Let $X$ be a locally compact, Hausdorff, path connected and locally path connected space. Assume a group $G$ acts freely and properly discontinuously on $X$, which means $\forall K^{compact},~~\{g\in G:g.K\cap K \not= \emptyset\}$ is finite and $g.x\not=x~~\forall g\not=1$.
I want to show the orbit space $X/G$ is locally compact, path connected and locally path connected.
For this, I considered a natural projection from $X$ onto $X/G$. It is obvious that the path connectednes is directly from the continuity. Hence I guess it is actually a covering map. But in my text book, the definition of covering space assumed the locally path connectedness, so I'm not sure how to approach this problem.
Any help will be appreciated.
Hint: Let $x\in X$ consider a neighborhood $U$ of $x$ whose adherence is compact. it exists since $X$ is locally compact. Thus $H_x=\{g\in G:g(U)\cap U\neq \phi\}$ is finite. You can shrink $U$ such that $H_x$ is empty since the action is free and $X$ is separated. To see this, consider $g_1,...,g_n$ the elements of $H_x$, since $X$ is separated, consider neighborhoods $U_i$ of $g_i(x)$ such that $U_i\cap U_j$ is empty, without restricting the generality, we can also suppose that $U_i\cap U$ is empty. Write $V=\cap_{i=1}^{i=n}{g_i}^{-1}(U_i)\cap U$. $V$ is open contains $x$ and $g(V)\cap V$ is empty for every $g\in G$. the restriction of the covering map $p:X\rightarrow X/G$ to $V$ is injective, thus $V$ and $p(V)$ are homeomorphic. Thus $p(V)$ is locally compact and locally path connected properties inherited from $V$.