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The probem is:

Suppose $M$ is connected, orientable, smooth manifold and $\Gamma$ is a discrete group acting smoothly, freely, and properly on $M$. We say that the action is orientable-preserving if for each $\gamma \in \Gamma$, the diffeomorphism $x \rightarrow \gamma \cdot x$ is orientation-preserving. Show that $M/ \Gamma $ is orientable if only if the action of $\Gamma$ is orientable-preserving.

Notes: I tried using that $\pi\colon M\to M/\Gamma$ is a covering map. I also tried to use that for any $\gamma$ we have two disjoint open sets $U,V\subset M$ such that $\pi|_{U}$ and $\pi|_{V}$ are diffeomorphisms and $(\pi|^{-1}_{U} \circ \pi|_V)(x)=\gamma.x$ but I didn't prove this result too.

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Dear Kelson, I noticed that you undid some edits I made yesterday, which is fine, but I just wanted to explain them. (1) Placing the > for block quotes in the middle of a math environment results in $\in>$, which I don't think is what you want (2) I think it's different in other languages, but in English I've only ever seen the spelling "diffeomorphism" (3) In specifying the source and target of a morphism it seems to me that \colon gives better spacing than :. This one is more of a preference, though. (4) "tried use" seems off. Cheers, – Dylan Moreland Jun 28 '12 at 19:31

If the action is orientation-preserving, construct explicitly an orientation on the quotient. Fix an orientation on $M$. If $\bar x\in M/\Gamma$ is a point and $x\in M$ is such that $\pi(x)=\bar x$, the differential $d\pi_x:T_xM\to T_{\bar x}(M/\Gamma)$ is an isomorphism, so you can push the orientation of $T_xM$ given by the orientation of $M$ to one on $T_{\bar x}(M/\Gamma)$. Check that this depends only on $\bar x$ and not on the preimage $x$ chosen: this is where the hypothesis comes in. Finally, check that this way of orienting the tangent spaces to $M/\Gamma$ is in fact an orientation of $M/\Gamma$.

Can you do the converse?

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Could this result holds if we drop "smooth"? – Henry Jun 3 '15 at 11:26

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