Does a free action always induce a diffeomorphism?

Suppose that $G$ is a Lie group with a transitive action on a smooth manifold $M$. The regular theory of Lie groups tells us that $G$ and $M$ are diffeomorphic if the isotropy group is trivial.

The proof that I know goes, as usual, by proving that the canonical map $G/H \rightarrow M$ is bijective and its differential is injective. This implies that the canonical map is a diffeomorphism, but heavily relies on the fact that both manifolds are finite dimensional.

Suppose now that both $G$ and $M$ are infinite dimensional manifolds. Is it still true that $G$ and $M$ are diffeomorphic if the isotropy group is trivial? Or asked differently, is there a direct proof that the differential of $G/H \rightarrow M$ is surjective?

I just realized that the proof I know also depends on the SMOOTH exponential map, which can be defined in the infinite dimensional case, but according to "INFINITE DIMENSIONAL LIE GROUPS WITH APPLICATIONS TO MATHEMATICAL PHYSICS" by Rudoph Schmid is not always differential (the counterexample is the group of $s$-Sobolev diffeomorphisms).

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Do you have a particular kind of infinite dimensional manifold in mind? Hilbert, Banach, Frechet etc – Michael Murray Jul 4 '13 at 13:41
Whatever you like. I'd prefer Banach for generality, but if there is a proof for the Hilbert or Frechet case, I'd take that, too. The same goes for the second countability axiom. If you've got an idea how to do it without, great. If you can do it only with, great anyway. – user35946 Jul 4 '13 at 13:43
@user35946: Second countability is important even in the finite dimensional case: Let $G= \mathbb{R}\times\mathbb{R}_{disc}$, where $\mathbb{R}_{disc}$ is $\mathbb{R}$ with the discrete topology. Then $G$ is a $1$-dimensional Lie group that is not second countable. $G$ has an obvious smooth, transitive action on $\mathbb{R}^2$ with trivial isotropy, but $G$ is not diffeomorphic to $\mathbb{R}^2$. – Robert Bryant Jul 4 '13 at 14:51
See mathoverflow.net/questions/57015/… for a related question. – Igor Belegradek Jul 4 '13 at 15:12
@RobertBryant Why don’t you just use $\mathbb{R}_{disc}$ acting on the line? – The User Jul 4 '13 at 16:16

The answer is no, in general, even for both $G$ and $M$ connected. A simple example is the following: Take $G=Diff(S^1)$, the diffeomorphism group with the the $C^\infty$-topology, which is a regular Frechet Lie group. Let $M=Diff(S^1)$ (smooth diffeomorphisms), but now with the Sobolev $H^1$-topology, which is a topological group modelled on pre-Hilbert spaces. Then $G$ acts smoothly on $M$ but $G$ and $M$ are not diffeomorphic. Of course you can say, that $M$ carries the wrong topology, and that I just forced this outcome. But this is the possibility in infinite dimensions.