TITLE: Freeing a sphere from within a sphere
QUESTION [5 upvotes]: 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 $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?

REPLY [2 votes]: Edit: This is meant to answer the question of why we can't have an embedding $\mathbb S^{n-1}\times I\hookrightarrow\mathbb R^n$ such that the boundary is two side-by-side spheres rather than two nested spheres. I said "the second question" but it changed.
It seems to me that if you have an embedding of $S^{n-1}\times I$ into $\mathbb R^n$, this is the same as an ambient isotopy of one copy of $S^{n-1}$ to another.  (I found this easier to visualize when thinking of the fact that you can't embed a cylinder in $\mathbb R^2$ except as an annulus of concentric circles.)
In particular, you're trying to get one sphere to bound the same disk as the other sphere does, by a homotopy whose image at any time never intersects the image at another time.  You've already noted that two concentric spheres bound the same disk: so you're trying to get the first sphere $A$ to surround the other sphere $B$. Since $B$ cuts off a component of $\mathbb R^n$, we can remove that component and know the isotopy will not pass through it without first passing through $B$, which is not allowed.  Therefore remove a ball $C$ or a point from the interior of $B$.  $A$ is contractible in $\mathbb R^n-C$, and $B$ is not: $B$ generates the $n-1$st homology group of the resulting manifold.  This also is a nice way to see why it will work in $\mathbb R^{n+1}$, though of course in that case you really do have a "cylinder" (take $I$ to run in the $n+1$st dimension with each $S^{n-1}$ in an $n$-hyperplane).