TITLE: Homotopy equivalence of diffeomorphism groups
QUESTION [23 upvotes]: Let $M$ be a closed connected smooth manifold and let ${\rm Diff}^r(M)$ be the group of $C^r$-diffeomorphisms equipped with the compact-open $C^r$-topology. I am looking for a reference to the fact that the natural inclusion ${\rm Diff}^r(M)\to {\rm Diff}^1(M)$ is a homotopy equivalence for $1\leq r\leq \infty$.

REPLY [10 votes]: I am not aware of a detailed reference but here is a sketch. 

$\mathrm{Diff}^r(M)$ is a Hilbert manifold (i.e., it is locally homeomorphic to a separable Hilbert space). This can be found e.g., in section 10 of Michor's "Manifolds of differentiable mappings". EDIT: Instead of referring to Michor who indeed focuses on the $C^\infty$ case let me give a direct proof. Let $\exp$ be the exponential map of some smooth Riemannian metric on $M$. Then any self-diffeomorphism $\phi$ of $M$ defines a vector field $X$ given by $\phi(p)=\exp_p X(p)$. Conversely, any vector field on $M$ can be exponentiated to a self-map of $M$ so that the vector field is close to zero if and only if the corresponding diffeomorphism is close to the identity. Also $\phi$ is $C^r$ if and only if $X$ is $C^r$. This defines a homeomorphism of a neighborhood of the identity in $\mathrm{Diff}^r(M)$ onto a neighborhood of the origin in the separable Frechet space of $C^r$ vector fields on $M$. Since $\mathrm{Diff}^r(M)$ is topologically homogeneous we get a chart in every point (not just at the identity). Finally any separable Frechet space is homeomorphic to $\ell^2$, so $\mathrm{Diff}^r(M)$ is a Hilbert manifold.
Any Hilbert manifold is an ANR (because $\ell^2$ is an ANR, and any local ANR is an ANR). Also any ANR it is homotopy equivalent to a CW complex. In particular, there is Whitehead's theorem and to show that the inclusion $\mathrm{Diff}^r(M)\to\mathrm{Diff}^1(M)$ is a homotopy equivalence it is enough to prove that it induces an isomorphism on homotopy groups. 
Build a continuous smoothing operator that takes as an input a $C^r$ self-map of $M$, and instantly makes it $C^\infty$. This is a good elementary exercise in differential topology. (Sketch: $C^\infty$ embed the compatible $C^\infty$ structure on $M$ into some $\mathbb R^n$, use methods of Chapter 2 section 2 of Hirsch's "Differential topology" to continuously approximate $C^r$ maps $M\to\mathbb R^n$ by $C^\infty$ maps, and then use tubular neighborhood projection to push the maps to $M$).
The smoothing operator instantly pushes a continuous map from a disk to $\mathrm{Diff}^1(M)$ into $C^\infty(M, M)$, and hence into $\mathrm{Diff}^\infty(M)$ because by the inverse function theorem diffeomorphisms form an open subset in smooth maps. Thus any singular sphere in $\mathrm{Diff}^1(M)$ can be deformed to $\mathrm{Diff}^r(M)$, and if a singular sphere $\mathrm{Diff}^r(M)$ contracts in $\mathrm{Diff}^1(M)$, then the null-homotopy can be pushed into $\mathrm{Diff}^r(M)$. 

In summary,
the inclusion $\mathrm{Diff}^r(M)\to \mathrm{Diff}^1(M)$ induces an isomorphism on homotopy groups, and hence a homotopy equivalence.