TITLE: Diffeomorphism group and covering spaces
QUESTION [6 upvotes]: Let $M$ be a smooth manifold and Diff($M$) be its diffeomorphism group. Let $N\to M$ be a finite cover (for example an orientable double cover if $M$ is not orientable). Then what is the relation between Diff($N$) and Diff($M$? 
Is it true for example that Diff($N$) injects in Diff($M$)? If the cover is canonical in some sense (e.g. universal), is it true that the action of any group on $M$ lifts to $N$?

REPLY [4 votes]: Just to be a bit contrary, group actions in sense do lift.  Of course, Lee is right that if $G$ acts on $M$ nontrivally, there is no reason to assume $G$ acts on $N$ non-trivially (or vice versa).  But you were allowed to replace $M$ with $N$ so it's only fair that you allow $G$ to be replaced as well.
For ease, I will assume $N\rightarrow M$ is the universal covering.  Suppose a topological group $G$ acts on $M$.
This is defined by some kind of map $G\times M\rightarrow M$ satisfying certain conditions.   We are looking for a map $H\times N\rightarrow N$ where $H$ is a topological group which could, in some sense, be called a lift of the $G$ action on $M$.  To that end, let $H$ denote the universal cover of $G$.
Then we have a canonical map $H\times N\rightarrow G\times M\rightarrow M$.  Since $H$ and $N$ are simply connected, the lifting criteria for coverings guarantees we can lift the map.  If we fix $n\in N$, we know we want $(e,n)$ (where $e\in H$ is the identity) to map to $n$.  If we specifiy this, then we get a unique lift $H\times N\rightarrow N$.
Now one can verify that this lift does in fact define a group action of $H$ on $N$.  Further, if $\pi:H\rightarrow G$ and $\rho:N\rightarrow M$ are the universal covering maps, then for any $h\in H$, $n\in N$, one can verify that $\rho(h\ast n) = \pi(h)\ast\rho(n)$, so the actions are compatible in this sense.