We can embed $S^2\times I$ into $\mathbb{R}^3$ by taking a compact 3-ball and removing an open 3-ball from its interior. Taking the boundary gives an embedding $i: S^2\sqcup S^2\hookrightarrow\mathbb{R}^3$, as "a sphere contained inside another sphere". Now it's intuitively clear that this embedding is not ambient-isotopic to the embedding $j$ given by putting these two spheres "side-by-side" in $\mathbb{R}^3$. That is, there is no isotopy $F\_t:\mathbb{R}^3\to \mathbb{R}^3$ with $F\_0=1\_{\mathbb{R}^3}$ and $F\_1\circ i=j$. At least, this looks visually obvious. Assuming my intuition isn't betraying me and this isn't false, what's an elegant way to prove this? 


Also, (why) does there (not) exist an embedding $S^2\times I\hookrightarrow\mathbb{R}^3$ whose boundary *is* ambient-isotopic to the "side-by-side" embedding? How about for $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?



Freeing a sphere from within a sphere
**For your first question**, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "[knot group](http://en.wikipedia.org/wiki/Knot_group)"):


Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.


$\Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$\Big)$


**For your third (and second) question**, in generality, here's an argument that there is no embedding from $M=S^{n-1}\times I$ to $\mathbb{R}^n$ with boundary ambient isotopic to the "side-by-side" embedding of   
$S\_0\sqcup S\_1$. By any letter $S$ I denote a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:


**(1)** Up to ambient isotopy, there are only three embeddings of $S\_0\sqcup S\_1$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S\_0$ inside $S\_1$, the "avocado$^-$ type" embeddings $A^-$ with $S\_1$ inside $S\_0$, and the "side-by-side type" embeddings $B$ with $S\_0$ and $S\_1$ "next to" each other. 


$\Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S\_1$ embeddings match up, and then being in the same case means the two $S\_2's$ are embedding in the same component of the compliment of $S\_1$, so (b) you can move them around in that component until they match up, too.$\Big)$


**(2)** $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.


**(3)** An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy. 


$\Big($ Why? Here's one proof, which I found kind of fun: Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial M$ must lie outside both spheres of $\partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S\_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of   
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S\_1$ or $S\_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$\Big)$


**(4)** By process of elimination, $\partial M$ must be ambient isotopic to $A^\pm$, which by (2) is *not* ambient isotopic to the "side-by-side" embedding $B$, which is what you wanted.


