This is homework so no answers please.
Consider the quotient map $\pi:\mathbb{R}^{2}\to \mathbb{R}^{2}/\sim$ ,where $(x_{1}y_{1})\sim (x_{2},y_{2})$ iff $x_{1}-x_{2},y_{1}-y_{2}\in \mathbb{Z}$. Then show that $\pi|_{(a,b)\times (c,d)}$, where $b-a,d-c\in (0,1)$, is a coordinate chart for $\mathbb{R}^{2}/\sim$.
Any mistakes:
I already showed homeomorphism.
Smooth compatibility: Because distinct points in $(a,b)\times (c,d)$ get mapped identicaly to M, the map $\pi|_{(a,b)\times (c,d)}=Id_{(a,b)\times (c,d)}$ and thus smooth. So their transition maps will also be smooth.
But how can the above make sense since: the inverse is $\pi^{-1}(x,y)=\bigcup_{s,t\in \mathbb{Z}}(x+t,y+s)$ and not the identity?
Thanks