If $\pi:X\to Y$ is a surjective local homeomorphism and $X$ is a smooth manifold, then does $Y$ become a smooth manifold?

Question

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 $M$.

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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?

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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.

Answer 1

$\newcommand{\res}[2]{\left.#1\right|_{#2}}$ $\newcommand{\id}{{\rm id}}$ $\newcommand{\vphi}{\varphi}$ $\newcommand{\vare}{\varepsilon}$ The proof is divided into two parts.

(i) $\pi$ is a local homeomorphism. With $G$ being properly discontinuous, for any $x\in M$ we can find a neighborhood $U_0$ such that $\{g\in G:gU_0\cap U_0\neq\varnothing\}$ is a finite set. If it contains only ${\rm id}$ then we are done. If not, let the elements be $$g_1={\rm id},\ g_2,\cdots,\ g_n$$ Now by the Hausdorff property of $M$ and the fact that $G$ is free from fixed points we find nonintersecting neighborhoods $$U_1,\ \cdots,\ U_n\quad\text{of}\quad x,\ g_2x,\ \cdots,\ g_nx$$ respectively. Finally let $U_x=U_0\cap(\bigcap_{k=1}^ng_k^{-1}U_k)$. Then $U_x$ is a neighborhood of $x$ such that $g(U_x)\cap U_x=\varnothing$ for all $g\neq\id$. From this we conclude $\pi|_{U_x}:U_x\to\pi(U_x)$ is injective and hence bijective, and apparently $\pi^{-1}(U_x)=\bigcup_{g\in G}g(U_x)$ is open, it follows that $\pi|_{U_x}:U_x\to\pi(U_x)$ is a homeomorphism (the continuity of $(\pi|_{U_x})$ and $(\pi|_{U_x})^{-1}$ are easy to verify). Therefore, $\pi$ is a local homeomorphism.

(ii) $M/G$ has a structure of the same type as $M$. For each $x\in M$, from (i) there exists a neighborhood $U_x$ of $x$ such that $\res{\pi}{U_x}$ is a homeomorphism. By taking an intersection if necessary, we may assume $U_x$ is a coordinate neighborhood with the corresponding chart $\varphi_x$. We claim that the set $$\{\varphi_x(\res{\pi}{U_x})^{-1},\ x\in M\}$$ is an atlas on $M/G$. The domains of them obviously constitute an open cover of $M/G$, hence it remains to consider the transition maps, which are of the form (here, $\pi(U_x)\cap\pi(U_y)$ is assumed to be connected, as we can discuss each connected component separately) $$\vphi_x(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})\vphi_y^{-1},\quad\pi(U_x)\cap\pi(U_y)\neq\varnothing$$ It suffices to show that the middle part satisfies $$(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})=g,\quad\text{in}\quad(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$$ for some $g\in G$ because each $g$ preserve the structure of $M$. First, we choose some $x_0\in U_x$ and $y_0\in U_y$ with $\bar x_0=\bar y_0\in\pi(U_x)\cap\pi(U_y)$, hence $$x_0=g_0y_0=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_0\text{ for some }g_0$$ Since $\pi(U_x)\cap\pi(U_y)$ is connected so are $(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$ and $(\res{\pi}{U_x})^{-1}(\pi(U_x)\cap\pi(U_y))$, we claim that $$g_0y=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y\text{ for all }y\in(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$$ Let the path $\gamma:[0,1]\to(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$ have $y_0$ and $y$ as its initial and terminal points respectively. Let $$S=\{T\in[0,1]:g_0\gamma(t)=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})\gamma(t)\text{ for all }t\in[0,T]\}$$ Obviously $0\in S$. Then let $$T_0=\sup S$$ By the continuity of $g_0,(\res{\pi}{U_x})^{-1}$ and $\gamma$ we have $T_0\in S$. We claim $T_0=1$. If not, suppose $T_0<1$ and let $y_0'=\gamma(T_0)$, then there is a sequence $y_k=\gamma(T_0+\vare_k)$ such that $$y_k\to y_0'$$ $$(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_k=g_ky_k\neq g_0y_k\text{ with }g_k\neq g_0$$ By the continuity of $(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})$ we have $$g_ky_k=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_k\to(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})y_0'=g_0y_0'$$ that is, $$g_kg_0^{-1}(g_0y_k)\to g_0y_0'$$ On the other hand, the continuity of $g_0$ also gives $$g_0y_k\to g_0y_0'$$ Since $G$ acts properly discontinuously without fixed points, $g_0y_0'$ has a neighborhood $U$ such that $gU\cap U=\varnothing$ for all $g\neq\id$. Hence we have $$g_0y_k\to g_0y_0'\\ \implies g_0y_k\in U\text{ for all sufficiently large }k\\ \implies g_kg_0^{-1}(g_0y_k)\notin U\text{ for all sufficiently large }k\text{ because }g_k\neq g_0$$ contradicting $g_kg_0^{-1}(g_0y_k)\to g_0y_0'$. This means $T_0=1$ and by the definition of $T_0,\gamma$ and $S$ we obtain $$g_0=(\res{\pi}{U_x})^{-1}(\res{\pi}{U_y})\text{ in }(\res{\pi}{U_y})^{-1}(\pi(U_x)\cap\pi(U_y))$$ It follows that the transition maps have the form $$\varphi_xg\varphi_y^{-1}$$ with $g$ being an automorphism that preserves the structure of $M$. It follows that $M/G$ admits an atlas and hence a structure of the same type as $M$.

Answer 2

The type of group action in the updated proposition is also known as a "covering space action". (A term, I think, coined by Hatcher.)

TFAE for an isometric group action of a locally compact group $G$ on a locally compact Hausdorff metric space $M$ (e.g. any manifold):

1. $G$ acts properly discontinuously and freely (=without fixed points);
2. $G$ acts totally discontinuously (every $x$ has a nbh $U$ with $gU \cap U \neq \varnothing \implies g=e$);
3. The map $M \to M/G$ is a covering map. (And hence $M/G$ inherits the structure from $M$ if $G$ preserves that structure.)
4. $G$ acts freely with discrete orbits;

and they imply that $G$ is discrete. Metrizability is only needed for 4 $\implies$ 1,2,3.

See e.g. Proposition 4 in these notes on Fuchsian groups by Pete L. Clark.

Answer 3

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 $0$.