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.
Now comes the answer. For $M$ a non-compact manifold without boundary, it seems the homotopy fibre of $\iota:\text{Diff}(M)\to\text{Emb}(M,M)$ at $\text{id}_M$ is weakly contractible. As hinted implicitly by Tom Goodwillie in his answer, it is not hard to conclude that all the homotopy fibres of $\iota$ are then either empty or weakly contractible.
Before I continue, I will clarify the structure of the homotopy fibre $F$. As a set, $F$ consists of the isotopies $(\varphi_t:M\to M)_{t\in I}$ through embeddings $M\to M$ for which $\varphi_0 = \text{id}_M$, and $\varphi_1$ is a diffeomorphism of $M$. An isotopy $(\varphi_t)_{t\in I}$ can also be written as a map $\varphi: M\times I\to M$. The topology on $F$ makes it a subspace of the space of paths in $\text{Emb}(M,M)$, where $\text{Emb}(M,M)$ has the compact-open $C^1$ topology.
Further, let me state that I will need a strong form of the isotopy extension lemma: when $X$ is a manifold without boundary, and $Y$ is a compact manifold, the map
$$ (\text{eval},\text{proj}):\text{Diff}_c(X)\times\text{Emb}(Y,X) \longrightarrow \text{Emb}(Y,X)\times\text{Emb}(Y,X) $$
is a fibration, where $\text{Diff}_c(X)$ denotes the space of compactly supported diffeomorphisms of $X$. In fact, the above map is a locally trivial fibre bundle, as one can give local trivializations using tubular neighborhoods; thus the above map is 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 sketch the argument that the homotopy fibre $F$ of $\iota:\text{Diff}(M)\to\text{Emb}(M,M)$ at $\text{id}_M$ is weakly contractible. The argument below 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 $b=(\text{id}_M)_{t\in I}$ of $F$. First we construct an exhaustion of $M$ by compact submanifolds $M_i$ for $i\in {\mathbb N}$ such that:
- $\bigcup_{i\in {\mathbb N}} M_i = M$;
- $M_0=\emptyset$;
- $M_i \subset \text{int } M_{i+1}$ for all $i\in {\mathbb N}$;
- $\varphi(M_i \times I)\subset \text{int } M_{i+1}$ for all $\varphi$ in the image of $f$.
For example, we may use a proper smooth function $\rho: M\to [0,+\infty)$ and take $M_i=\rho^{-1}(x_i)$ for a suitable 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 I use compactness of $K$, 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 for all $k < n$ we have constructed $f_{k+1}:K\to F$ and a homotopy $H_k : f_k \simeq f_{k+1}$. Furthermore, assume that for all $\varphi$ in the image of $H_k: K\times I\to F$, and for all $\psi$ in the image of $f_{k+1}:K\to F$ the following conditions hold:
- $\psi_t|_{M_k} = \text{id}_{M_k}$ for all $t\in I$;
$\varphi_t|_{M_{k-1}} = \text{id}_{M_{k-1}}$ for all $t\in I$;- $\varphi(M_l \times I) \subset \text{int } M_{l+1}$ for each $l\in {\mathbb N}$.
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 $\varphi(M_n \times I) \subset \text{int } M_{n+1}$. Therefore, $\varphi|_{M_n \times I}$ is an isotopy through embeddings $M_n \to \text{int } M_{n+1}$. Since $\varphi_0 = \text{id}_M$, the isotopy extension lemma implies that $\varphi|_{M_n \times I}$ extends to a compactly supported diffeotopy $\widetilde{\varphi}=g(x): \text{int } M_{n+1}\times I \to \text{int } M_{n+1}$ with $\widetilde{\varphi}_0=\text{id}_{\text{int } M_{n+1}}$.
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
$$ \begin{matrix}
K & \xrightarrow{\ (a,b)\ } & \text{Diff}_c(\text{int } M_{n+1})\times\text{Emb}(M_n,\text{int } M_{n+1}) \\
\llap{\scriptstyle \text{incl}_0}\Big\downarrow & &\Big\downarrow\rlap{\scriptstyle (\text{eval},\text{proj})} \\
K\times I & \xrightarrow[\ (c,d)\ ]{} & \text{Emb}(M_n,\text{int } M_{n+1})\times\text{Emb}(M_n,\text{int } M_{n+1})
\end{matrix} $$
where the components of the horizontal maps are defined by
$$ a(x) = \text{id}_{\text{int } M_{n+1}} \qquad b(x) = [f_n(x)]_0 = ( \text{incl}:M_n\hookrightarrow \text{int } M_{n+1} ) $$
$$ c(x,t) = [f_n(x)]_t|_{M_n} \qquad d(x,t) = b(x) = ( \text{incl}:M_n\hookrightarrow \text{int } M_{n+1} ) $$
(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
$$ (e,d) : K\times I \to \text{Diff}_c(\text{int } M_{n+1})\times\text{Emb}(M_n,\text{int } M_{n+1}) $$
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 $M_n \times I$;
$\widetilde{\varphi}_0=[g(x)]_0=\text{id}_{\text{int } M_{n+1}}$.
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 $\text{int } M_{n+1}$ (depending continuously on $x\in K$) which extends $\varphi|_{M_n \times I}$ and such that $\widetilde{\varphi}_0 = \text{id}_{\text{int } M_{n+1}}$. Finally, extend $\widetilde{\varphi} = g(x)$ to be the identity outside of $\text{int } M_{n+1}$ --- note that this operation preserves continuity on $x\in K$. Then $\widetilde{\varphi} = g(x)$ becomes a compactly supported diffeomorphism of the whole manifold $M$ depending continuously on $x\in K$.
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 $\varphi|_{M_n \times I}$.
We further define the homotopy $H_n : K\times I \to F$ between $f_n$ and $f_{n+1}$ as follows:
$$ [H_n(x,s)]_t = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t = ([g(x)]_{\min\{s,t\}})^{-1} \circ [f_n(x)]_t $$
It is easy to check that $H_n$ takes values in $F$: $[H_n(x,s)]_0 = (\widetilde{\varphi}_0)^{-1} \circ \varphi_0 = \text{id}_M$, 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
$\varphi_t|_{M_{n-1}} = [f_n(x)]_t|_{M_{n-1}} = \text{id}_{M_{n-1}}$ (which is condition (1) for $f_n$). Since $\widetilde{\varphi}$ extends $\varphi|_{M_n \times I}$, we also have that $\widetilde{\varphi}_t|_{M_{n-1}} = \text{id}_{M_{n-1}}$. We calculate
$$ [H_n(x,s)]_t|_{M_{n-1}} = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t|_{M_{n-1}} = (\widetilde{\varphi}_{\min\{s,t\}})^{-1}|_{M_{n-1}} = \text{id}_{M_{n-1}} $$
- Condition (3): Fix $l\in {\mathbb N}$. When $l < n$, condition (2) implies condition (3). If $l\geq n$, we know that $\varphi(M_l\times I)\subset \text{int}(M_{l+1})$ (either by condition (3) above if $n > 0$, or by the last condition for the submanifolds $M_i$ if $n=0$). Hence
$$ [H_n(x,s)]_t (M_{l}) = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t (M_l) \subset (\widetilde{\varphi}_{\min\{s,t\}})^{-1} (\text{int } M_{l+1}) = \text{int } M_{l+1} $$
where the last equality is a consequence of $\widetilde{\varphi}_u$ being a diffeomorphism which is the identity outside $\text{int } M_{n+1}$ (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 from $0$ to $\frac 1 2$;
- $H_1$ at four times the speed from $\frac 1 2$ to $\frac 3 4$;
- $H_2$ at eight times the speed from $\frac 3 4$ to $\frac 7 8$;
- in general, $H_n$ at $2^{n+1}$ times the speed from $\frac{2^n-1}{2^n}$ to $\frac{2^{n+1}-1}{2^{n+1}}$.
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 the above condition (2) for the homotopies $H_k$, together with the fact that we are considering the compact-open $C^1$ topology on the space of embeddings $\text{Emb}(M,M)$. This concludes the proof.
