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?

  • 3
    $\begingroup$ How exactly is that sphere "joined" with three disks? Where are the disks "joined" to the sphere...? $\endgroup$
    – DonAntonio
    Jan 16, 2017 at 17:36
  • $\begingroup$ The three disks are the coordinate disks joined to the sphere in the points $(1,0,0)$, $(-1,0,0)$ and so on... $\endgroup$ Jan 16, 2017 at 18:23
  • $\begingroup$ "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... $\endgroup$
    – DonAntonio
    Jan 16, 2017 at 20:41

1 Answer 1


Note that

\begin{align*} &\ \{(x, y, z) \in \mathbb{R}^3 \mid xyz = 0\}\\ =&\, \underbrace{\{(x, y, z) \in \mathbb{R}^3 \mid x = 0\}}_{yz-\text{plane}}\cup\underbrace{\{(x, y, z) \in \mathbb{R}^3 \mid y = 0\}}_{xz-\text{plane}}\cup\underbrace{\{(x, y, z) \in \mathbb{R}^3 \mid z = 0\}}_{xy-\text{plane}}. \end{align*}

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:

\begin{align*} U &= \{(x, y, z) \in X \mid |x| < 0.1, |y| < 0.1,\ \text{or}\ |z| < 0.1\}\\ V &= \{(x, y, z) \in X \mid 0.9 < \|(x, y, z)\| < 1.1\}. \end{align*}

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).


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