Rack colorings are framed Reidemeister invariants

At an R2 move, hold the overarc color $y$ fixed. The underarc label is changed by $R_y$ at one crossing and by $R_y^{-1}$ at the other, so the two boundary colorings determine one another uniquely. Reversing the move gives the inverse correspondence.

For a braid-like R3 move, use the crossing map $B(x,y)=(y,x\ast y)$. Its inverse is by right invertibility. The two sides send $(x,y,z)$ respectively to $(z,y\ast z,(x\ast y)\ast z)$ and $(z,y\ast z,(x\ast z)\ast (y\ast z))$; the last coordinates agree by rack self-distributivity. For every $z$, $R_z(x\ast y)=(x\ast y)\ast z=(x\ast z)\ast (y\ast z)$, so $R_z$ preserves $\ast$. If , then $u\ast b=a$; applying $R_z$ gives $R_z(u)\ast R_z(b)=R_z(a)$, and uniqueness of right division yields . Thus $R_z$ and its inverse preserve both operations. For either operation ${\star }_1,{\star }_2\in \{\ast ,\bar{\ast }\}$, this gives $(x{\star }_{1}y){\star }_{2}z=(x{\star }_{2}z){\star }_1(y{\star }_{2}z)$, the corresponding mixed R3 equation. The bijective crossing maps then give the extension bijection for every orientation variant. A planar isotopy transports arc labels.

Write the under-tangent as $a{v}_{\mathrm{o}\mathrm{v}\mathrm{e}\mathrm{r}}+b{n}_o$. Its determinant sign is the sign of $b$, so at a positive crossing the oriented underpass goes from source to target. In a positive curl with incoming label $x$ and outgoing label $y$, either the overarc is the incoming arc and $y=x\ast x$, or it is the outgoing arc and $y=x\ast y$. In the latter case self-distributivity gives $R_y{R}_x=R_{x\ast y}{R}_y=R_y{R}_y$, hence $R_x=R_y$; therefore $y=x\ast y=x\ast x=\pi (x)$. Conversely $y=\pi (x)$ solves the equation since $R_{\pi (x)}=R_x$ by the kink-map proof. Thus both positive curl shapes have unique transport $y=\pi (x)$. In a negative curl the same equations are read backwards and give $y={\pi }^{-1}(x)$. This checks both choices of which boundary arc is the overarc, independently of a picture.

For a single unframed curl with boundary color $x$, extension requires $x=\pi (x)$; if $\pi$ is the identity, this condition holds uniquely for every $x$. Conversely, if every boundary color extends uniquely, every $x$ is fixed by $\pi$, so $x\ast x=x$ for all $x$ and the rack is a quandle.

The kink shift propagates through any crossing. Self-distributivity gives $\pi (x\ast y)=(x\ast y)\ast (x\ast y)=(x\ast x)\ast y$, hence $\pi R_y=R_y\pi$. It also gives $(x\ast y)\ast y=(x\ast y)\ast (y\ast y)$ for every $x$. Since $R_y$ is surjective, this implies $R_{\pi (y)}=R_y$. By [F2], $\pi$ is bijective, so these identities hold for its inverse as well: ${\pi }^k{R}_y^{\delta }=R_y^{\delta }{\pi }^k$ and $R_{{\pi }^{\mathrm{\ell }}(y)}^{\delta }=R_y^{\delta }$ for all $k,\mathrm{\ell }\in \mathbb{Z}$ and $\delta \in \{1,-1\}$. In particular, if a crossing obeys $x_t=R_y(x_s)$, independently shifting its two under-colors by the same ${\pi }^k$ and its over-color by ${\pi }^{\mathrm{\ell }}$ preserves it, because $R_{{\pi }^{\mathrm{\ell }}(y)}({\pi }^k{x}_s)={\pi }^k{R}_y(x_s)={\pi }^k{x}_t$. The reversed crossing relation is preserved by the same calculation with $R_y^{-1}$.

A compensated pair creates two curls on the same oriented component, or removes their inverse configuration. If its blackboard framing is preserved, [F4] forces the two created crossing signs to sum to zero; the same holds when opposite signed framing changes are specified directly. Choose either of the two component intervals between the sites and traverse it in the component orientation, writing $ϵ\in \{1,-1\}$ for the sign of its first curl; the second has sign $-ϵ$. Starting from an old coloring, retain every color off this interval and replace every old color on it by ${\pi }^{ϵ}$ of that color, splitting an old under-arc at the two sites if needed. Insert the unique local curl labels from step1.3. At the first site the transport changes the old incoming label to its ${\pi }^{ϵ}$-shift; at the second the transport ${\pi }^{-ϵ}$ restores the old outgoing label. Thus the prescribed labels agree along every new maximal under-arc.

At any old crossing, the selected component interval contains both local halves of the underpassing branch or neither, since its endpoints are away from crossings. Thus the source and target under-colors have the same shift exponent $k\in \{0,ϵ\}$. Independently the overpassing branch has shift exponent $\mathrm{\ell }\in \{0,ϵ\}$. Step1.5 preserves the crossing relation for all four possibilities, including a self-crossing at which the interval contains exactly one of the two visits. The old crossing relations therefore all hold after the recoloring. Conversely, start with any coloring of the twice-curled diagram, undo the ${\pi }^{ϵ}$ shift on that same marked component interval, and remove the two uniquely determined curl portions using step1.3; step1.5 again preserves every intervening relation. The endpoint transports now cancel in the inverse order. These constructions are inverse on every arc label, so give the required bijection for arbitrarily separated opposite curls, not only adjacent curls. The removal move uses the inverse bijection.

Steps 1.1–1.2 give the R2, R3 and planar-isotopy bijections; steps 2.1, 3.1 give the compensation bijection. A finite rack admits only finitely many assignments on a finite diagram, so the cardinalities are preserved. Step 1.4 proves exactly when the individual unframed R1 bijection is available. The algebraic assertions use no choice; only the smooth framing interpretation uses [A1].