Diagram fundamental quandles and adjoint groups from Wirtinger data

Under R1, the new underarc generator $y$ satisfies either $y=x\ast x$ or $x=y\ast x$ for the unchanged arc color $x$. In the first case idempotency gives $y=x$; in the second, $x=x\ast x$ and right invertibility give the unique solution $y=x$. Thus the new generator is eliminated. Under R2, each added underarc generator is uniquely solved from its neighboring arc and the overarc by $\ast$ or $\bar{\ast }$; the reverse move gives the inverse substitution. Thus these moves are quandle Tietze transformations.

Under R3, assigning boundary colors $x,y,z$ gives the two possible iterated labels $(x\ast y)\ast z$ and $(x\ast z)\ast (y\ast z)$; these agree by right self-distributivity. With the over-colors fixed, each under-color substitution is invertible by right invertibility, so the reverse replacements give inverse substitutions. The same calculation for reversed crossings follows by applying the inverse right translations, which preserve both operations by self-distributivity.

For the knot-group comparison assume $L=K$. Let $\mathrm{\Gamma }$ be the finite planar graph obtained by taking the crossings as vertices and the strand pieces between consecutive crossings as edges. Choose disjoint small disks about the crossing vertices, each meeting $D$ only in its two transverse branches, and disjoint tubular strips around the compact edge pieces outside those disks. At a crossing, oriented local coordinates straighten the branches to two transverse axes. Choose a nearby rational point and two rational direction vectors, still transverse and in the same cyclic order as the original branches; the four rational half-segments give the coded crossing and its over/under choice. Extend each edge strip a short distance into its endpoint disks through the corresponding disjoint branch sectors, and choose rational bend ports there. In each strip choose a simple rational polygonal path that is a graph in strip coordinates and joins its ports; choose it fine enough that rounding its bends gives a small normal graph over the original edge. If there are no crossings, choose a simple rational polygon in a tubular annulus of the projected circle. Since there are finitely many compact edge pieces, nonincident pieces have positive separation, and each crossing is transverse; density of ${\mathbb{Q}}^2$ and a sufficiently fine finite subdivision therefore let all vertices and bends be rational while preserving the one-circle tracing, crossing cyclic orders, orientations, and over/under data, with no other intersections. The resulting record is a valid code by [F4]; smoothing its bends in disjoint disks gives a smooth regular diagram with the same crossing data by [F6]. To compare it with $D$, move each crossing point to its nearby rational point by an isotopy supported in its disk and the short branch sectors, fixed at the outer ends of the strips, and interpolate its four half-branches, within their disjoint sectors, through transverse direction pairs to the rational branches. This is an isotopy preserving the cyclic order and over/under choice. After these local adjustments, each rounded edge replacement is a sufficiently small normal graph $r=f(s)$ over the original edge in its strip, with $f=0$ near the ports. Choose the approximation so $\mathrm{\mid }f\mathrm{\mid }\mathrm{\parallel }{\chi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ for a cutoff $\chi$ that is $1$ near the edge and $0$ near the strip boundary. Then $(s,r)\mapsto (s,r+t\chi (r)f(s))$, $0\le t\le 1$, is a diffeomorphism in each strip, fixed near its boundary, and gives an ambient isotopy of that edge. Compose the finitely many disk and strip isotopies, using smooth time reparameterizations constant near their endpoints, to obtain a smooth plane isotopy $h_t$ supported in a fixed compact set $B$. Realize $D$ by $K\subset {\mathbb{R}}^2\times \mathbb{R}$ and choose $M$ with $K\subset {\mathbb{R}}^2\times (-M,M)$. Choose a smooth compactly supported function $\rho :\mathbb{R}\to [0,1]$ equal to $1$ on $[-M,M]$, and set $H_t(x,z)=(h_{t\rho (z)}(x),z)$. Its smooth inverse is $(x,z)\mapsto (h_{t\rho (z)}^{-1}(x),z)$, and its support lies in the compact set $B\times \mathrm{supp}\rho$. Hence $H_t$ extends by the identity near $\mathrm{\infty }$ to a smooth ambient isotopy of $S^3$. On $K$ it agrees with $(x,z)\mapsto (h_t(x),z)$, so it preserves all crossing heights and carries $K$ to a realization of the smoothed code diagram. Thus the code represents the same oriented knot $K$. The induced bijection on under-arcs preserves source, target, and overarc at each crossing because the positive over-normal and branch orientations are preserved. This finite construction uses no choice axiom.

The conjugation operation on $W(D)$ is idempotent because $g\ast g=g^{-1}gg=g$. Its right translation by $h$ is conjugation by $h^{-1}$, with inverse conjugation by $h$. It is right self-distributive because both $(g\ast h)\ast k$ and $(g\ast k)\ast (h\ast k)$ reduce to $k^{-1}{h}^{-1}ghk$. Thus $\mathrm{Conj}(W(D))$ is a quandle. Its generators $g_a$ satisfy the crossing equations by the relators of $W(D)$, so recursive evaluation of terms gives a quandle homomorphism $f:Q(D)\to \mathrm{Conj}(W(D))$. Define $\psi$ on adjoint generators by $\psi (e_q)=f(q)$. Since $f(q\ast r)=f(r{)}^{-1}f(q)f(r)$, every adjoint relator maps to the identity; [F2] therefore makes $\psi$ a group homomorphism.

For the knot-group comparison define $\varphi :W(D)\to \mathrm{Adj}(Q(D))$ on generators by $\varphi (g_a)=e_{x_a}$. The code from step 1.3 has the same under-arcs and crossing relations as $D$, so the constructor's relators in [F5] are exactly those of $W(D)$. At each crossing, $x_v=x_u\ast x_o$ in $Q(D)$, and the adjoint relation gives $e_{x_v}=e_{x_o}^{-1}{e}_{x_u}{e}_{x_o}$. Thus every relator of $W(D)$ maps to the identity; [F2] makes the homomorphism from the free group on its arc generators descend to $W(D)$.

By [F3], diagrams of $L$ are related by a finite sequence of these moves, and step 1.1 and step 1.2 identify the corresponding quandle presentations at each move. Their composites give an isomorphism between the two diagram quandles. Countable choice is used only for [F3]; these local presentation transformations use no choice.

The composites $\psi \varphi$ fix each $g_a$. The other composite fixes each generator $e_q$: every $q\in Q(D)$ is represented by a term in the arc generators using $\ast$ and $\bar{\ast }$, and the adjoint relations give $e_{u\ast v}=e_v^{-1}{e}_u{e}_v$ and $e_{u\bar{\ast }v}=e_v{e}_u{e}_v^{-1}$, so induction on the term gives $\varphi \psi (e_q)=e_q$. Hence $\varphi$ and $\psi$ are inverse isomorphisms. By [F5] and step 1.3, $W(D)$ is the group of the Wirtinger constructor for a code of $K$, and is isomorphic to ${\pi }_1(S^3\setminus K)$.