Fundamental group of $S^2\cup\{xyz = 0\}$ 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?
 A: 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).
