ScalingStacks

Proof. The proof would be easy if we knew that $f$ were isotopic to the identity, but $\pi_{0}(\mathrm{Diff}^{+}({\mathbb{R}}^{3}\times{[0,1]},{\mathbb{R}}^{3}\times\{0,1\}))$ is unknown. We can, however, replace $f$ with a diffeomorphism $f^{\prime}$ which is isotopic (rel boundary) to $f$, or replace $f$ with a diffeomorphism $f^{\prime}$ which coincides with $f$ in a neighborhood of $\Sigma$. In both cases, proving that $f^{\prime}(\Sigma)$ is isotopic to $\Sigma$ easily implies that $f(\Sigma)$ is isotopic to $\Sigma$.

Choose a point $p\in{\mathbb{R}}^{3}$ such that $p\times{[0,1]}$ is disjoint from $\Sigma$. There is no obstruction to modifying (post-composing) $f$ by an isotopy which takes $f(p\times{[0,1]})$ to $p\times{[0,1]}$, so we may assume that $f$ restricts to the identity on $p\times{[0,1]}$. (Note that this modification changes $f(\Sigma)$ as well as $f(p\times{[0,1]})$, so there is no issue of the image of $p\times{[0,1]}$ getting “caught" on the image of $\Sigma$.)

Next consider the tangent map of $f$ along $p\times{[0,1]}$. We would like to deform the tangent map to the identity, but there is an obstruction living in $\pi_{1}(SO(3))\cong{\mathbb{Z}}/2$. We can modify (precompose) $f$ in a neighborhood of $p\times{[0,1]}$ (and away from $\Sigma$) so that this obstruction vanishes. (Specifically, let $f:[0,1]\to[0,1]$ be a smooth function such that $f(t)=0$ for $t$ near 0 and $f(t)=1$ for $t$ near 1. Let $\gamma:[0,1]\to SO(3)$ be a representative of the nontrivial element of $\pi_{1}(SO(3))$, with $\gamma(0)=\gamma(1)=\mathbf{1}$. Let $B^{3}$ be the unit ball in ${\mathbb{R}}^{3}$, and for $p\in B^{3}$, let $|p|$ denote the distance from $p$ to the origin. Define a diffeomorphism of $[0,1]\times B^{3}$ by

$$(s,\>p)\mapsto(s,\>\gamma(f(s\cdot(1-|p|)))(p)).$$

This diffeomorphism is the identity near $[0,1]\times\partial B^{3}$ and it effects a full twist on the tangent space along $[0,1]\times\{0\}$.)

Once the above obstruction vanishes we can isotope $f$ to a map which is the identity on a neighborhood $N$ of $p\times{[0,1]}\cup{\mathbb{R}}^{3}\times\{0,1\}$.

Choose a family of diffeomorphisms $g_{t}:{\mathbb{R}}^{3}\times{[0,1]}\to{\mathbb{R}}^{3}\times{[0,1]}$, with $t\in{[0,1]}$, such that $g_{0}$ is the identity, $g_{t}$ restricted to ${\mathbb{R}}^{3}\times\{0,1\}$ is the identity for all $t$, and $g_{1}(\Sigma)\subset N$. The family of surfaces $f(g_{t}(\Sigma))$ provides an isotopy from $f(\Sigma)=f(g_{0}(\Sigma))$ to $f(g_{1}(\Sigma))$. But $f$ is the identity on $N$ and $g_{1}(\Sigma)\subset N$, so $f(g_{1}(\Sigma))=g_{1}(\Sigma)$. The family of surfaces $g_{t}(\Sigma)$ provides an isotopy from $g_{1}(\Sigma)$ to $g_{0}(\Sigma)=\Sigma$. Composing these two isotopies provides the desired isotopy from $f(\Sigma)$ to $\Sigma$. ∎