TITLE: Is the space of diffeomorphisms homotopy equivalent to a CW-complex?
QUESTION [30 upvotes]: Clarification: My question concerns the homotopy type of the space of $C^k$ diffeomorphisms with the compact-open $C^k$ topology, where $0< k \leq\infty$. I have stated my question below with $k=1$ for definiteness and simplicity. I am not particularly interested in any specific value of $k$.$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\RR}{\mathbb{R}}$
It is fairly well-known that the space $\Diff(M)$ of $C^1$ diffeomorphisms of a closed smooth manifold $M$ is homotopy equivalent to a CW-complex. Here are the relevant facts:

$\Diff(M)$ is a Banach manifold modelled locally on the space of $C^1$ vector fields on $M$.
Metrizable Banach manifolds have the homotopy type of CW-complexes, as shown by Palais.

[Remark: When $M$ is closed, the spaces of $C^k$ diffeomorphisms of $M$ (for $0 < k \leq\infty$) are all homotopy equivalent to each other via the natural inclusions. This can be shown by embedding $M$ smoothly in $\RR^N$, and then using smoothing operators defined by taking convolution with a mollifier.]
When we allow $M$ to not be compact, there are several common topologies on $\Diff(M)$. I am interested in the compact-open (or weak) $C^1$ topology. 

Questions: Is it known whether the space $\Diff(M)$ of $C^1$ diffeomorphisms with the compact-open $C^1$-topology is homotopy equivalent to a CW-complex when $M$ is a smooth manifold without boundary? Is there a known counter-example? Are there particular cases where the answer is known, for example if $M$ is the interior of a compact manifold?

Feel free to use instead $C^k$ diffeomorphisms and/or the compact-open $C^k$ topology for any $0 < k \leq\infty$.
I would also be interested in hearing about any known results related to this question: e.g. for spaces of embeddings of manifolds in the compact-open/weak topology when the source is not the interior of a compact manifold.
Edit: Allen Hatcher gave a very nice answer to my question. Afterwards, I also posted an answer very similar to Allen's, which I was writing when Allen posted his. A pertinent question still remains: (1) Does the result hold for the interior of a compact manifold? Here is a perhaps less pertinent question: (2) For $M$ without boundary, do the path components of $\Diff(M)$ have the homotopy type of a CW-complex?

REPLY [28 votes]: 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.

REPLY [6 votes]: [Edit: Allen Hatcher posted an answer while I was writing this one. Both answers seem to use similar ideas. I will leave my answer here anyway.]
$\newcommand{\Diff}{\operatorname{Diff}}$$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\CC}{\mathbb{C}}$$\newcommand{\connsum}{\mathbin{\#}}$$\newcommand{\id}{\mathrm{id}}$$\newcommand{\set}[1]{\lbrace #1 \rbrace}$Reading the answer by Peter Michor gave me an idea for a counter-example, which I explain below. The question still remains as to whether the result is valid for the interior of a compact manifold.
Claim: The path components of a space homotopy equivalent to a CW-complex are all open.
A homotopy equivalence induces a bijection on path components. Consequently, if a space $X$ is homotopy equivalent to a space $Y$, and the path components of $Y$ are open in $Y$, then the path components of $X$ are open in $X$. Finally, observe that any CW-complex is locally path connected, and thus its path components are open.
Construction of the counter-example
It thus suffices to present a smooth manifold $M$ without boundary such that the path components of $\Diff(M)$ are not open. Define $M$ to be the open submanifold of $\RR \times S^1$ given by
$$ M = (\RR \times S^1) \setminus (\ZZ \times \set{1}) $$
where $1\in S^1 \subset\CC$. One can also view $M$ as a connected sum of infinitely many punctured spheres $P=S^2\setminus\set{(1,0,0)}$:
$$ M \cong \, \cdots \connsum P \connsum P \connsum P \connsum \cdots $$
Proof that the path component of $\id_M$ is not open in $\Diff(M)$
Pick a neighbourhood $U$ of $\id_M$ in $\Diff(M)$. I will describe a diffeomorphism $\varphi\in U$ which is not in the path component of $\id_M$ in $\Diff(M)$, thus concluding the proof. By definition of the compact-open topology on $\Diff(M)$, there exists a compact subspace $K$ of $M$ such that a given diffeomorphism of $M$ is in $U$ if it is the identity on $K$. Let $n$ be a positive integer large enough so that $K\subset [-n,n]\times S^1$. Now we define the required diffeomorphism $\varphi$ of $M$:
$$ \varphi(t,x) = \bigl( t , e^{i\cdot\theta(t-n)} x \bigr) $$
where $\theta:\RR\to\RR$ is any smooth function which is identically zero on $(-\infty,0]$, and equals $2\pi$ on $[1,+\infty)$. Viewing $M$ as the connected sum of infinitely many punctured spheres, the diffeomorphism $\varphi$ is the result of applying a Dehn twist at the junction cylinder joining the spheres numbered $n$ and $n+1$.
It is easy to see that $\varphi:M\to M$ is not even homotopic to the identity: in fact, the homomorphism induced by $\varphi$ on $\pi_1(M)$ (based at the point $(n,-1)$) is not conjugate to the identity. Here is a simple way to see this:

The group $\pi_1 M$ is free on infinitely many generators.
The homomorphism $\pi_1 \varphi$ coincides with the identity on all loops in $M$ contained in $(-\infty,n]\times S^1$. In particular, $\pi_1 \varphi$ fixes two distinct (actually, infinitely many) free generators of $\pi_1 M$.
The homomorphism $\pi_1 \varphi$ is not the identity homomorphism on $\pi_1 M$: a loop going once around the puncture $(n+1,1)\in\RR\times S^1$ is not fixed by $\pi_1 \varphi$.
In a free group, conjugation by a non-identity element can fix at most one of the free generators.