While working on some problem in differential topology, I had to prove the lemma below. It seems to me like I have not needed all the requirements in the lemma, leading me to think that I have missed some of the finer points.
Question
Given a group of $G$ of diffeomorphisms acting freely and properly discontinuous on a connected orientable smooth manifold $M$, show that the orbit space $M/G$ is orientable iff each $g \in G$ is an orientation preserving diffeomorphism of $M$.
Some Working
Firstly, let $\pi : M \rightarrow M/G$ be the natural homomorphism.
Forward: Let $M/ G$ be orientable; we want to show that $g \in G$ is orientation preserving. Let $\omega$ be an orientation form on $M / G$, then we can consider the pullback $\pi^* \omega$ which is an orientation form on $M$. We want to show that $g^* \pi^* \omega$ has the same orientation as $\pi^* \omega$. We note that $\pi \circ g = \pi$; therefore $g^* \pi^* \omega = \pi^* \omega$ and we conclude that $g$ is orientation preserving.
Converse: Suppose every $g \in G$ is orientation preserving, then given some orientation form $\omega$ on $M$, $g^* \omega$ is also consistently oriented with respect to $\omega$. Consider some $[x] \in M/G$, we can push $\omega$ to $M/G$ through $\mathrm{d} \pi_x : T_x M \rightarrow T_{[x]} (M/G)$. The resulting orientation is well defined since $\omega$ at each element in $\pi^{-1}([x])$ is consistently oriented.
The Issue
The question requires the elements of $G$ to be acting freely and properly discontinuous. That is, for all $p \in M$ there exists a neighbourhood $U$ of $p$ such that $g(U) \cap U = \varnothing$.
I do not seem to use that property in my proof, leading me to think that my proof is incorrect. Can someone explain where this property comes into play? I'm suspecting that it comes into play in the proof of the converse, but I'm not sure how as it seems to me like I just need the property that $g$ is diffeomorphic.
