I'm trying to solve the following problem from Hatcher (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 orientable.
Here are some relevant definitions as Hatcher mentions that "covering space action" is non-standard terminology.
For our manifold $M$ and our group $G$ define covering space action as an action satisfying: $\forall x \in M$, $\exists U$ neighborhood of $x$ s.t. all images $g(U)$ for varying $g \in G$ are disjoint. (In other words $g_1(U) \cap g_2(U) \neq \emptyset \implies g_1 = g_2$.)
Also we note that a point in $M/G$ is $Gx = \{y \in M : g(x) = y$ for some $g \in G$ an orbit of $x$. (Yes it's true the notation is a bit abusive as really g is an element of g which is assigned to some homeomorphism, but I believe it is standard).
We also know since $M$ is orientable that there exists a global orientation function $\mu_M: M \longrightarrow \{\pm 1\}$ which takes $x$ to $\mu_x$ the preferred generator for the local homology group of $x$ such that there exists a neighborhood $U \cong D^n$ of $x$ such that $\forall y \in U$, $\mu_y = \mu_x$. (A local compatibility condition, i.e. global function respects local orientation).
Here's my start of a proof:
Claim: $\mu_{M/G}: M/G \rightarrow \{\pm 1\}$ which takes $xG$ to $\mu_M(x)$ is a global orientation function satisfying the proper conditions.
Now let $U \cong D^n$ be a neighborhood of $x$. Let $y \in U$, then $y = g(z)$ for some $z \in M$ and $g \in G$.
From here I'm not really sure how to get to the compatibility conditions, any help would be very appreciated.
Edit: Perhaps we just choose $U$ a neighborhood of $x$ such that its equivalence classes are the elements in an open neighborhood $V$ of $z$ in $M$ satisfying the proper compatibility conditions, then it should follow from that fact?
You have seen a couple of exercises concerning orientability so I suggest you should be able do solve it yourself with the following hint:
Hint: You have to try to push down the orientation $\mu$ to $M/G$. To do this you have to use, that there are no orientation reversing covering homeomorphisms, by noting that the preferred generators of any orbit in the covering gives you the same preferred generator at the image point in $M/G$. For this just consider the commutative diagram of local homology groups, corresponding to a covering translation of $G$ together with the projection. Now note that under a orientation preserving homeomorphism, preferred generator is mapped to preferred generator.
(Now the hint actually turned out to be a whole proof.)