Personal comment: It seems the discussion in this question finally led me to understand how to modify Agol's argument to answer the present question. In fact, my motivation when asking that question answered by Agol was mostly to resolve the present question.
Answer: For $M$ a non-compact manifold without boundary, it seems the homotopy fibre of the inclusion $\iota:\textrm{Diff}(M)\hookrightarrow\textrm{Emb}(M,M)$ at $\textrm{id}_M$ is weakly contractible. As hinted implicitly by Tom Goodwillie in his answer, it is not hard to conclude that every homotopy fibre of $\iota$ is then either empty or weakly contractible.
Before continuing, I will clarify the structure of the homotopy fibre, $F$, of $\iota:\textrm{Diff}(M)\to\textrm{Emb}(M,M)$ at $\textrm{id}_M$. Recall from the question above that $\textrm{Diff}(M)$ and $\textrm{Emb}(M,M)$ are both endowed with the compact-open (or weak) $C^1$-topology. As a set, $F$ consists of the continuous isotopies $(\varphi_t:M\to M)_{t\in I}$ through embeddings $M\to M$ for which $\varphi_0 = \textrm{id}_M$, and $\varphi_1$ is a diffeomorphism of $M$. An isotopy $(\varphi_t)_{t\in I}$ will also be written as a map $\varphi : M\times I \to M$. As a topological space, $F$ is the subspace of the space of paths in $\textrm{Emb}(M,M)$ consisting of the paths which start at $\textrm{id}_M$ and end at a diffeomorphism of $M$.
Further, let me state a strong form of the isotopy extension lemma which I will require later: when $X$ is a manifold without boundary, and $Y$ is a compact manifold, the map
$$ (\textrm{eval},\textrm{proj}):\textrm{Diff}_c(X)\times\textrm{Emb}(Y,X) \longrightarrow \textrm{Emb}(Y,X)\times\textrm{Emb}(Y,X) $$
given by $(\textrm{eval},\textrm{proj})(u,v)=(u\circ v,v)$ is a fibration. Here $\textrm{Diff}_c(X)$ denotes the space of compactly supported diffeomorphisms of $X$, equipped with the strong (or Whitney) $C^1$ topology --- the use of the strong topology is essential in the present answer, as remarked at a later point. The map $(\textrm{eval},\textrm{proj})$ is a locally trivial fibre bundle, as one can give local trivializations using tubular neighbourhoods; thus the above map is also a fibration, since the base is paracompact (even a metrizable space). The above statement implies most usual forms of the isotopy extension lemma.
I will now describe the argument that the homotopy fibre $F$ of $\iota:\textrm{Diff}(M)\to\textrm{Emb}(M,M)$ at $\textrm{id}_M$ is weakly contractible. It is a modification of Agol's ingenious answer linked above.
Fix a map $f:K\to F$ where $K$ is compact. I will describe a homotopy between $f$ and the constant map $b:K\to F$ equal to the basepoint $b=(\textrm{id}_M)_{t\in I}$ of $F$. First we construct an exhaustion of $M$ by compact submanifolds $M_i$ for $i\in {\mathbb N}$ such that:
- $\bigcup_{i\in {\mathbb N}} M_i = M$;
- $M_0=\emptyset$;
- $M_i \subset \textrm{int}\ M_{i+1}$ for all $i\in {\mathbb N}$;
- $\varphi(M_i \times I)\subset \textrm{int}\ M_{i+1}$ for all $\varphi$ in the image of $f$.
For example, we may use a proper smooth function $\rho: M\to [0,+\infty)$, and take $M_i=\rho^{-1}([0,x_i])$ for a suitable unbounded, strictly increasing sequence $(x_i)_{i\in{\mathbb N}}$ of regular values of $\rho$ (which are dense in $[0,+\infty)$ by Sard's theorem). The final condition above is where the compactness of $K$ is required, and I do not know how to avoid using compactness there.
With the compact submanifolds $M_i \subset M$ at hand, we can inductively construct a homotopy $f\simeq b$. Set $f_0=f$, and let $n\in {\mathbb N}$. Assume inductively that we have constructed, for each $k < n$, a map $f_{k+1}:K\to F$, and a homotopy $H_k : f_k \simeq f_{k+1}$. Furthermore, assume that for all $\phi$ in the image of $f_{k+1}:K\to F$, and for all $\psi$ in the image of $H_k: K\times I\to F$, the following conditions hold:
- $\phi_t|_{M_k} = \textrm{id}_{M_k}$ for all $t\in I$;
- $\psi_t|_{M_{k-1}} = \textrm{id}_{M_{k-1}}$ for all $t\in I$;
- $\psi(M_l \times I) \subset \textrm{int}\ M_{l+1}$ for each $l\in {\mathbb N}$ with $l\geq k$.
Now we construct a homotopy $H_n : f_n \simeq f_{n+1}$ based on the above data. Let $\varphi = f_n(x)$ for a fixed $x\in K$. By condition (3) (and $M_0=\emptyset$), we have $\varphi(M_n \times I) \subset \textrm{int}\ M_{n+1}$. Therefore, the restriction $\varphi|_{M_n \times I}$ is an isotopy through embeddings $M_n \to \textrm{int}\ M_{n+1}$. Since $\varphi_0 = \textrm{id}_M$, the isotopy extension lemma implies that $\varphi|_{M_n \times I}$ extends to a compactly supported diffeotopy $\widetilde{\varphi}=g(x): (\textrm{int}\ M_{n+1})\times I \to \textrm{int}\ M_{n+1}$ with $\widetilde{\varphi}_0=\textrm{id}_{\textrm{int}\ M_{n+1}}$.
To guarantee that $\widetilde{\varphi}=g(x)$ depends continuously on $x\in K$, we now take a detour in which we apply the strong form of the isotopy extension lemma stated earlier. Consider the commutative square
$$ \begin{matrix}
K & \xrightarrow{\ (a,b)\ } & \textrm{Diff}_c(\textrm{int}\ M_{n+1})\times\textrm{Emb}(M_n,\textrm{int}\ M_{n+1}) \\
\llap{\scriptstyle \textrm{incl}_0}\Big\downarrow & & \llap{\scriptstyle (\textrm{eval},\textrm{proj})}\Big\downarrow \\
K\times I & \xrightarrow{\smash{\ (c,d)\ }} & \textrm{Emb}(M_n,\textrm{int}\ M_{n+1})\times\textrm{Emb}(M_n,\textrm{int}\ M_{n+1})
\end{matrix} $$
where the components of the horizontal maps are defined by
$$ a(x) = \textrm{id}_{\textrm{int}\ M_{n+1}} \qquad b(x) = [f_n(x)]_0|_{M_n} = ( \textrm{incl}:M_n\hookrightarrow \textrm{int}\ M_{n+1} ) $$
$$ c(x,t) = [f_n(x)]_t|_{M_n} \qquad d(x,t) = b(x) = ( \textrm{incl}:M_n\hookrightarrow \textrm{int}\ M_{n+1} ) $$
(so only the map $c$ is not constant). Since the vertical map on the right of the square is a fibration, there exists a diagonal lift
$$ (e,d) : K\times I \longrightarrow \textrm{Diff}_c(\textrm{int}\ M_{n+1})\times\textrm{Emb}(M_n,\textrm{int}\ M_{n+1}) $$
and the map $g:K\to F$ (such that $\widetilde{\varphi}=g(x)$) is determined by
$$ \widetilde{\varphi}_t=[g(x)]_t = e(x,t) $$
It is easy to check the required conditions:
- $\widetilde{\varphi}=g(x)$ and $\varphi=f_n(x)$ coincide on $M_n \times I$;
- $\widetilde{\varphi}_0=[g(x)]_0=\textrm{id}_{\textrm{int}\ M_{n+1}}$.
as follows from the fact that $(e,d)$ is a diagonal lift for the square above.
In conclusion, we have a compactly supported diffeotopy $\widetilde{\varphi} = g(x)$ of $\textrm{int}\ M_{n+1}$ (depending continuously on $x\in K$) which extends $\varphi|_{M_n \times I}$. Extend $\widetilde{\varphi} = g(x)$ to all of $M$ by making it the identity outside $\textrm{int}\ M_{n+1}$. Fundamentally, observe that this extension operation gives a continuous map $\textrm{Diff}_c(\textrm{int}\ M_{n+1}) \to \textrm{Diff}_c(M) \to \textrm{Diff}(M)$ thanks to the strong (or Whitney) $C^1$ topology we placed on the space of compactly supported diffeomorphisms, so we are preserving continuity on $x\in K$. Then $\widetilde{\varphi} = g(x)$ becomes a compactly supported diffeotopy of the whole manifold $M$ depending continuously on $x\in K$. Moreover, $\widetilde{\varphi}_0 = \textrm{id}_M$.
It is now easy to define $f_{n+1} : K \to F$:
$$ [f_{n+1}(x)]_t = (\widetilde{\varphi}_t)^{-1} \circ \varphi_t = ([g(x)]_t)^{-1} \circ [f_n(x)]_t $$
Condition (1) above then follows from the fact that $\widetilde{\varphi}$ extends $\varphi|_{M_n \times I}$. Note that $f_{n+1}$ is continuous since mapping a diffeomorphism to its inverse is continuous in the compact-open $C^1$ topology ($\textrm{Diff}(M)$ is a topological group); alternatively, we can instead use that $g$ is actually continuous with respect to the strong $C^1$ topology on $\textrm{Diff}_c(M)$, and that inversion is also continuous in the strong $C^1$ topology.
We further define the homotopy $H_n : K\times I \to F$ between $f_n$ and $f_{n+1}$ as follows:
$$ [H_n(x,s)]_t = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t = ([g(x)]_{\min\{s,t\}})^{-1} \circ [f_n(x)]_t $$
As was done for $f_{n+1}$, we can show that $H_n$ is continuous. It is also easy to check that $H_n$ takes values in $F$: $[H_n(x,s)]_0 = (\widetilde{\varphi}_0)^{-1} \circ \varphi_0 = \textrm{id}_M$, and $[H_n(x,s)]_1 = (\widetilde{\varphi}_s)^{-1} \circ \varphi_1$ is a diffeomorphism of $M$. Moreover, the conditions (2) and (3) above are also straightforward:
- Condition (2): Note that $\varphi_t|_{M_{n-1}} = [f_n(x)]_t|_{M_{n-1}} = \textrm{id}_{M_{n-1}}$ (which is condition (1) for $f_n$). Since $\widetilde{\varphi}$ extends $\varphi|_{M_n \times I}$, we also have that $\widetilde{\varphi}_t|_{M_{n-1}} = \textrm{id}_{M_{n-1}}$. We calculate
$$ [H_n(x,s)]_t|_{M_{n-1}} = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t|_{M_{n-1}} = (\widetilde{\varphi}_{\min\{s,t\}})^{-1}|_{M_{n-1}} = \textrm{id}_{M_{n-1}} $$
- Condition (3): Fix $l\in {\mathbb N}$ with $l\geq n$. We know that $\varphi(M_l\times I)\subset \textrm{int}\ M_{l+1}$ (either by condition (3) above if $n > 0$, or by the last condition for the submanifolds $M_i$ if $n=0$). Hence
$$ [H_n(x,s)]_t (M_{l}) = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t (M_l) \subset (\widetilde{\varphi}_{\min\{s,t\}})^{-1} (\textrm{int}\ M_{l+1}) = \textrm{int}\ M_{l+1} $$
where the last equality is a consequence of $\widetilde{\varphi}_u$ being a diffeomorphism which is the identity outside $\textrm{int}\ M_{n+1}$ (and $l\geq n$).
This completes the inductive construction of $H_n : f_n \simeq f_{n+1}$.
To finish the proof, consider the infinite concatenation of all the homotopies $H_n$:
$$ H = H_0 \ast ( H_1 \ast ( H_2 \ast ( H_3 \ast \cdots ) ) ) $$
i.e. $H$ is:
- $H_0$ at twice the speed on the interval $[0,\frac 1 2]$;
- $H_1$ at four times the speed on the interval $[\frac 1 2,\frac 3 4]$;
- $H_2$ at eight times the speed on the interval $[\frac 3 4,\frac 7 8]$;
- in general, $H_n$ at $2^{n+1}$ times the speed on the interval $[\frac{2^n-1}{2^n},\frac{2^{n+1}-1}{2^{n+1}}]$.
Finally, we declare $H(-,1)=b$. Then $H$ is the desired homotopy $H : f \simeq b$. The fact that $H$ is continuous at time $1$ is a consequence of:
- most importantly, the above condition (2) for the homotopies $H_k$;
- the submanifolds $M_i$ form a compact exhaustion of $M$ (their union is $M$, and $M_i \subset \textrm{int}\ M_i$);
- together with the fact that we are considering the compact-open $C^1$ topology on the space of embeddings $\textrm{Emb}(M,M)$.
This concludes the proof.
Best Answer
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.