When reading Lee's book, I encountered the following exercise:
Let $\mathcal{P}\colon M\rightarrow G\backslash M$ be the covering arising from a free and proper discrete group action of $G$ on $M$ and suppose $M$ is connected. Let $\Gamma_G:=\{l_g\in \text{Diff}(M):g\in G\}$, then $G$ is isomorphic to $\Gamma_G$ by the obvious map $g\mapsto l_g$ and furthermore $\Gamma_G= \text{Deck}(\mathcal{P})$
I have two questions:
(1) Why do we need $M$ to be connected? I think the conclusion is obvious and does not involve connectedness of $M$.
(2) There is no assumption that the discrete group action is smooth, so isn't $\Gamma_G$ the empty set? Or we can deduce smoothness of the action from the connectedness of $M$?