Let $D^2\subseteq\mathbb{R}^2$ be the closed disc with center at the origin and radius 1 and let $S^1$ be its boundary. Also, let $F_n=\{1,2,\ldots,n\}$ be a $n$-point set with the discrete topology. Define, for $n\geq 2$ $$ X_n = \left(D^2\times F_n\right)/\sim$$ where $\sim$ is the equivalence relation described by $$ \left(x,a\right)\sim\left(y,b\right)\iff x=y\in S^1$$ Prove (by induction on $n$) that $X_n$ is simply connected for all $n\geq 2$.
I've quite easily proven the base case ($n=2$), since $X_2$ is homeomorphic to the $S^2$ sphere, which is simply connected. For the induction step, proving that $X_{n+1}$ is path-connected is also fairly straight-forward. However, I'm struggling to prove that $X_{n+1}$ has the trivial fundamental group. I've tried using the Seifert–van Kampen theorem, yet I'm unable to find two path-connected open set that would satify van Kampen's conditions.
Any hint would be appreciated.
The idea here is that $D^2 \times F_n = \bigsqcup_{i=0}^n D^2$ are $n$ disjoint disks, and the quotient glues them all at their boundary. This is a bit hard to imagine when $n > 2$, but we can extrapolate the following idea.
The sphere $S^2$ can be thought as two semi-spheres seamed at their boundary; that being the sphere's equator. Each semi-sphere is homeomorphic to a disk.
One way to see that a - path connected, say - space $X$ satisfies $\pi_1(X) = 0$ is to show that it consists of the union of two simply connected open sets $U,V$ with path connected intersection. In that case, by van Kampen we have a surjection
$$ 1 = \pi_1(U) \ast \pi_2(V) \to \pi_1(X) $$
and so $X$ is again simply connected.
In the case of the sphere, we can take away the north pole and set $U := S^2 \setminus \{N\}$, and then put $V = B$ an open disk around $N$. The latter is easily seen to be contractible. Now we are left to show that $U$ is simply connected, but this follows from the fact that it deformation retracts to a disk (moreover, it is contractible, but that won't generalize as smoothly).
Can you make this work for higher $n$? Try to translate this 'movements' in terms of the "open disk presentation" for $S^2$.
Consider $U = D^2 \times F_{n+1} \setminus \{p\}$, where $p \in D^2 \times \{n+1\}$ is a point in one of the disks, and $V$ a small disk around it. You can define the open sets in the disjoint union and check that they descend to open sets $\widetilde{U}$ and $\widetilde{V}$ in the quotient.
In the same fashion, the hotomopy that contracts $V$ to a point induces a homotopy that proves $\widetilde{V} \simeq \ast$, and the homotopy equivalence $D^2 \times \{n+1\} \setminus \{p\} \times \{n+1\}\simeq S^1 \times \{n+1\}$ shows that $\widetilde{U} \simeq X_n$ and $\pi_1(U) = \pi_1(X_n) = 1$.