Why is $\mathbb{R}^2/G$ homeomorphic to the Klein bottle?

Question

Let $G$ be the group of transformation generated by $a,b:\mathbb{R}^2\to \mathbb{R}^2$ where $a(x,y)=(x+1,y-1)$ and $b(x,y)=(x,y+1)$.

We note than $bab=a$ and that $G$ acts properly discontinuously on $\mathbb{R}^2$. Then $\pi:\mathbb{R}^2\to\mathbb{R}^2/G$ is a covering map.