Kauffman bracket R2 and R3 identities

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. For the opposite crossing, rotating the source chart by a positive quarter-turn exchanges the two endpoint pairings, so its A/B smoothings are respectively $e_i$ and $I$. By [F1] these expansions are $C_i=AI+A^{-1}{e}_i$ and $C_i^{-}=A^{-1}I+A{e}_i$. Using [F2] and [F3], $C_i{C}_i^{-}=(AI+A^{-1}{e}_i)(A^{-1}I+A{e}_i)=I+(A^2+A^{-2}+\delta )e_i=I\mathrm{.}$ The same multiplication in the other order gives $C_i^{-}{C}_i=I$, since both factors are polynomials in $e_i$. Thus the two opposite crossings in an R2 pair cancel in either order.

Put $x=e_i$ and $y=e_{i+1}$. Expanding and using [F3] gives $C_i{C}_{i+1}{C}_i=A^{3}I+A(2x+y)+A^{-1}(xy+yx+\delta x)+A^{-3}x,$ while interchanging $x$ and $y$ gives $C_{i+1}{C}_i{C}_{i+1}=A^{3}I+A(2y+x)+A^{-1}(yx+xy+\delta y)+A^{-3}y\mathrm{.}$ Their difference is $A^{-1}(A^2+\delta +A^{-2})(x-y)=0$. Hence the two sides of the braid-like R3 move have equal bracket operators.

Near the three-strand event, choose a linear longitudinal coordinate nonzero on each of the three tangent lines, shrink the event disk, and orient the unoriented local tracks in its increasing direction. All three tracks are then monotone; these temporary arrows do not affect the bracket. Label the tracks $1,2,3$ at the incoming end. The two R3 sides meet their pairs in orders $(12,13,23)$ and $(23,13,12)$, and a strict height order permits exactly the exponent triples $+++,---,++-,-++,+--,--+$. The triples $+-+$ and $-+-$ would require a cyclic height order. Put $a=C_i$ and $b=C_{i+1}$. Besides $aba=bab$ from step 1.2 and its inverse from step 1.1, the four needed identities are $ab{a}^{-1}=b^{-1}ab,a^{-1}ba=ba{b}^{-1},a{b}^{-1}{a}^{-1}=b^{-1}{a}^{-1}b,a^{-1}{b}^{-1}a=b{a}^{-1}{b}^{-1}\mathrm{.}$ The first follows by multiplying $aba=bab$ on the left by $b^{-1}$ and on the right by $a^{-1}$; the second follows by multiplying on the left by $a^{-1}$ and on the right by $b^{-1}$; the last two are their inverses. Thus each of the six height orders gives equality with exactly its actual over/under choices. For R2, choose a longitudinal coordinate along the common tangent at the bigon event, shrink the disk, and temporarily orient both tracks increasingly; the two local crossings have opposite types and cancel by step 1.1. Reversing any original strand arrow leaves all smoothing weights unchanged. Adjoining the unchanged exterior tangle and closing its boundary preserves these equalities: expand each crossing into its two smoothings, use bilinearity, and evaluate each resulting crossingless closed diagram with one factor $\delta$ per circle as in [F2]. This gives equal sums ${\sum }_S{A}^{a(S)-b(S)}{\delta }^{\mathrm{\mid }S\mathrm{\mid }}$ on the two sides. Every complete smoothing of the nonempty diagram has $\mathrm{\mid }S\mathrm{\mid }\ge 1$, and $\delta =-A^2-A^{-2}$ is nonzero in the integral domain $\mathbb{Z}[A^{\pm 1}]$; hence these sums are $\delta ⟨D⟩$ and $\delta ⟨D^{\mathrm{\prime }}⟩$. Cancelling $\delta$ proves $⟨D⟩=⟨D^{\mathrm{\prime }}⟩$, so every R2/R3 variant preserves the bracket.