Let us consider the topological space given by the quotient of the boundary of $[0,1] \times [0,1]$ by the following equivalence relation:
$$ (0,t) \sim (1,t) \, \forall t \in [0,1] $$ and $$ (\frac{1}{2},0)\sim (\frac{1}{2},1) $$
It is the cylinder with two points identified. If I represent the square and connect $(\frac{1}{2},0)$ with $(\frac{1}{2},1)$, the straight line become a circle. So $S^1$ is homeomorphic to $\{\frac{1}{2}\} \times [0,1]$ with the extremes identified. How can I use Van Kampen's theorem to find the fundamental group? In other terms, I want to find $U,V$ open sets such that $X=U \cup V$, $U \cap V \neq \emptyset$ and to compute their fundamental groups. If we see the cylinder as a square with the lateral sides identified, I take $U=$cilynder minus an inner point $r$ (not belonging to the segment connecting the points of the boundary identified by $\sim$, i.e. $(\frac{1}{2},0)$ and $(\frac{1}{2},1)$); $V$ the interior of the cylinder, so that $U \cap V=$interior of the cilynder minus a point. It is easy to see that $\pi_1(V)$ is trivial, while $\pi_1(U)=\mathbb{Z} *\mathbb{Z} *\mathbb{Z}$ and $\pi_1(U \cap V)=\mathbb{Z}$. Let $\delta$ be the generator of $\pi_1(U \cap V)$. After doing computations, I see that $\delta$ is homotopic to $\alpha \gamma \beta \gamma^{-1}$ (where $\alpha,\beta,\gamma$ are the generators of $\pi_1(U)$ and correspond to the sides of the squares, with $c$ is the identified. The problem is that $<\alpha,\beta,\gamma|\alpha \gamma \beta \gamma^{-1}>$ is not the free group on two generators as I expect. I don't know if going from one vertex of the square, moving along $a$ until we reach the identified point, then moving along $b$ and then along $c^{-1}$ gives a loop homotopic to $\delta$.
Can you provide a detailed resolution of the exercise, please?
When you identify those two points, note that this is the same as joining them with a line and moving the line round you can get that this is equivalent to a cylinder $C$ wedged with $S^1$. Note also that $C$ can be deformation retracted to $S^1$ as well so we can say:
$$ \frac{[0,1]\text{x}[0,1]}{\sim} \sim C\vee S^1 \sim S^1\vee S^1 $$
Now applying Van Kampen to the wedge of two circles is nice. Take your U and V to be each circle, 'expanded' to be open, so will take a little of the other circle too. Then we get $U\cap V \sim \{ * \}$ a single point. Therefore by van kampen we know:
$$\pi_1\left(\frac{[0,1]\text{x}[0,1]}{\sim}\right) \cong \pi_1(S^1 \vee S^1) \cong \mathbb{Z} \star \mathbb{Z}$$ Which makes sense intuitively as we would have two generarting loops, one that runs around the cylinder and another that runs around the circle.