Fundamental group and covering space of Möbius band wedged with $2$-sphere Let $M$ be a Möbius band and let $X = M \vee S^2$, the space from M by gluing one point on its boundary to a point on $S^2$.
a) Find the fundamental group and covering space of $X$. Explain your reasoning.
b) Draw two sheeted and three sheeted covering space of $X$.
I tried to use the Seifert–Van Kampen theorem where $U=X/p$ and $V=X/q$ where $q$ in $M$ and $p$ in $S^2$ are different from gluing point but got stuck on intersection... Is it right approach? Any help will be appreciated.
 A: As mentioned in the other answer, the fundamental group is $\pi_1(S^1) \cong \mathbb{Z}$ and it essentially only counts how many times a given loop goes around the Möbius band. To go into more detail: let $p \in M \vee S^2$ be the basepoint (where the sphere is glued onto the band). Let $N$ denote a contractible open neighborhood of $p$. Let $U = M \cup N$, and let $V = S^2 \cup N$. Then by applying the Seifert–Van Kampen theorem to the pair $(U,V)$ we see that $$\pi_1(M \vee S^2) = \pi_1(U \cup V) \cong \pi_1(U) * \pi_1(V) \cong \pi_1(M) *\pi_1(S^2) \cong \pi_1(M) \cong \pi_1(S^1).$$
(The latter isomorphism holds because the band deformation retracts onto any inner circle.)
Now imagine the Möbius band $M$ as a quotient of the fundamental polygon $[0,1]^2 \subset \mathbb{R}^2$ by the relation $(0,y) \sim (1,1-y)$. This relation is in fact an instance of a $\mathbb{Z}$-action on the infinite band $\mathbb{R}\times [0,1]$, where the positive generator of $\mathbb{Z}$ sends $(x,y)$ to $(x+1,1-y)$; and the Möbius band can equivalently be described as the quotient $(\mathbb R \times[0,1])/\mathbb{Z}$ of the infinite band by this action. This is easily seen to be a covering, so the infinite band $\mathbb R \times[0,1]$ is the universal cover of the Möbius band $M$. It follows that $U = (\mathbb R \times[0,1]) \vee \bigcup_{i \in \mathbb{Z}} S^2$ is the universal cover of $M \vee S^2$. (Here I am abusing notation: all the spheres $S^2$ are glued at different points, namely, at each point in the fiber of the gluing point in the base $M \vee S^2$ by the covering map.)
From covering space theory, we know that any $n$-sheeted covering space of $M \vee S^2$ arises as a quotient of its universal cover $U$ by an index-$n$ subgroup of the group of deck transformations of the universal cover, which is isomorphic to $\pi_1(M \vee S^2) \cong \mathbb{Z}$. In this case, the only possibility is that the subgroup is $n\mathbb{Z}$, and this subgroup is generated by the deck transformation that maps $(x,y)$ to either $(x+n,y)$ (if $n$ is even) or $(x+n,1-y)$ (if $n$ is odd) on the infinite band. Therefore, to obtain an $n$-sheeted cover of $M \vee S^2$, take the quotient of the fundamental polygon $[0,n] \times [0,1]$ by the relation $(0,y) \sim (n,y)$ (if $n$ is even) or $(0,y) \sim (n,1-y)$ (if $n$ is odd), and glue $n$ spheres $S^2$ on the $n$ boundary points that map to the special boundary point on the base $M \vee S^2$. In particular, note that the cover you obtain is a cylinder with added spheres if $n$ is even, and a Möbius band with added spheres if $n$ is odd.
A: $X=M\vee S^2$ ?
$\pi_1(M\vee S^2)\simeq \pi_1(M)\oplus\pi_1(S^2)\simeq \pi_1(M)\simeq \pi_1(S^1)\cong \mathbb{Z}$.
Note that the Mobius band $M$ is homotopy equivalent to its middle circle $S^1$, then we have $X\simeq S^1\vee S^2$.
