ScalingStacks

Proof. These facts follow from [CS93, CRS97]. ∎

Proof. Lemma 4.1 (1) implies that we have such parallel transport maps $\mathrm{KhR}_{N}(\phi_{t})$ between the fibers over any pair of points $\phi_{0}$ and $\phi_{1}$ in the base. Note that $\pi_{1}(M(B),\phi_{0})\cong\pi_{1}(SO(3))\cong{\mathbb{Z}}/2{\mathbb{Z}}$ is generated by the class of the loop obtained from $\phi_{0}$ by rotating through $360$ degrees around the $z$-axis, to which $\mathrm{KhR}_{N}$ assigns the identity map, see also Section 6.4. Lemma 4.1 (2) and Theorem 2.4 thus imply that the parallel transport maps between the fibers do not depend on the choice of the path $\phi_{t}$. ∎

Proof. We first show independence of $\phi$, given a fixed choice of $\phi_{\mathrm{in}}$ and $\phi_{\mathrm{out}}$. Suppose that $\phi^{\prime}\colon W\to{\mathbb{R}}^{3}\times{[0,1]}$ is another diffeomorphism restricting to $\phi_{\mathrm{in}}$ and $\phi_{\mathrm{out}}$ on $W_{\mathrm{in}}$ and $W_{\mathrm{out}}$ respectively.

Lemma 4.7, proved below, implies that the link cobordisms $\phi(\Sigma)$ and $\phi^{\prime}(\Sigma)$ are isotopic rel boundary in ${\mathbb{R}}^{3}\times{[0,1]}$ and we have $\mathrm{KhR}_{N}(\phi(\Sigma))=\mathrm{KhR}_{N}(\phi^{\prime}(\Sigma))$ by Theorem 2.4.

Next we show independence of $\phi_{\mathrm{in}}$, given a fixed choice of $\phi_{\mathrm{out}}$. Let $\phi^{\prime}_{\mathrm{in}}\colon W_{\mathrm{in}}\to{\mathbb{R}}^{3}$ be another diffeomorphism such that $\phi^{\prime}_{\mathrm{in}}(\Sigma_{\mathrm{in}})$ is generic and blackboard-framed, and $\phi^{\prime}$ another diffeomorphism $W\to{\mathbb{R}}^{3}\times{[0,1]}$ restricting to $\phi^{\prime}_{\mathrm{in}}$ on $W_{\mathrm{in}}$ but still to $\phi_{\mathrm{out}}$ on $W_{\mathrm{out}}$. Then, by Lemma 4.1 (1), we can find a family $\phi_{\mathrm{in},t}$ connecting $\phi_{\mathrm{in}}$ to $\phi^{\prime}_{\mathrm{in}}$. By definition of parallel transport, we have:

Now we obtain a new diffeomorphism $\phi^{\prime}\circ\phi_{\mathrm{in},t}\colon W\to{\mathbb{R}}^{3}\times{[0,1]}$ and by the previous independence result and Theorem 2.4, we have:

Thus, the definition was independent of the choice of $\phi_{\mathrm{in}}$. An analogous argument also establishes independence of the choice of $\phi_{\mathrm{out}}$. ∎

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

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 $\Sigma$. ∎