Vtyjkyuk

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

Using $mathbbR^2$ as a covering space, with the quotient map as a covering maps when defining the the quotient space: $mathbbR^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 $mathbbR^2$ by $p$ to $tildegamma : Irightarrow mathbbR^2$ such that $tildegamma(0) , tildegamma(1) in p^-1(pt)$, So $tildegamma$ is a path between two "red points" in the demonstrated grid. Now I claim that $tildegamma$ is homotopic to a finite concatenation of $"red - purple -red"$ paths in $mathbbR^2 $ cause one might continuously stretch $gamma$ to the grid.

Denoting $"red-purple-red"$ paths in $mathbbR^2$ 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_*([tildegamma]) = 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 $(mathbbZ,+)$.