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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? And (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 $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?

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 $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?