(Hatcher Exercise 3.3.4) Given a covering space action of a group $G$ on an orientable manifold $M$ by orientation-preserving homeomorphisms, show that $M/G$ is also orientable.
I first assumed $\{\mu_x\}$ be the local orientations of $M$. And I showed that if I denote $\bar{x}$ be the orbit space of $x$ under the action $G$, $H_n(M,M-\{x\})\to H_n(M/G,M/G-\{\bar{x}\})$ is an isomorphism with $\mu_x\mapsto \mu_{\bar{x}}$ and this map is well-defined as $G$ consists of orientation preserving homeomorphisms. But I don't know how to show the local consistency condition i.e., there is an open ball $\bar{x}\in B_{\bar{x}}\subset M/G$ such that $H_n(M/G,M/G-B_{\bar{x}})\to H_n(M/G,M/G-\{\bar{x}\})$ maps $\mu_{B_{\bar{x}}}\mapsto \mu_{\bar{y}}$ for every $\bar{y}\in B_{\bar{x}}$. How can I show this?
Note. There is a post that asks the same question but I don't think that answer contains local consistency part.
The definition for covering space action by Hatcher is that, for all $x\in M$ there exists an open set $U \subset M$ with $x \in U$, such that $\{g \in G \; | \; g(U) \cap U \neq \emptyset\} = \{e\}$.
If $\pi:M\to M/G$ denotes the quotient map, this says, that $\pi(U) \subset M/G$ is open and $\pi|_U :U \to \pi(U)$ is a homeomorphism. Now choose a neighbourhood $V$ of $x$ such that $\bar{V} \subset U$. So $\pi$ restricts as $\pi: U\setminus V \to \pi(U) \setminus \pi(V)$.
Now we have a commuting diagram
\begin{align*}\require{AMScd} \begin{CD} H_n(M,M\setminus\{x\}) @>{\pi_*}>> H_n(M/G,M/G\setminus\{\bar{x}\})\\ @AAA @AAA \\ H_n(M,M\setminus V) @>>> H_n(M/G,M/G\setminus\pi(V))\\ @AAA @AAA\\ H_n(U,U\setminus V) @>{\pi_*}>> H_n(\pi(U),\pi(U)\setminus \pi(V)) \end{CD} \end{align*}
Where the two lower verticals are excision isomorphisms, which can be used to define the horizontal map in the middle.