Suppose that $X$ is a compact, finite dimensional manifold and $Y$ is an infinite dimensional, second countable ($C^\infty$-)Banach manifold. Let $\nu \in \mathbb{N}$.
Question: Is the space $C^\nu(X,Y)$ a ($C^\infty$ or $C^k$ for some $k \in \mathbb{N}$) Banach manifold which is Lindelöf (i.e. every open cover has a countable subcover)?
Any help, as well as references are very much appreciated.
Some motivation:
My goal is to apply a parametric transversality theorem, where my parameter space is $C^\nu(X,Y)$. In order to so it is required that the parameter space is Lindelöf and a ($C^k$-) Banach manifold. I don't have much experience with function spaces and their topology, especially when the target is infinite dimensional. So I apologize if it turns out that my question is "well known".
Best Answer
See the paper
for a description of the manifold structure on $C^\nu(X,Y)$. But you need somthing like a Riemannian exponential mapping on $Y$ which is a diffeomorphism from a neighborhood of the 0-section in $TY$ to a neighborhood of the diagonal in $Y\times Y$; you can forget some properties of the exponential mapping, then this is called a local addition (see here). If $Y$ is nice enough (smoothly paracompact or at least $C^1$-paracompact allows to construct a linear connection on $TY$; or you may be given a linear connection on $TY$) then you have this.
If $X$ is compact and $Y$ is a separable Banach space (added in second edit), then $C^\nu(X,Y)$ is separable; In fact it is $C^\nu(X)\hat{\hat\otimes} Y$ for the completed inductive tensor product. (see here, e.g) Then it is obvious that the tensor product is separable.
In "Kriegl, Michor: The convenient setting ..." (see my homepage) 14.11.1 states that there does not exist any Frechet differentiable bump function. You can embed an interval in any manifold, thus $C^0(M)$ does not admit differentiable bump functions. Furthermore, $C^0$ sits at the top of $C^k(M)$, so also $C^k(M)$ does not admit differentiable bump functions. On the other hand, $C^\infty(M)$ is nuclear Frechet, so has smooth partitions of unity. This is the reason why I vastly prefer $C^\infty$, and Sobolev completions, if I have to solve equations.