# Calculating the fundamental group of Möbius bend using covering space

I wish for proof verification, and would like to get other ideas without using van Kampen theorem.

Using $\mathbb{R}^2$ as a covering space, with the quotient map as a covering maps when defining the the quotient space: $\mathbb{R}^2 / \{(x,y)\sim(x+2,y) ,(x,y)\sim(-x,y) ,(x,y)\sim (x,y+2), (x,y)\sim(x,-y) \}$

A sketch:

so $p([(x,y)]) = (x,y)$ is the covering map.

For the fundamental group, denote $pt$ the bottom-left point of the polygonal representation of $M$ (red point), $\gamma : (I,\partial I) \rightarrow (M,pt)$, is lifted to $\mathbb{R}^2$ by $p$ to $\tilde{\gamma} : I\rightarrow \mathbb{R}^2$ such that $\tilde{\gamma(0)} , \tilde{\gamma(1)} \in p^{-1}(pt)$, So $\tilde{\gamma}$ is a path between two "red points" in the demonstrated grid. Now I claim that $\tilde{\gamma}$ is homotopic to a finite concatenation of $"red - purple -red"$ paths in $\mathbb{R^2}$ cause one might continuously stretch $\gamma$ to the grid.

Denoting $"red-purple-red"$ paths in $\mathbb{R}^2$ as $rpr_i$, and $p(rpr_i) = rpr_M$ when $rpr_M$ is the loop formed from the left and the top edges of $M's$ polygonal representation. by the induced p: $p_*([\tilde{\gamma}]) = p_*([rpr_1]) * [rpr_2]) * ... * [rpr_n]) \Rightarrow [\gamma] = [rpr_M]^n$ ($p_*$ is one to one)

We learn that $\pi_1(M,pt)$ is generated by one element and is infinite, so $\pi_1(M,pt)$ is isomorphic to $(\mathbb{Z},+)$.

• Another proof: the band deformation retracts to its central circle, so its fundamental group is $\mathbb Z$ too. – Pedro Tamaroff Aug 26 '18 at 13:19