why the space of smooth mappings has the homotopy type of a CW complex?

Question

let $N$ be a smooth compact manifold. I want to know why $C^\infty (N,\mathbb{R})$ has the homotopy type of a CW complex? I know that if Y has the homotopy type of a CW complex and X is a finite CW complex, then the function space $Y^X$ with the compact-open topology has the homotopy type of a CW complex, Milnor, but i can't proof a compact manifold is a finite CW complex.

Answer 1

Since $C^\infty(N,\mathbb{R})$ is a topological vector space, it is contractible (just use the contraction $(f,t)\mapsto tf$).

(It is much less trivial to say whether $C^\infty(N,M)$ has the homotopy type of a CW-complex when $M$ is an arbitrary smooth manifold. Since smooth manifolds are triangulable, $M$ and $N$ are CW-complexes, so by Milnor the space of continuous maps $C(N,M)$ has the homotopy type of a CW-complex. I believe you can then show that the inclusion map $C^\infty(N,M)\to C(N,M)$ is a homotopy equivalence, but I don't know the details off the top of my head.)