I have this question, Let $M$ be a manifold and let $G \subset Homeo(M)$ be a group acting on $M$. Suppose that this group action is properly discontinuous and free prove that the quotient space $M/G$ is a manifold.
This is the definition of properly discontinuous: For this problem properly discontinuous means that if $K\subset M$ is compact then the set $\{g\in G|g(K)\cap K\not= \emptyset\}$ is finite and free means the only element of $g$ that fixes any point of $M$ is the identity.
Here is my proof, it is very long and would like to see if anyone would be willing to parse through it and find any possible error. Thanks
We know that $G\subset Homeo(G)$ is some group of homeomorphism, whose action on the manifold $M$ is properly discontinuous and free. Suppose $q: M \to M/G$ is a quotient map. We observe that following: Suppose that $x \in M$ and we choose some locally euclidean neighborhood $x \in U\subset M.$ Sow say for some $1\not= g\in G$, we get $g(U)\cap U \not= \emptyset.$ Since the action is free, we have that $x\not=gx$. Since $M$ is hausdorff, shrinking the neighborhood $U$ if necessary, we can assume that $gx\not\in U$. Now, since $M$ is locally euclidean, it is locally compact, so we can find another euclidean neighborhood $x\in V \subset U$ so that $\bar{V}$ is compact and $\bar{V} \subset U$. Then we get that
$$gx\in G(U)\setminus \bar{V}$$
which is open since $\bar{V}$ is closed. Then we can find some euclidean neighborhood $gx \in W_1 \subset g(U)\backslash \bar{V}$. Consider the open set $g^{-1}(W_1)$. Clearly, $x \in V\cap g^{-1}(W_1)$ and we can find some euclidean neighborhood $x \in U_1\subset V\cap g^{-1}(W_1)$, Then by construction we have that $gx \in g(U_1) \in g(U_1)\subset W_1 \subset g(U)\backslash \bar{V}$ and a $U_1 \subset V \subset\bar{V}$, we get that $g_1(U_1)\cap U_1 = \emptyset$.
Now we need to prove that $M/G$ is locally euclidean and Hausdorff. Let $y \in M/G$. Then $y$ is in the equivalence class of some $x \in M$. Let $U$ be some locally euclidean neighborhood $U$ of $x$ contained in $M$ so that $\bar{U}$ is compact. Since the action is properly discontinuous, we have that there are finitely many elements $\{g_1,\cdots, g_m\} \subset G$ such that $g_i(\bar{U})\cap\bar{U} \not=\emptyset.$ Now, if $g_i(U)\cap U\not=\emptyset$, by the observation above, we can get some euclidean neighborhood $U_i$ of $x$ contained in $U$ such that $g(U_i)\cap U_i =\emptyset.$ If $g_i(U)\cap U=\emptyset$ then we set $U_i=U$, take the intersection $U' = \bigcap_{n=1}^{m}U_n$ since this is a finite intersection, it is open. Then take some euclidean neighborhood $V\subset U'$ containing $x$. Then by construction $g_i(V)\cap V = \emptyset$ for $i=1,\cdots,m$. $g(V)$ are disjoint from each other for every $g\in G$. Let $W = q(V)$. Then $q^{-1}(W) = \bigsqcup g(V)$ is an open set. So, $W$ is an open neighborhood of $y$. Also, $q|_V : V\to W$ is a homeomorphism and hence $W$ is in fact a eucldiean neighborhood of $y$. Thus, $M/G$ is locally euclidean.
Now suppose $y_1\not= y_2$ in $M/G$. Let $y_1 = [x_1]$ and $y_2 = [x_2]$ for $x_1$ and $x_2$ in $M$. Clearly, $x_2\not=gx_1$ for any $g \in G$. As before we can find neighborhoods $y_1\in W_1 = \bigsqcup g(V_1)$ and $y_2\in W_2 = \bigsqcup g(V_2)$, where $x_1\in V_1$ and $x_2\in V_2$. Since $M$ is Hausdorff, we can find $x_1\in B_1 \subset V_1$, $x_2\in B_2 \subset V_2$ such that $B_1\cap B_2 =\emptyset$. Then $q(B_1)\cap q(B_2) = \emptyset$ and they are open, since $q^{-1}(q(B_i)) = \bigsqcup g(B_i)$. Thus, $M/G$ is Hausdorff. Thus $M/G$ is a manifold.