Update: Thanks to some comments below, I realized that the properties of $M$ and $\Gamma$ are also important, taking which into consideration I have obtained a new similar proposition below. I'll provide my proof as an answer. Welcome to point out any mistake or comment on other aspects!
Some notations:
Let $M$ be a manifold with a certain structure. Let $G$ be a group of transformations that preserves this structure (for example, if $M$ is a topological manifold, then $G$ consists of homeomorphisms; if $M$ is a smooth manifold, then $G$ consists of diffeomorphisms; if $M$ has a metric, then $G$ consists of isometries). $G$ is said to act on $M$ properly discontinuously if for all $x\in M$ there is a neighborhood $U_x$ of $x$ such that $\{g\in G:gU_x\cap U_x=\varnothing\}$ is a finite set.
Proposition: Let $M$ be as above. Let $G$ be a group of transformations that preserves the structure of $G$. If $G$ acts properly discontinuous and without fixed points, then the natural projection ($\bar x\in M/G$ is the equivalence class of $x\in M$)
$$\pi:M\to M/G$$
$$x\mapsto\bar x$$
is a local homeomorphism. In particular, for every $x\in M$, there is a coordinate neighborhood $U_x$ of $x$ such that $\pi|_{U_x}:U_x\to\pi(U_x)$ is a homeomorphism. Moreover, if we denote the corresponding chart of $U_x$ by $\varphi_x$, then the maps $\varphi_x(\pi|_{U_x})^{-1}$ constitute an atlas of $M/G$ that assigns to $M/G$ the same type of structure of $M$.
Original question:
I am trying to determine whether this proposition is true.
Let $X$ be an $n$-dimensional smooth manifold, $Y$ a topological space and $\pi:X\to Y$ a local homeomorphism. Then we can assign to $Y$ a differentiable structure such that $\pi$ is a smooth map.
My idea is to define an atlas on $Y$ as follows. For any $y\in Y$, take any $x\in \pi^{-1}(y)$. Since $\pi$ is a local homeomorphism, there is a neighborhood $U_x$ of $x$ such that
$$\pi|_{U_x}:U_x\to\pi(U_x)$$
is a homeomorphism. By taking an intersection if necessary, we can assume $U_x$ is a coordinate chart $\varphi_x$. Apparently $\pi(U_x)$ is a neighborhood of $y$, hence we can define a chart near $y$ as
$$\psi_y=\varphi_x(\pi|_{U_x})^{-1}$$
The problem is, I cannot verify that transition maps are smooth. Suppose for the same $y$, we have two different $x_1,x_2\in \pi^{-1}(y)$. Then by the reasoning above there are two coordinate neighborhoods $U_{x_1},U_{x_2}$. By the Hausdorff property of $X$ we may assume $U_{x_1}$ and $U_{x_2}$ are disjoint, then there is at least one transition map of the form
$$\varphi_{x_1}(\pi|_{U_{x_1}})^{-1}(\pi|_{U_{x_2}})\varphi_{x_2}^{-1}$$
However, since $U_{x_1}$ and $U_{x_2}$ are disjoint, the middle part $(\pi|_{U_{x_1}})^{-1}(\pi|_{U_{x_2}})$ does not cancel, and I cannot conclude that the transition map is smooth.
Questions:
(1) Can I fix this by removing some charts of the form above?
(2) If not, can I impose some more conditions to make the proposition true? In particular, I want to apply this to quotients like $\mathbb C/M$ and $\mathbb H/\Gamma$ and conclude that they are Riemann surfaces. Is there anything special about $\mathbb C$, $\mathbb H$, $M$ or $\Gamma$ that I fail to include in the assumptions of the suggested proposition?
Some clarification:
$M$ is a lattice of rank 2 in $\mathbb C$ and $\Gamma$ is a discrete subgroup of $PSL(2,\mathbb R)$. What I am interested in is, are the properties of $M$ and $\Gamma$ necessary for $\mathbb C/M$ and $\mathbb H/\Gamma$ to become a Riemann surface? In a textbook, the argument is made by showing the natural projection is a local homeomorphism, so I was wondering whether a (surjective) local homeomorphism is enough.
Best Answer
This is only an answer to the original question.
Of course the minimal assumption is that $\pi$ is a surjection because $Y \setminus \pi(X)$ could be everything.
In general $Y$ need not even be Hausdorff. Let $X = \mathbb R \times \{1, 2\}$ with the obvious differentiable structure and let $Y$ be the line with two origins (call them $p_1,p_2$) which is the standard example of a "non-Hausdorff manifold" (see The Line with two origins). Define $\pi : X \to Y$ by $p(x,i) = x$ for $x \ne 0$ and $\pi(0,i) = p_i$.
So let us assume that $Y$ is Hausdorff. Since $\pi$ is a local homeomorphism, it is an open map and $Y$ is locally Euclidean. Since $X$ is a manifold, it has a countable base $\mathcal B$. It is then easy to see that $\pi(\mathcal B) = \{ \pi(B) \mid B \in \mathcal B \}$ is a (trivially countable) base for $Y$. Therefore $Y$ is a topological manifold. However, we cannot expect that there exists a differentiable structure on $Y$ such that $\pi$ is a local diffeomorphism (but note that this is a stronger requirement than $\pi$ smooth).
Let $X = \mathbb R \times \{1, 2\}$ and $Y = \mathbb R$. Define $\pi : X \to Y$ by $\pi(x,1) = x$ and $\pi(x,2) = \sqrt[3]{x}$. Next define $\pi_i : \mathbb R \to \mathbb R, \pi_i(x) = \pi(x,i)$. These maps are homeomorphisms (in fact, $\pi_1 = id$ and $\pi_2 =$ cubic root). Assume that there exists a differentiable structure $\mathcal D$ on $Y = \mathbb R$ such that $\pi$ is a local diffeomorphism. Then so are the maps $\pi_i$ and hence also $$\pi_2 = (\pi_1)^{-1} \circ \pi_2.$$ But $\pi_2$ is not even differentiable in $0$.