Show that a smooth manifold modulo diffeomorphism group is a smooth manifold

Would like help in starting this exercise:

Suppose $\Gamma$ is a group of diffeomorphisms of a manifold $\left( {X,C_X^\infty } \right)$.

Suppose that the action of $\Gamma$ is fixed-point-free and properly discontinuous in the sense that every point possesses a neighborhood $N$ s.t. $\gamma(N)\cap N = \oslash$ unless $\gamma = id$.

Give $Y=X/\Gamma$ the quotient topology and let $\pi :X \to Y$ denote the projection.

Define $f \in C_Y^\infty \left( U \right)$ iff ${\pi ^*}f \in C_X^\infty \left( {{\pi ^{ - 1}}U} \right)$ .

Show $\left( {Y,C_Y^\infty } \right)$ is a smooth manifold.

Hint: Show first that the quotient map $\pi:X \to X/ \Gamma$ is a local homeomorphism where $X / \Gamma$ is given the quotient topology. From this conclude $X/ \Gamma$ is locally homeomorphic to $\mathbb{R}^{\textrm{dim}(X)}$. This should also give you second countability and lastly the properly discontinuous action will be used to show Hausdorff.
Remark: I know this seems like a restatement to what you have, but here I just wanted to outline what each assumption will be used to show i.e the three main characterizations for $X/ \Gamma$ to be a manifold.
I can start you off so this hint won't be completely useless. Let $\tilde{x} \in X/\Gamma$ and consider $\pi^{-1}(\tilde{x})$. For each $x,y \in \pi^{-1}(\tilde{x})$ we can get disjoint neighborhoods $U_x,U_y$ about them due to the hausdorffness of $X$. Thus the map $\pi|_{U_x}: U_x \to \pi(U_x) \subset X/ \Gamma$ is a homeomorphism.
• Actually, the OP's definition of properly discontinuous is not sufficient to conclude that $Y$ is a manifold. See my answer to this question. – Jack Lee May 24 '16 at 1:35