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

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.

• What are $M$ and $\Gamma$ in your current context? (I presume $\mathbb C$ are the complex numbers and $\mathbb H$ are the quaternions?). If $M$ and $\Gamma$ play nice (discrete groups acting freely by isometries on $\mathbb C$ and $\mathbb H$ respectively or something akin to that), we can assign $Y$ a very natural metric by just pulling back the quotient map and letting things factor nicely through the group action. There is a more general theory for any $A/B$ where $B$ acts on $A$ by isometries, but I less familiar with it Aug 28, 2019 at 3:40
• $M$ 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. Aug 28, 2019 at 3:46
• I believe this is not true in general. Every covering map is a surjective, local homeomorphism (by definition) but this answer shows not every covering map can induce a differentiable structure. I believe the properties of $M$ and $\Gamma$ are necessary, though I am not sure how much you need... in some sense, it seems it "almost" works without assuming anything about the spaces you are quotienting by Aug 28, 2019 at 3:51
• To clarify my prior comment on defining a structure such that $\pi$ is a local isometry see this link. Aug 28, 2019 at 4:16
• I don't thnink that what you need is restrictions on the spaces involved. In the examples of $\mathbb C$ and $\mathbb H$, the subgroups act by diffeomorphisms on the spaces and this action is transitive on each fiber of $q$. This avoids the problems about chart changes you encountered. I would expect that the theorem you propose holds if you assume that for points $x_1,x_2\in X$ with $\pi(x_1)=\pi(x_2)$ there is a diffeomorphism of $X$ that maps fibers of $q$ to fibers of $q$ and sends $x_1$ to $x_2$. Aug 28, 2019 at 8:06

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 $$0$$.

$$\newcommand{\res}[2]{\left.#1\right|_{#2}}$$ $$\newcommand{\id}{{\rm id}}$$ $$\newcommand{\vphi}{\varphi}$$ $$\newcommand{\vare}{\varepsilon}$$ 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 $$M$$.

• Great looking answer to your question! I'm glad you were able to work out a working hypothesis; as "punctured disk" notes in their answer, your assumption of being properly discontinuous is equivalent to what I was proposing (I proposed $G$ act freely with discrete orbits, which allows you to show we get a covering map and pull back the structure) so I am glad to see my comments were not too misguided :) I look forward to reading your answer more in-depth when I have the time later. Aug 29, 2019 at 0:19

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.

• Sadly, Pete (who, in general, is a very careful writer) stated and "proved" something which is false. It was discussed at MSE many times that the four properties are not equivalent. But they are equivalent if the group acts isometrically, which is the case that OP and Pete are interested in. See math.stackexchange.com/questions/2183326/…, math.stackexchange.com/questions/1082834/… Sep 3, 2019 at 21:03
• Thank you, I fixed the statement. Sep 3, 2019 at 21:06
• @Moishe: Thanks for the compliment and for informing me. It seems I really can't be too careful... Dec 9, 2020 at 3:30
• @PeteL.Clark: You may want to add a correction to your notes on Fuchsian groups: (b) does not imply (c) (or, equivalently, (a)), unless you assume that the action is isometric. Dec 9, 2020 at 23:29