# 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 – Brevan Ellefsen Aug 28 '19 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. – trisct Aug 28 '19 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 – Brevan Ellefsen Aug 28 '19 at 3:51
• To clarify my prior comment on defining a structure such that $\pi$ is a local isometry see this link. – Brevan Ellefsen Aug 28 '19 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 $x_2$. – Andreas Cap Aug 28 '19 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 $$\pi(x,2) = \sqrt{x}$$. 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 $$0$$.

$$\newcommand{\res}{\left.#1\right|_{#2}}$$ $$\newcommand{\id}{{\rm id}}$$ $$\newcommand{\vphi}{\varphi}$$ $$\newcommand{\vare}{\varepsilon}$$ 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. – Brevan Ellefsen Aug 29 '19 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.