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I would like to find an example, if one exists, of manifolds $M$ and $N$ with embeddings $f:M\to N$ and $g:N\to M$ such that $f\circ g$ and $g\circ f$ are both isotopic (i.e. homotopic through embeddings) to the respective identities, yet the interiors of $M$ and $N$ are not diffeomorphic. Obviously, $M$ and $N$ cannot be closed. You may assume that the manifolds have no boundary, but I would also be interested in compact examples.

By the way: in the other direction, are there conditions under which $M$ and $N$ are necessarily diffeomorphic if $f$ and $g$ as above exist?

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Would small exotic $\mathbb R^4$s work? –  Zack Jun 24 '12 at 2:33
Does your edit mean you invite examples with boundary? A closed interval and an open interval. For a compact example, take manifolds that are $h$-cobordant but not diffeomorphic. Crossing with an open interval gives diffeomorphic manifolds and crossing with a closed interval gives an example of your property. –  Ben Wieland Jun 24 '12 at 18:05
@Lee: The image of $Whitehead\subset Ball\subset Whitehead$ doesn't contain any nonlocal knots. –  Zack Jun 24 '12 at 22:40
A knot in a $3$-manifold is called nonlocal if it is not contained in any closed ball. An open $3$-manifold containing no nonlocal knots is homeomorphic to $\Bbb R^3$, ams.org/journals/proc/1971-028-01/S0002-9939-1971-0271919-1 so Zack's remark suffices to answer the 3-dimensional case of the question. To see an explicit nonlocal knot in the Whitehead manifold $W$ it suffices to know that the $n$th Whitehead link is nontrivial (and this is how Whitehead originally proved that $W\not\cong\Bbb R^3$). –  Sergey Melikhov Jun 26 '12 at 22:19
Ricardo: by the isotopy extension theorem (see Agol's answer). A non-local knot in $W$ is smoothly isotopic to a knot in $h(W)\subset B\subset W$; by the isotopy extension theorem this isotopy is covered by a smooth isotopy of $W$. Less trivially, Smythe proved that a knot in a $3$-manifold, PL isotopic to the unknot (by a possibly non-locally-flat PL isotopy), is contained in a ball (see Lemma 2.2 in arxiv.org/abs/math.GT/0103114 ). By the way, I of course meant to say that Zack's remark suffices to answer the case $M=\Bbb R^3$ of your question, not the full 3D case. –  Sergey Melikhov Jun 27 '12 at 2:16
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up vote 23 down vote accepted

Here I'll prove that $M$ and $N$ must be diffeomorphic under your hypotheses (I'll assume that $M$ and $N$ are open, i.e. no boundary).

Consider the direct limit (see below) $X= M \overset{f}{\to} N \overset{g}{\to} M \overset{f}{\to} N \overset{g}{\to}\cdots$. By this, I mean the union $M \times \mathbb{N} \sqcup N\times \mathbb{N}$ modulo the identifications $(m,k)\sim (f(m),k), (n,k)\sim (g(n),k+1)$, for $m\in M, n\in N, k\in \mathbb{N}$. Then $X$ is a smooth manifold, since it is a nested union of smooth manifolds. Similarly, we have the direct limit $Y=N \overset{g}{\to} M \overset{f}{\to} N \overset{g}{\to}\cdots$. Clearly, $X \cong Y$, since the direct limit depends only on the tail of the sequence. Also, $X$ (and therefore $Y$) is diffeomorphic to the direct limits $ M \overset{g\circ f}{\to} M \overset{g\circ f}{\to} M \overset{g\circ f}{\to}\cdots$ and $N\overset{f\circ g}{\to} N\overset{f\circ g}{\to} N\overset{f\circ g}{\to}\cdots$.

Claim: If $F:M\to M$ is isotopic to the identity, then the direct limit $M\overset{F}{\to} M\overset{F}{\to} M\overset{F}{\to} \cdots \cong M$.

By the above discussion, this implies that $M\cong N$.

To prove the claim, let $F_t:M\to M, t\in [0,1]$ be an isotopy of $F$ to the identity, so $F_1=F$, and $F_0=Id$. Take an exhaustion of $M$ by smooth compact submanifolds with boundary $K_1\subset K_2\subset K_3 \subset \cdots$, so that $F([0,1]\times K_i)\subset int(K_{i+1})$, $K_i\subset int(K_{i+1})$, and $\cup K_i=M$. The we see that $X_t = M\overset{F_t}{\to} M\overset{F_t}{\to}M\overset{F_t}{\to}\cdots$ is diffeomorphic to $Y_t= K_2 \overset{F_t}{\to} K_4 \overset{F_t}{\to} K_6 \overset{F_t}{\to} \cdots$ for all $t$ (where we really meant $F_{t| K_{2i}}$ in the maps of this direct limit and topologized by the inclusion of interiors). To see this, note that $X_t$ is a quotient of $M\times \mathbb{N}$, which is exhausted by $\{(K_k,i), k,i\in\mathbb{N}\}$. Since $Y_t\subset X_t$, we need only show that each point of $X_t$ is in $Y_t$. Suppose $x\in X_t$, then $x$ is the image of some $x'\in (K_k,i)$ for some $k,i$. If $k\leq 2i$, then $x'\in (K_{2i},i)$, so we see that $x\in Y_t$. Otherwise, $k>2i$ and we see that $(K_k,i) \overset{F_t}{\subset}(K_{k+1},i+1) \overset{F_t}{\subset} \cdots \overset{F_t}{\subset} (K_{2k-2i},k-i)$ after taking the quotient, so $x\in Y_t$.

By the Isotopy Extension Theorem, one may see that $Y_0 \cong Y_1$. This gives the desired diffeomorphism $M\cong M\overset{F}{\to} M\overset{F}{\to} M\overset{F}{\to} \cdots $.

We actually use the isotopy extension theorem by induction. We want to find a sequence of diffeomorphisms $G^i: K_{2i}\to K_{2i}$ so that $G^{i+1}_{|K_{2i}} = F\circ G^i$. This immediately gives the diffeomorphism

$$\begin{matrix} Y_1= K_2 & \overset{F}{\to} & K_4 & \overset{F}{\to} & K_6 & \overset{F}{\to} & \cdots\\ G^1 \uparrow & &G^2\uparrow & &G^3\uparrow & & \\ Y_0= K_2 & \hookrightarrow & K_4 & \hookrightarrow & K_6 & \hookrightarrow & \cdots \end{matrix}$$

We will actually construct a diffeotopy $G^i_t:K_{2i}\to K_{2i}, t\in [0,1]$ such that $G^{i+1}_{t|K_{2i}}=F_t\circ G^i_t$, where $G^i_0=Id$, and then set $G^i=G^i_1$ for the desired diffeomorphism.

Let $G^1_t=Id$. Suppose we have constructed $G^i_t$ by induction, so that $G^i_0=Id$, $G^i_{t|K_{2i-2}}=F_t \circ G^{i-1}_t$, $t\in [0,1]$. Then $F_t\circ G^i_t: K_{2i}\to int(K_{2i+2})$ is an isotopy from $F_0\circ G^i_0=Id$ to $F_1\circ G^i_1=F\circ G^i_1$. By the isotopy extension theorem, there exists a diffeotopy $G^{i+1}_t: K_{2i+2}\to K_{2i+2}$ such that $G^{i+1}_0=Id$, and $G^{i+1}_{t|K_{2i}}=F_t\circ G^i_t$, with compact support in $int(K_{2i+2})$. This completes the proof.

Addendum: I'll add some explanation of direct limits. Consider a sequence of smooth embeddings of open manifolds $X_i \overset{f_i}{\to} X_{i+1}$. By the direct limit $X$ of $X_1\overset{f_1}{\to} X_2 \overset{f_2}{\to} X_3 \cdots$, I mean the quotient of $X_1\sqcup X_2 \sqcup X_3 \sqcup \cdots$ with respect to the equivalence relation generated by $x_i \sim f_i(x_i)$ for all $x_i \in X_i, i\in \mathbb{N}$. Then there is a natural embedding of $X_i \hookrightarrow X$ for all $i$, in such a way that $X= \cup_i X_i$, and one gives $X$ the structure of a smooth manifold by taking the atlas of charts generated by the charts of each $X_i$.

The space $X$ is determined by the direct limit of any subsequence $\{ X_{i_j}\}$, with maps $F_j: X_{i_j}\to X_{i_{j+1}}$ defined by $F_j=f_{i_{j+1}-1}\circ \cdots \circ f_{i_j}$. One may therefore verify that two direct limits are diffeomorphic if there are compatible diffeos. between subsequences, which justifies some of the arguments above.

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Do you consider direct limits in the category of topological spaces? If so, I don't quite follow in what sense each of your homeomorphisms between direct limits is supposed to be a diffeomorphism. You can transfer the smooth structure from a finite stage to the limit when they are homeomorphic, but is that finite stage even or odd? –  Sergey Melikhov Jun 26 '12 at 21:46
@Agol: Thanks for the answer. It looks good to me. I will wait a little while to see what others say. It is certainly a beautiful idea. I was trying to make an analogy with the Schroeder-Bernstein theorem but kept coming short because the manifolds were not compact (so I could not apply some isotopy extension). Your method seems to cleverly realize that vague intuition. Just a couple of silly remarks: (1) I think you meant $(K_{2k-2i},k-2i)$ at the end of the fifth paragraph. (2) Perhaps you should redefine $Y_t$ to be the colimit of the interiors of $K_{2i}$, ... –  Ricardo Andrade Jun 27 '12 at 0:09
@ Ricardo: Addressing your remarks (1) I think I meant $k-i$, since it is $i+ k-2i$, where by hypothesis $k-2i>0$ (2) since $K_{2i} \subset int(K_{2i+2})$, I think it doesn't matter, but maybe I should add a remark about this. –  Ian Agol Jun 27 '12 at 0:13
... or at least make it clear that is how you obtain the differential structure on $Y_t$. –  Ricardo Andrade Jun 27 '12 at 0:13
@Agol: You are right about my remark (1)... It seems I cannot count... Sorry. –  Ricardo Andrade Jun 27 '12 at 0:15
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