What is the fundamental group of $X= \left \{(x,y) \in \mathbb{C^2} \mid x\ne0, y\ne0 \right\} \cup \left \{(0,1) \right\} $? What is the fundamental group of $X= \left \{(x,y) \in \mathbb{C^2} \mid x\ne0, y\ne0 \right\} \cup \left \{(0,1) \right\} $?
What teorems I have to use to find this fundamental group?
Thanks
 A: As a warmup, consider $Y = \{(x,y) \in \mathbb{C}^2 \bigm| x \ne 0, y \ne 0\}$. Consider also the unit sphere $S^3 \subset \mathbb{C}^2$. There is a deformation retraction $Y \mapsto Y \cap S^3$ defined $\vec v \mapsto \vec v \, / \, |\vec v|$ for each $\vec v = (x,y)$. Thinking of $S^3$ as being equal to the one point compactification of $\mathbb{R}^3$ (and using $(u,v,w)$ as coordinates for $\mathbb{R}^3$), the set $Y \cap S^3$ is equal to the complement in $\mathbb{R}^3$ of the $w$ axis union the $(u,v)$ unit circle:
$$\mathbb{R}^3 - \bigl( \{(0,0,w) \in \mathbb{R}^3\} \cup \{(u,v,0) \in \mathbb{R}^3 \bigm| u^2 + v^2 = 1\} \bigr)
$$ 
This set, in turn, deformation retracts onto a standard embedded torus $T^2 \subset \mathbb{R}^3$. So the conclusion of the warmup is that $\pi_1(Y) \approx \mathbb{Z}^2$. Probably there is a good "one-step" description of a deformation retraction $Y \mapsto T^2$.
Now consider $X = Y \cup \{(0,1)\}$. The point $(0,1)$ lies in $S^3$ and, under the correspondence with the one-point compactification of $\mathbb{R}^3$, it corresponds to the point $(0,0,1) \in \mathbb{R}^3$. If we simply add that point back in, the deformation retraction from $Y$ to $Y \cap S^3$ extends to a deformation retraction from $X$ to $(Y \cap S^3) \cup \{(0,0,1)\}$. That set is, in turn, a deformation retraction of the complement in $\mathbb{R}^3$ of the unit $(u,v)$ circle:
$$\mathbb{R}^3 - \{(u,v,0) \bigm| u^2 + v^2 = 1\}
$$
So to conclude, that last set is a homotopy equivalent to $Y$ and its fundamental group is clearly isomorphic to $\mathbb{Z}$.
