Induced action is proper discontinuous Let $f:X\to Y$ be a surjective map, and let $G$ act on $X$ such that for each $g\in G$ and $x,x'\in X$  $f(x')=f(x)$ implies $f(g\cdot x)=f(g\cdot x')$. Further assume that the group action on $X$ is proper discontinuous.
In the above situation, we get an induced action on $Y$. But is this action again proper discontinuous?
My attempt so far: Let $U\subset X$ with $(g\cdot U \cap U \neq \emptyset \Rightarrow g=e)$.
Set $V=f(U)$ and let $y\in g\cdot V \cap V$, i.e. $y=g\cdot y'$ for some $y'\in V$. As $V=f(U)$, we can find $x,x' \in U$ such that $y=f(x)$ and $y' =f(x')$. Then we have $f(g\cdot x)=f(x')$ by the definition of the action on $Y$.
However, I need $g\cdot x=x'$ in order to get a contradiction.
Did I miss something? Do I need more assumptions? 

Consider $\mathbb{R}\times S^1$ where $S^1$ is the quotient of $\mathbb{R}$ by the action of $\mathbb{Z}$ defined by $n.x=x+n$ We denote by $p:\mathbb{R}\rightarrow S^1$ the quotient map. Consider $\mathbb{R}\times S^1$ endowed with the action of $\mathbb{Z}$ defined by $n.(x,y)=(x+n,p(y+nc))$ where $c$ is irrational number.
Let $f:\mathbb{R}\times S^1\rightarrow S^1$ defined by $f(x,y)=y$, the action of $\mathbb{Z}$ is proper on $\mathbb{Z}\times S^1$ but not on $S^1$.
