ScalingStacks

Proof. Consider an isotopy that brings the tangle diagram $T$ into almost braid closure form $T^{\prime}$ and denote its image under the Khovanov invariant as $\phi$. Furthermore, let the maps associated to the sweep-around for $T$ and $T^{\prime}$ be denoted by $\mathrm{sw}_{T}$ and $\mathrm{sw}_{T^{\prime}}$ respectively. Now, note that $\mathrm{sw}_{T}\simeq\phi^{-1}\circ\mathrm{sw}_{T^{\prime}}\circ\phi$ because the underlying link cobordisms are isotopic in ${\mathbb{R}}^{3}\times{[0,1]}$. By assumption $\mathrm{sw}_{T^{\prime}}\simeq\textrm{id}_{T^{\prime}}$ and thus also $\mathrm{sw}_{T}\simeq\textrm{id}_{T}$. ∎