# homotopy type of embeddings versus diffeomorphisms

Previously, I asked a question on mathoverflow comparing smooth embeddings and diffeomorphisms, which received a very interesting and somewhat unexpected answer by Agol. I now ask a further question about the relation between the homotopy type of embedding spaces and diffeomorphism spaces.

Assume $M$ is a non-compact smooth manifold without boundary (although I would also be interested in the case where $M$ is compact with non-empty boundary). I would like to find an explicit counter-example --- or a proof --- for the following statement: The inclusion $\iota:\text{Diff}(M)\to \text{Emb}(M,M)$ of the space of diffeomorphisms of $M$ into the space of smooth self-embeddings is a weak equivalence on each component of $\text{Diff}(M)$. In other words, every homotopy fibre of the inclusion $\iota$ is either empty or weakly contractible. I am considering the compact-open $C^1$-topology on the above spaces.

Obviously, I would also be very interested in any known results relating the homotopy type of $\text{Diff}(M)$ with that of $\text{Emb}(M,M)$.

Edit: Tom Goodwillie has provided an answer for the case when $M$ is compact or the interior of a compact manifold. I placed a bounty for the case in which $M$ is open (and not necessarily the interior of a compact manifold).

Edit 2: I managed to adapt Agol's nice idea from my previous question to apparently resolve the present question. Since nobody has raised any major issues so far, and people are upvoting my answer, I will likely accept it. As a curious aside, I edited my answer so many times --- out of persnicketiness --- that it turned into community wiki. I was unaware that would happen. Oh well...

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FYI, if the bounty falls through, the highest-voted answer will automatically be accepted. –  HJRW Jan 28 '13 at 15:20
I have a hunch that Agol's method for answering that other question can be refined to answer this one. –  Tom Goodwillie Jan 28 '13 at 17:42

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:

1. $\phi_t|_{M_k} = \textrm{id}_{M_k}$ for all $t\in I$;
2. $\psi_t|_{M_{k-1}} = \textrm{id}_{M_{k-1}}$ for all $t\in I$;
3. $\psi(M_l \times I) \subset \textrm{int}\ M_{l+1}$ 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.

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EDIT Ricardo has fully answered his own question, so I think people should upvote his answer to give him the bounty.

When $M$ is compact with nonempty boundary the answer is in general no. I think always no if the dimension is not too small. Fix an embedding $M\to M$ isotopic to the identity whose image, say $M_0$, is the complement of a collar. The restriction map $Emb(M,M)\to Emb(M_0,int M)$ is a homotopy equivalence, but the fiber of the restriction map $Diff(M)\to Emb(M_0,int M)$ (over the image of the identity element) is essentially what is called the pseudoisotopy space of the boundary $B=\partial M$, namely the space of diffeomorphisms $B\times I\to B\times I$ fixing $B\times 0$ pointwise. The latter is generally noncontractible. For example, if $M$ is the closed $n$-disk, so that $B$ is the $(n-1)$-sphere, and $n$ is big enough, then this pseudoisotopy space has an element of infinite order in $\pi_3$ related to the algebraic $K$-group $K_5(\mathbb Z)$.

When $M$ is the interior of a compact manifold then the answer is yes. Reasoning much as above, you are led to the space of all diffeomorphisms $B\times [0,\infty)\to B\times [0,\infty)$ fixing $B\times 0$ pointwise (where $B$ is the boundary of the compact manifold whose interior is $M$), which is contractible by pushing to infinity, i.e. by conjugating by translations in the $\mathbb R$ direction.

I don't know about cases where there is no collar at infinity like that.

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Thank you very much, Tom. There is an interesting relation between your answer and Agol's answer to my previous question. Perhaps one can merge the two ideas (yours and Agol's) to do the open case? Seems hard... By the way, by "a swindle" do you mean "pushing off to infinity"? –  Ricardo Andrade Jan 23 '13 at 6:08
Yes, I've rewritten it. –  Tom Goodwillie Jan 28 '13 at 13:07
@Tom: Thanks for your endorsement. I am happy you found no major issues with my answer. Actually, out of pickiness, I edited my answer so many times it turned into community wiki. –  Ricardo Andrade Jan 30 '13 at 19:42
By the way, it seems that the person placing the bounty cannot claim it. If the person who placed the bounty also gets the accepted answer, then the bounty is discarded. –  Ricardo Andrade Jan 31 '13 at 0:03
I can't honestly say that I read your answer through all that carefully, although I plan to. I'd like to really understand this. –  Tom Goodwillie Jan 31 '13 at 1:02