Define an action of $\mathbb{Z}$ on $\mathbb{R}^{2}$ by $$ n \cdot(x, y)=\left(x+n,(-1)^{n} y\right) $$
Show that the action is smooth, free, and proper.
this action is free :
$n \cdot(x, y)=\left(x+n,(-1)^{n} y\right)=(x, y) $ then $n=0$.
How we can show that the action is smooth and proper ?
The action is said to be proper if the map $ \phi :G \times M \rightarrow M \times M$ given by $(g, p) \mapsto(g \cdot p, p)$ is a proper map (i.e., the preimage of any compact set is compact). (Note that this is not the same as requiring that the map $G \times M \rightarrow M$ defining the action be a proper map.)
Now let $K$ be compact subset of $\mathbb{R}^{2}×\mathbb{R}^{2}$ then there exist $A,B \subseteq \mathbb{R}^{2}$ such that $K=A×B$. Now we must show $\phi ^{-1}(A,B)$ is compact. We have $\phi ^{-1}(...,B)=(...,B)$ so we must show compactnes for $A$.