Observe however that a prebraiding is not (!) required to satisfy an analog of the braid relation (a.k.a. third Reidemeister move) of the form \[\begin{gathered} b_{F(z),F(y),F(x)}\circ (\beta_{y,z}\boxtimes \mathrm{id})\circ b^{-1}_{F(y),F(z),F(x)}\circ(\mathrm{id}\boxtimes \beta_{x,z}) \circ b_{F(y),F(x),F(z)}\circ (\beta_{x,y}\boxtimes \mathrm{id})\\ = (\mathrm{id}\boxtimes \beta_{x,y})\circ b_{F(z),F(x),F(y)}\circ(\beta_{x,z}\boxtimes \mathrm{id}) \circ b^{-1}_{F(x),F(z),F(y)}\circ (\mathrm{id}\boxtimes \beta_{y,z})\circ b_{F(x),F(y),F(z)} \end{gathered}\]
Original source: arXiv:2401.02956v2
Original source · 2401.02956v2