U$, to be a connected open neighbourhood of the three planes, and the other, $V$, to be a connected open neighbourhood of the sphere; $U\cap V$ is an open neigbourhood of three great circles on $S^2$ which, if $U$ and $V$ are chosen correctly, is also connected. However, $U$ is contractible (again, if chosen correctly), so it has trivial fundamental group and therefore $X$ has trivial fundamental group.
An explicit choice of suitable $U$ and $V$ is as follows:
Alternatively, by 'shrinking' the planes to the origin, one can see that $X$ is homotopy equivalent to a bouquet of eight spheres (i.e. a wedge product of eight copies of $S^2$). It follows from the Seifert-van Kampen Theorem that $\pi_1(Y\vee Z) = \pi_1(Y)*\pi_1(Z)$, so $X$ is simply connected (i.e. has trivial fundamental group).