I'm currently studying the fundamental group and the applications of Seifert van Kampen's theorem. I'm doing several exercises but I really have difficulties in understanding how to choose the relations of the presentation of the fundamental groups. In this exercise the relation was (I think) trivial but I'd really like to learn a "standard" way to choose the relations on the presentation of the fg.
I have to calculate the fundamental group of of $X:=S^2\cup (S^1\times [-1,1]\cup D^2\times \{\pm 1\})$.
First I observed that $X$ is path connected and I chose two open sets in $X$ such that:
$$U=X\cap \Big\{z<\dfrac{1}{2}\Big\},V=X\cap \Big\{z>-\dfrac{1}{2}\Big\}.$$
The union $U\cup V$ is the set $X$, and the intersection is $$U\cap V=X\cap \Big\{z<\dfrac{1}{2}\Big\}\cap\Big\{z>-\dfrac{1}{2}\Big\}=X\cap\Big\{-\dfrac{1}{2}<z<\dfrac{1}{2}\Big\}.$$
Let $C$ be the set $S^1\times [-1,1]\cup D^2\times \{\pm 1\}$, then we have that $U\cong V$ and $U\cap C\underset{h}{\sim}D^2\times\{-1\}$, which is its deformation retract. Then I said that $U\underset{h}{\sim}S^2\vee S^1$. My (intuitive) idea is that we can consider a function $\rho$ that maps the points of the inferior surface of the cylinder to the surface of the inferior semi-sphere, while this last surface is mapped to the semisphere with $z>0$.
Then, using the fact that a segment is contractible, I can extend the "south" point $S$ to the segment $\overline{SN}$ on the whole sphere $S^2$, where $N$ is the "north" pole.
As loop for the fundamental group of $U$, I chose a generic path $\alpha$ that contains $\overline{SN}$ on the surface of the sphere obtained by the trasformation I described before.
So, $\Pi_1(U)\cong\Pi (S^2\vee S^1)=\langle \alpha \rangle\cong \mathbb Z$ and similarly $\Pi_1(V)=\langle \beta \rangle \cong \mathbb Z$.
Retracting the intersection $U\cap V$, we have that $U\cap V\underset{h}{\sim} S^1\implies \Pi_1(U\cap V)=\langle \gamma \rangle \cong \mathbb Z$.
If $i_{1}*$ is the inclusion $U\cap V\hookrightarrow U$ and $i_2*$ is the inclusion $U\cap V\hookrightarrow V$, then $i_1*\gamma=i_2*\gamma=1$ (constant path).
The relation on the intersection is trivial and, since $U\cap V$ is path connected, I concluded that $$\Pi_1(U\cap V)=\langle\alpha,\beta|\emptyset\rangle=\langle \alpha,\beta\rangle\cong\mathbb Z*\mathbb Z.$$
Best Answer
$X$ is the union of the unit sphere (presumably) and the capped unit cylinder (presumably). (I say "presumably" because $S^2$ can be any sphere and $S^1$ can be any circle.) It looks like
The intersections of the sphere and capped cylinder are highlighted in green. There are three of them -- two points at the centers of the caps, $(0,0,\pm 1)$, and one circle, $S^1 \times \{0\}$.
$U$ is all of that up to $z < 1/2$, the lower half of $X$ with two collars attached to $S^1 \times \{0\}$ on the exterior of the volume enclosed by the lower half of $X$. Picture:
The collars deformation retract onto the circle intersection, leaving the lower half of $X$. It should be easy to see that this is a sphere with two of its points identified.
Another name for this shape is horn torus. Normally, a torus has two generators, but in a horn torus, the longitudinal cycle (homotopic to the green circle) is nullhomotopic, specifically homotopic to the identified points in the sphere. The meridional cycle survives (shown in red).
Then $U \cap V$ is the circle intersection with four collars attached.
The collars deformation retract onto the circle. So $\pi_1(U \cap V) \cong \Bbb{Z}$ and $S^1 \times \{0\}$ (with an orientation and one of its points chosen to be the basepoint) generates $\pi_1(U \cap V)$.
$\pi_1(U) \cong \pi_1(V) \cong \Bbb{Z}$, where a generator is the red path (with an orientation and, in order to be convenient for all three fundamental groups, the basepoint being the intersection of the red path and green circle). Siefert-van Kampen allows us to realize the one generator in $U \cap V$ is nulhomotopic in each of $U$ and $V$ (although it would have been sufficient for it to be nulhomotopic in either since any homotopic version of it in $U$ can be moved to $U \cap V$ and from there to $V$ and vice versa). So of the three generators, for $\pi_1(U)$, $\pi_1(V)$, and $\pi_1(U \cap V)$, only the two meridional generators survive to the fundamental group of $X$.