Is the space of diffeomorphisms homotopy equivalent to a CW-complex?

Here is an example where ${\rm Diff}(M)$ with the compact-open topology is not homotopy equivalent to a CW complex. Take $M$ to be a surface of infinite genus, say the simplest one with just one noncompact end. I will describe an infinite sequence of diffeomorphisms $f_n:M\to M$ converging to the identity in the compact-open topology and all lying in different path-components of ${\rm Diff}(M)$. Assuming this, suppose $\phi:{\rm Diff}(M) \to X$ is a homotopy equivalence with $X$ a CW complex. The infinite sequence $f_n$ together with its limit forms a compact set in ${\rm Diff}(M)$, so its image under $\phi$ would be compact and hence would lie in a finite subcomplex of $X$, meeting only finitely many components of $X$. Thus $\phi$ would not induce a bijection on path-components, a contradiction.

To construct $f_n$, start with an infinite sequence of disjoint simple closed curves $c_n$ in $M$ marching out to infinity, and let $f_n$ be a Dehn twist along $c_n$. The $f_n$'s converge to the identity in the compact-open topology since the $c_n$'s approach infinity. We can choose the $c_n$'s so that they represent distinct elements in a basis for $H_1(M)$ and then the $f_n$'s will induce distinct automorphisms of $H_1(M)$. If two different $f_n$'s were in the same path-component of ${\rm Diff}(M)$ they would have to induce the same automorphism of $H_1(M)$ since any path joining them would restrict to an isotopy of any simple closed curve in $M$ (see the next paragraph below) and a basis for $H_1(M)$ is represented by simple closed curves.

If $g_t$ is a path in ${\rm Diff}(M)$ then the images $g_t(c)$ of any simple closed curve $c$ vary by isotopy since this is true as $t$ varies over a small neighborhood of a given $t_0$, so since the $t$-interval $[0,1]$ is compact, a finite number of these neighborhoods cover $I$ and the claim follows.

Remark: The $f_n$'s were chosen to be Dehn twists just for convenience. Many other choices of diffeomorphisms would work just as well. One can easily see how to generalize to higher dimensions.

Check the paper Antonelli, P. L.; Burghelea, D.; Kahn, P. J. The non-finite homotopy type of some diffeomorphism groups. Topology 11 (1972), 1–49.

There they mention an old result of Palais (Homotopy theory of infinite dimensional manifolds. Topology 5 (1966), 1-16) which states that the identity component of ${\rm Diff}_0(M)$ has the homotopy type of a countable $CW$-complex.

In their paper, Antonelli, Burghelea and Kahn prove that for many smooth manifolds (including spheres of dimension $\geq 7$) the group ${\rm Diff}_0(M)$ does not have the homotopy type of a finite $CW$-complex. (This is highly nontrivial.)

Above, by diffeomorphisms they mean smooth diffeomorphisms and the topology is the $C^\infty$-topology.

$\newcommand{\Diff}{\operatorname{Diff}}$ First let $M$ be closed. Then $\Diff^\infty(M)$ is locally modelled on the space of smooth vector fields in $M$ which is a Frechet space. All (infinite dimensional separable) Frechet spaces are homeomorphic. So this is reduced to the case you described. The same is true for $\Diff^k(M)$.

Let $M$ be open (non-compact without boundary). Then $\Diff^\infty(M)$ is open in $C^\infty(M,M)$ in the Whitney $C^\infty$ topology. But it is not locally contractible in this topology, and the arc connected component of the identity is contained in the group $\Diff^\infty_c(M)$ of diffeomorphisms which differ from the identity only on a compact set. The group $\Diff^\infty_c(M)$ is a regular Lie group locally modeled on the space of $\mathfrak X_c(M)$ smooth vector fields with compact support which is a nuclear (LF)-space. Similar for $\Diff^k(M)$. I do not remember whether $\mathfrak X_c(M)$ is a CW-complex.

See [Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158 pp., MR 83g:58009, ZM 433.58001]
for lots of details on this, including the case of manifolds with boundary and with corners. See also section 41 of [here].

Edit: Vidit Nanda pointed out that the compact open $C^1$ or $C^\infty$ topology was asked for. This is not a good topology: In general it is not locally contractible, so no manifold modelled on topological vector spaces. Also $\Diff^k(M)$ is not open in $C^k(M,M)$ for any $k\ge 1$ if $M$ is not compact.

I once tried to describe a setting where $\Diff^\infty(M)$ with the compact $C^\infty$ topology would be a Lie group: smooth manifolds based on smooth curves instead of charts; but you need a lot of other structures like a geodesic structure, locally convex spaces as tangent spaces, and parallel transport. The resulting category of manifolds is monodially closed, and the the manifolds with finite dimensional (or even Banach) tangent spaces are exactly the usual ones. The topology is the final topology with respect to the smooth curves, and for $\Diff(M)$ it is indeed the compact $C^\infty$-topology. The theory is horrendibly complicated, and nobody ever used it. I have no idea whether this helps for the quest for CW-complexes. See:

• Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178. (pdf of part 1) (pdf of part 2)

Liviu Nicolaescu's answer seems to be the best for your question.