0NA5 Corollary 3.4. For every almost braid closure diagram TT, we have swT=𝟏T\mathrm{sw}_{T}=\mathbf{1}_{T}.
0NA6 Proof. We have swT=(sw−)−1∘sw+=(sw−)−1∘sw−=𝟏T\mathrm{sw}_{T}=(\mathrm{sw}_{-})^{-1}\circ\mathrm{sw}_{+}=(\mathrm{sw}_{-})^{-1}\circ\mathrm{sw}_{-}=\mathbf{1}_{T}. ∎