Wirtinger presentations and knot groups

For each $p\in K$, [F2] supplies a local inverse neighborhood for $E$. Form the set of all tuples consisting of a local trivialization, an open base set $O$ containing its center, and a radius $r>0$ such that $O\times \{v:\mathrm{\parallel }v\mathrm{\parallel }<r\}$ lies in that injectivity neighborhood. These base sets cover $K$. Compactness gives finitely many such tuples whose base sets $O_i$ cover $K$; their recorded radii have a positive finite minimum $r_0$. The tuples are collected as a set of all witnesses and only a finite subfamily is extracted, so no point-indexed choice is used.

Apply [F3] to the finite cover $\{O_i\}$ of the metric space $K$, obtaining $\delta >0$. Choose $0<\epsilon <\min \{r_0,\delta \mathrm{/}3\}$. If $E(p,v)=E(q,w)$ with $\mathrm{\parallel }v\mathrm{\parallel },\mathrm{\parallel }w\mathrm{\parallel }<\epsilon$, then $\mathrm{\parallel }p-q\mathrm{\parallel }=\mathrm{\parallel }w-v\mathrm{\parallel }<2\epsilon <\delta$. Thus $\{p,q\}$ lies in one $O_i$, and both $(p,v)$ and $(q,w)$ lie in the same recorded neighborhood where $E$ is injective; hence $(p,v)=(q,w)$. Every point of the radius-$\epsilon$ tube also lies in one of those local-diffeomorphism neighborhoods, so the restriction is an injective local diffeomorphism.

An injective local diffeomorphism is open and has smooth local inverses that agree on overlaps. Therefore $E$ is a diffeomorphism onto its open image $N_{\epsilon }$, which contains $K$ by [F1].

For $x=E(p,v)\in N_{\epsilon }\setminus K$, define $H_t(x)=E\text{\hspace{0.17em}⁣}(p,\frac{(1-t)\mathrm{\parallel }v\mathrm{\parallel }+t\epsilon }{\mathrm{\parallel }v\mathrm{\parallel }}v),0\le t\le 1,$ and set $H_t(x)=x$ for $x\mathrm{\notin }{N}_{\epsilon }$. The normal coordinates are unique by step 3.1, and the formula approaches the identity at $\mathrm{\parallel }v\mathrm{\parallel }=\epsilon$, so these clauses paste continuously. It fixes the exterior pointwise and at $t=1$ sends every point in the punctured tube to its radius-$\epsilon$ boundary. Thus it is a deformation retraction of ${\mathbb{R}}^3\setminus K$ onto ${\mathbb{R}}^3\setminus N_{\epsilon }$. Only finitely many local neighborhoods and a finite minimum were used.

Cut out disjoint small underpass intervals, one in each crossing box. The remaining projected arcs are embedded and pairwise disjoint: an interior double point would have an underpassing preimage, which has been removed. Put these arcs at height $1$, their ends in the plane $z=0$, and join each pair of ends of a removed underpass below that plane. This normalization preserves the complement pair. Indeed, inside a crossing box the two branches retain a strict height inequality while their heights interpolate; away from these boxes projection is injective and vertical interpolation cannot introduce intersections. A change of height $d(q)$ on a strand extends to a local ambient map $(q,z)\mapsto (q,z+\chi (z)d(q))$, subdividing the interpolation so that $1+{\chi }^{\mathrm{\prime }}(z)d(q)>0$. Finitely many disjoint strip charts, smooth cutoffs and parameter subdivisions suffice. Planar bends can likewise be straightened in their disjoint strip charts. All these maps are the identity in a common high collar.

For $c=0$, first compare the actual smooth knot with a polygon, rather than identifying embeddings merely from the code. Its projection is a smooth embedded circle $\gamma$, so the knot is $(\gamma (s),g(s))$. Apply the finite compact-tube construction [F2] to the flat curve $(\gamma (s),0)$: the planar slice of its normal tube is an injective strip $\mathrm{\Psi }(s,r)=\gamma (s)+rn(s)$, with $n$ the positive unit planar normal. In this strip the ambient map $(\mathrm{\Psi }(s,r),z)\mapsto (\mathrm{\Psi }(s,r),z-t\chi (r)g(s))$, with $\chi (0)=1$ and $\chi$ zero near the strip boundary, is a homeomorphism for each $t$ (it is a translation on every vertical fibre), and is the identity outside the strip. It flattens the actual knot. Uniform continuity of the tangent and the finite tube bounds then give a sufficiently fine inscribed polygon: each chord stays in the strip and its projection to the circle is strictly increasing, since its tangent has positive scalar product with the nearby oriented circle tangents. The polygonal projection has one full turn, so it is bijective, and the polygon is the normal graph $r=f(s)$. Choose the subdivision so $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\chi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$. The fibre maps $r\mapsto r+t\chi (r)f(s)$ are strictly increasing homeomorphisms fixing the strip boundary; extended by the identity, they give an ambient planar isotopy carrying $\gamma$ to this polygon. Its product with the identity in $z$ is an ambient three-dimensional isotopy. All subdivisions and bounds use a finite cover and one sufficiently large subdivision number. This relates the actual embedding to the polygon without any later diagram-code supplier.

Here is a precise convenient tunnel model for the normalized knot. In each crossing box choose a small convex polygonal footprint $F_i$ and a convex lower cap $C_i=\{(q,z):q\in F_i,h_i(q)\le z\le 0\},$ where $h_i$ is piecewise linear, negative inside $F_i$ and zero on its boundary. Take the lower arc to be the two straight segments from its two interior top-face endpoints to one interior point of $C_i$. This replacement of the underpass is realized by vertical interpolation of its negative graph, fixing its endpoints. Each $C_i$ is a product neighborhood of that arc: radial coordinates from the interior vertex identify $C_i$ with a ball; on its boundary send two disjoint disks around the endpoints of a diameter to small disks around the two top endpoints, and extend radially. Such a boundary map is constructed by joining the disk centers by simple polygonal arcs, taking rectangular strips around those arcs, and using affine maps on a finite common triangulation; outside the strips it is the identity. Thus it identifies a standard cylinder core with the two-segment arc and its two end disks with the designated exits. Attach product tubes along the disjoint upper arcs to those exits. The resulting neighborhood $N_{\ast }$ has one cap and one upper tube per underpass, joined in the same single cycle as $K$. The disk fibers on the caps and tubes use the specified exit-disk maps as transition maps, so they fit along the finite cyclic core to give a disk bundle over the circle; this verifies the tube structure rather than assuming an arbitrary regular neighborhood. Corners and the core bends can be rounded inside disjoint smaller charts; retain the polygonal cap boundary, which is only used topologically.

The polygon's bounded side is supplied by [F6]. Triangulate it by diagonals: at a convex vertex use its neighbor triangle if it contains no other vertex; otherwise join it to a vertex in that triangle farthest from the opposite side. The smaller open triangle contains no intervening vertex, so the diagonal is internal; induction on the vertex count yields a finite triangulation. A leaf of its triangle-adjacency tree is an ear $uvw$, with ear tip $v$ and base $uw$. Remove any collinear vertices before this step. Put $q$ in the interior of $uw$. Choose $z_{+}=q+(1+\eta )(v-q)$ and $z_{-}=q-ϵ(v-q)$, with small positive $\eta ,ϵ$, so the convex quadrilateral $Q=(u,z_{+},w,z_{-})$ meets the rest of the polygon only at $u,w$. Such bounds exist because the closed ear has no other polygon points, nonadjacent edges have positive separation, and the two remaining incident edges have a positive angle away from the ear; finite determinant and distance comparisons select the bounds. Both $v$ and $q$ are interior to $Q$. Triangulate $Q$ by the fan from $v$. On each fan triangle map its apex $v$ to $v_t=(1-t)v+tq$, fix its two boundary vertices, and interpolate affinely. Since $Q$ is convex and $v_t$ is always interior, the image triangles are a nondegenerate fan partition of the same quadrilateral. The resulting map is therefore injective, surjective and piecewise linear, fixes the whole boundary, and extends by the identity to an ambient plane isotopy. At $t=1$ it sends the two ear edges $uv,vw$ onto $uq,qw$, while fixing all other polygon edges. This is an ambient straightening, not a collapse of a triangle to its base. Inductively remove the ears until a triangle remains. From an interior point of that triangle its boundary has a continuous positive radial-distance function $R(\theta )$; the plane homeomorphism multiplying each ray's radius by $1\mathrm{/}R(\theta )$ sends it to a round circle. The products of these plane maps with the identity in $z$ identify the actual knot complement with the round-circle complement.

For $c>0$, transport the rounded product disk-bundle neighborhood $N_{\ast }$ of step 2.1 back through the ambient normalization in step 1.1 and use it as $N(K)$. In each disk fiber of this tube write the normalized radial coordinate as $r\in [0,1]$. The homotopy $r\mapsto (1-s)r+s$ fixes the fiber boundary and sends every nonzero fiber point to that boundary at $s=1$; extend it by the identity outside the tube. The formulas agree continuously at $r=1$ and fix the target pointwise, so they give a strong deformation retraction of ${\mathbb{R}}^3\setminus K$ onto $E_K$. For $c=0$, the same radial formula in the Euclidean normal coordinates of [F2] gives the stated retraction.

Take a large ball $B_R$ containing $N_{\ast }$. Outside it, the homotopy $x\mapsto ((1-s)+sR\mathrm{/}\mathrm{\mid }x\mathrm{\mid })x$, with the interior fixed, retracts the exterior to its truncation $X$ in $B_R$. Its outer sphere is retained. Write $A=X\cap \{z\ge 0\}$. Below the plane, the exterior is exactly $B=\{(q,z):-\sqrt{R^2-\mathrm{\mid }q{\mathrm{\mid }}^2}\le z\le h(q)\},$ where $h=h_i$ on $F_i$ and is zero elsewhere; enlarge $R$ so every cap lies above the lower hemisphere. The top plates of the caps belong to $A$ as tube-boundary faces, except for the two exits occupied by upper tubes. Thus $X=A\cup B$ and $A\cap B$ is the equatorial disk with the interiors of the $F_i$ removed. On $B$ use $(q,z)\mapsto (q,(1-s)z+sh(q))$ and on $A$ the identity. These maps agree on the overlap, remain in the indicated vertical intervals, and paste continuously because the two sets are closed in $X$. They fix their target $Y_{\ast }=A\cup {\bigcup }_i{D}_i,D_i=\{(q,h_i(q)):q\in F_i\},$ pointwise. This proves the global strong deformation retraction, including the lower region and infinity.

The upper region $A$ is a ball with the neighborhoods of $c$ disjoint boundary-parallel arcs removed. The parallelism disks are the vertical curtains beneath their disjoint projected arcs. Their strip neighborhoods are disjoint. Cutting these strips off the upper half-ball leaves a ball with finitely many boundary indentations: flatten each indentation in its rectangular strip coordinate, fixing the other faces. The strip containing an arc is a ball with a straight diameter tube removed. In cylindrical coordinates this last space is $S^1\times \{(r,z):r\ge ϵ,r^2+z^2\le 1\},$ whose second factor is a convex disk. Hence $A$ is a boundary connected sum of $c$ solid tori, glued through specified disks to the remaining ball. For an actual bouquet spine, cut each solid torus across a meridian disk and absorb its remaining ball into the central ball. The result is a central ball with $c$ product one-handles. Join the centers of the $2c$ attaching disks to an interior point by disjoint star arms in that ball; after flattening the finitely many boundary disks, these are radial arms from their centers to the central point. Add the one-handle core intervals. The resulting embedded graph is a bouquet of $c$ circles, since every vertex other than the central one has degree two. To specify the CW structure for the relative contraction, use a finite polyhedral ball model for the central ball. Triangulate its boundary so that each attaching disk is a subcomplex and its center is a vertex, then cone that boundary triangulation to one interior vertex. Each radial star arm is an edge of this cone triangulation. The union $C$ of the attaching disks and these star edges is a finite CW subcomplex. It is contractible by contracting each disk to its center and then the star to the interior vertex. The central ball $B_0$ and $C$ are both nonempty and contractible, so the inclusion $C\hookrightarrow B_0$ is a homotopy equivalence. By [F3], $B_0$ strongly deformation retracts onto $C$, fixing $C$ pointwise; HEP alone is not being used to assert a contraction. Next contract every attaching disk linearly to its center, and contract the transverse disk factor of each product one-handle by the same linear factor. These formulas agree on the end disks and fix the star edges and handle cores, so they paste to a strong deformation retraction onto the stated bouquet. For each drilled arc, orient its graph loop to represent the based positive meridian. In an oriented normal-disk chart along the oriented knot tangent, choose the disk orientation so that tangent followed by disk orientation agrees with the ambient orientation, and traverse the disk boundary in its induced positive direction. In the complement product $S^1\times D^2$ of the drilled tube, this boundary is the $S^1$ factor; contracting the disk factor and using the retained collar access path carries it to the corresponding bouquet loop. Thus the chosen generator is the positive meridian, not its inverse. Each such loop crosses precisely its own curtain once. Consequently [F5] identifies these positively oriented meridians as the free basis, rather than merely normal generators.

Fix disk coordinates ${\varphi }_i:{\bar{D}}^2\to D_i$ and constants $0<r_0<r_1<1$. In $Y_{\ast }$ set $V_i={\varphi }_i(\{\mathrm{\mid }v\mathrm{\mid }<r_1\})$ and let $U$ be the complement of the union of the closed disks ${\varphi }_i(\{\mathrm{\mid }v\mathrm{\mid }\le r_0\})$. These sets are relative-open: the cap interiors have their ordinary disk topology, their closures away from their boundary are disjoint from $A$, and the omitted closed disks are closed in $Y_{\ast }$. They cover because $r_0<r_1$. Each $V_i$ is an open disk, different $V_i$ are disjoint, and their overlaps with $U$ are exactly the annuli $r_0<\mathrm{\mid }v\mathrm{\mid }<r_1$. On each outer collar, increasing its radius by $r\mapsto (1-s)r+s$ retracts $U$ to $A$, fixing $A$. Step 5.1 supplies its meridian basis. Transfer these sets and the composed homotopies back through the finite normalizations in steps 1.1, 2.1 and 3.1. They give the stated subspace and cover in the original exterior.

For the resulting round circle use $S^3\subset {\mathbb{C}}^2$, $K=\{z_1=0\}$, and infinity $p=(1,0)$. Choose the angular coordinate so that increasing $t$ gives the positively oriented meridian (replace $t$ by $-t$ if the carried knot orientation requires it). The map $(w,e^{it})⟼(\sqrt{1-\mathrm{\mid }w{\mathrm{\mid }}^2}{e}^{it},w)$ identifies $S^3\setminus K$ with $D^{\circ }\times S^1$, and $p$ with $(0,1)$. Choose a center-preserving homeomorphism from $D^{\circ }$ to $(-1,1{)}^2$. For $q\in (-1,1{)}^2$, put $r=\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}$ and define $c(q)=\{\begin{array}{ll}q,& r\le \frac{1}{2},\\ q\mathrm{/}(2r),& r>\frac{1}{2}\mathrm{.}\end{array}$ The homotopy $C_s(q)=(1-s)q+sc(q)$ stays in the open square, fixes $[-\frac{1}{2},\frac{1}{2}{]}^2$, and preserves $q=0$. Therefore $(q,z)\mapsto (C_s(q),z)$ is a strong deformation retraction of $(D^{\circ }\times S^1)\setminus \{(0,1)\}$ onto $([-1\mathrm{/}2,1\mathrm{/}2{]}^2\times S^1)\setminus \{(0,1)\}$: when $z=1$, every point in the punctured fiber has $q\mathrm{\ne }0$, and $C_s(q)$ is a positive scalar multiple of $q$, so the homotopy never reaches the puncture. Rescaling the square gives $([-1,1{]}^2\times S^1)\setminus \{(0,1)\}$. Write $S^1=[-1,1]\mathrm{/}(-1\sim 1)$ with the puncture at circle coordinate $0$. For $v\in [-1,1{]}^3\setminus \{0\}$, the homotopy $H_s(v)=(1-s+\frac{s}{\mathrm{\parallel }v{\mathrm{\parallel }}_{\mathrm{\infty }}})v$ retracts onto the cube boundary, fixes both identified end faces pointwise, and therefore descends to the product quotient. Its image $X_0$ is a finite CW complex: the lateral faces form the torus $T=\mathrm{\partial }[-1,1{]}^2\times S^1$, and the two square end faces become one disk $D^2$ attached along $\mathrm{\ell }=\mathrm{\partial }[-1,1{]}^2\times \{1\}$. Give $T$ its product CW structure with $\mathrm{\ell }$ as a subcomplex; then $D^2$ is a contractible CW subcomplex of $X_0$. By [F4], collapsing it is a homotopy equivalence, and the quotient is $T$ with $\mathrm{\ell }$ collapsed. Cutting $T$ along $\mathrm{\ell }$ gives a cylinder; after the two boundary copies of $\mathrm{\ell }$ are collapsed to their common point, the quotient is a sphere with its two poles identified. To identify its homotopy type explicitly, let $Z$ be $S^2$ with an interval attached at its two distinct poles. Give $Z$ the finite CW structure in which the interval is a 1-cell and its endpoints are 0-cells, so the interval is a contractible subcomplex. Collapsing it identifies the poles, so [F4] makes $Z$ homotopy equivalent to the sphere-with-poles-identified quotient. The two endpoint maps of the interval into $S^2$ are homotopic along a fixed path; choose this path to be the $e^{it}$ meridian cut at $t=0$. By the attaching-map clause of [F3], $Z$ is homotopy equivalent to the adjunction space with both endpoints attached at one pole, namely $S^2\vee S^1$, and the new circle is the retained positive meridian. For this wedge with wedge point $w$, choose $q\in S^2\setminus \{w\}$ and $r\in S^1\setminus \{w\}$, and set $A=S^1\cup (S^2\setminus \{q\}),B^{\mathrm{\prime }}=(S^1\setminus \{r\})\cup S^2\mathrm{.}$ These are open in the wedge topology and cover $S^1\vee S^2$. The first deformation retracts to $S^1$, the second to $S^2$, so both are path-connected; their overlap $(S^1\setminus \{r\})\cup (S^2\setminus \{q\})$ is nonempty and contracts to $w$ by contracting each punctured summand to $w$. Thus the sets and overlap satisfy the path-connected open-cover hypotheses of van Kampen [F5], which gives ${\pi }_1(S^1\vee S^2,w)\cong {\pi }_1(S^1,w)\ast {\pi }_1(S^2,w)\cong \mathbb{Z}\ast 1\cong \mathbb{Z}$, generated by the positive meridian circle. Finally step 3.1 gives the asserted retraction onto $E_K$ in both cases.

Take ambient coordinates $(x,y,z)$ with overstrand tangent $+{\mathrm{\partial }}_x$, viewing normal $+{\mathrm{\partial }}_z$, and positive projection basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$. Source has $y<0$ and target $y>0$, with the understrand below the overstrand. Parameterize the under-core from source to target by $u$, whether or not this agrees with the knot orientation. Choose its product tube $\mathrm{\Phi }(u,v)$ so that the oriented knot tangent followed by the ordered normal frame is ambient-positive; take the first normal-frame vector toward the positive-$z$ side of the understrand and the second to complete that orientation. Thus increasing angle from ${\theta }_0=0$ gives the positive meridian even when $u$ runs opposite to the knot orientation. Choose a small positive radius $r$. The path used here is the push-off $\beta (u)=\mathrm{\Phi }(u,r(\cos {\theta }_0,\sin {\theta }_0))$, from source to target; it never lies on the knot core. The two meridian loops are the boundaries ${\mu }_u(\tau )=\mathrm{\Phi }(u,r(\cos ({\theta }_0+2\pi \tau ),\sin ({\theta }_0+2\pi \tau )))$, based at the endpoints of $\beta$. For exterior paths take $r$ at least the removed tube radius; shrink that radius first if necessary. Choose the product tube disjoint from the overstrand and other arc pieces, and $r$ smaller than the crossing-chart separation bounds. Choose upper-collar paths ${\alpha }_s,{\alpha }_t$ to these actual boundary points. In the normal $(y,z)$ plane of the overstrand, their concatenation ${\alpha }_s\beta {\alpha }_t^{-1}$ follows a small perturbation of the polygon $(0,H),(-d,-h),(d,-h),(0,H)$, which has positive signed area. The polygon and a sufficiently small surrounding strip avoid the origin; the chosen radius makes the perturbation lie in that strip, so its winding is still $+1$. For explicit complement control, straighten the local under-core to $x=0,z=-h$ in the product chart and take its upper push-off at $z=-h+r$. The access triangle is then entirely above $z=-h$, while the over-core projects to the origin. Radial interpolation of its positive polar-distance function to a sufficiently small circle about the origin stays inside that triangle minus the origin, hence above the under-core as well. Retain the positive-radius angular access in the under-tube until reaching these straightened paths. Thus the resulting homotopy is to one counterclockwise over-meridian and avoids both strands throughout. Thus all paths and homotopies lie in the complement, the loop represents $m_o$, and ${\alpha }_t\simeq m_o^{-1}{\alpha }_s\beta$ rel endpoints there.

The positive-radius cylinder $(u,\tau )\mapsto \mathrm{\Phi }(u,r(\cos ({\theta }_0+2\pi \tau ),\sin ({\theta }_0+2\pi \tau )))$ transports the boundary meridians, with its basepoint edge equal to $\beta$. Its image misses the knot core and, by the separated tube choice, the overstrand and every other arc. Reading the boundary of this cylinder gives $\beta {\mu }_t{\beta }^{-1}\simeq {\mu }_s$ in the complement; its disk orientation is transported continuously even when the source-to-target direction opposes the knot orientation. Therefore $[{\alpha }_t{\mu }_t{\alpha }_t^{-1}]=m_o^{-1}[{\alpha }_s\beta {\mu }_t{\beta }^{-1}{\alpha }_s^{-1}]m_o=m_o^{-1}{m}_s{m}_o$. Equivalently the annular-core word $m_o^{-1}{m}_s{m}_o{m}_t^{-1}$ bounds the cap of [F1], consistently with [F3]. Reversing the understrand orientation inverts both its meridians, so inversion gives the same side-based equation, rather than a switch of the conjugating direction.

The under-tangent follows $n_o$ exactly when incoming is source and outgoing is target, giving the first incoming/outgoing equation. Otherwise solve the same equation for $m_s$ to obtain the second. A reversed meridian is the inverse loop by [F2]. If a common access path $\alpha$ is replaced by $\lambda \alpha$, the class $[\alpha \mu {\alpha }^{-1}]$ becomes $[\lambda ][\alpha \mu {\alpha }^{-1}][\lambda {]}^{-1}$; this holds for each of the three meridians. These observations prove all remaining assertions.

Since $K$ is compact, choose $R>0$ with $K$ and $b$ in the open ball $D=B_R(0)$. Put $U=S^3\setminus (K\cup \{\mathrm{\infty }\})={\mathbb{R}}^3\setminus K$ and $V=S^3\setminus \bar{D}$. Then $V$ is an open ball about infinity, disjoint from $K$, and $U\cap V=V\setminus \{\mathrm{\infty }\}$. First verify that $U$ is path-connected without assuming a diagram code. Parameterize the smooth embedded circle by $k:S^1\to K$. For $p\in U$, the cone $C_p=\{p+t(k(s)-p):s\in S^1,t\ge 1\}$ has three-dimensional outer volume zero: on each of finitely many smooth coordinate intervals of $S^1$ and each bounded interval $1\le t\le n$, the defining map is Lipschitz on a compact rectangle; subdividing into $O(N^2)$ squares of side $1\mathrm{/}N$ covers its image by balls of radius $O(1\mathrm{/}N)$, whose total three-dimensional volume is $O(1\mathrm{/}N)$. Thus each truncated image has outer volume zero. For each $n$, the finite chart and derivative bounds give an explicit cover volume $A_n\mathrm{/}N$ after mesh size $1\mathrm{/}N$; choosing $N_n>2^{n+2}{A}_n\mathrm{/}ϵ$ makes the $n$th truncated images of both cones have total cover volume below $ϵ2^{-n-1}$. These formula-defined mesh sizes give a cover of $C_p\cup C_q$ with total volume below $ϵ$ for every $ϵ>0$, so the union has outer volume zero without a choice of covers. An open ball contains a cube of positive volume, so choose $r$ in an open ball outside $C_p\cup C_q$. If the segment $[p,r]$ met $K$ at $k(s)$, then $r=p+t(k(s)-p)$ for some $t\ge 1$, contrary to $r\mathrm{\notin }{C}_p$; similarly $[q,r]$ misses $K$. The broken line from $p$ to $r$ to $q$ lies in $U$. Under inversion $J(x)=x\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}^2$, extended by $J(\mathrm{\infty })=0$, the set $V$ is the open ball of radius $1\mathrm{/}R$ about $0$. The radial homotopy $y\mapsto (1-t)y$ contracts it to $0$, so it is nonempty and path-connected. Since this ball is convex, every based loop contracts to its own basepoint by the straight-line homotopy inside the ball; [F3] therefore makes $V$ simply connected. The same inversion identifies $U\cap V$ with $0<\mathrm{\parallel }y\mathrm{\parallel }<1\mathrm{/}R$. Fix $\rho =1\mathrm{/}(2R)$; the puncture-preserving radial homotopy $y⟼((1-t)+t\frac{\rho }{\mathrm{\parallel }y\mathrm{\parallel }})y$ is a strong deformation retraction of this punctured ball onto the sphere $S_{\rho }^2$. The punctured ball is path-connected. Choose $y_0\in S_{\rho }^2$, let $i:S_{\rho }^2\hookrightarrow U\cap V$ be inclusion, and let $r$ be the radial retraction. Since the radial deformation fixes $S_{\rho }^2$ pointwise, [F7] and functoriality make $i_{\ast }$ and $r_{\ast }$ inverse isomorphisms at $y_0$. By [F2], ${\pi }_1(S_{\rho }^2,y_0)$ is trivial. For any other basepoint $y$ in the punctured ball, choose a path $\gamma$ from $y$ to $y_0$ and use $[\alpha ]\mapsto [\bar{\gamma }\ast \alpha \ast \gamma ]$. Concatenation and reparameterization preserve endpoint-fixed homotopies, so this is a well-defined homomorphism. The inverse uses $\bar{\gamma }$; the inserted pairs $\gamma \ast \bar{\gamma }$ and $\bar{\gamma }\ast \gamma$ contract by the piecewise retracing homotopy, so the homomorphism is an isomorphism. Thus every basepoint group of the overlap is trivial; its nonemptiness and path-connectedness with [F3] make $U\cap V$ simply connected.

Choose $b_0\in U\cap V$. The open sets $U,V$ cover $S^3\setminus K$; [F4] applies by the path-connectedness in step 1.1. Since $V$ and $U\cap V$ are simply connected by step 1.1, the special case [F5] shows that inclusion induces an isomorphism ${\pi }_1(U,b_0)\to {\pi }_1(S^3\setminus K,b_0)$.

The inclusion induces the based map by [F6]. Choose a path $\gamma$ in $U$ from $b$ to $b_0$, which exists by step 1.1. The formula $[\alpha ]\mapsto [\bar{\gamma }\ast \alpha \ast \gamma ]$ defines isomorphisms between the fundamental groups at these basepoints by the concatenation and retracing argument in step 1.1. Using the same path in $S^3\setminus K$ makes the square with inclusion commute, since postcomposing the concatenated path gives the same loop. Thus the inclusion map at $b$ is conjugate by isomorphisms to the isomorphism at $b_0$.

The input code has finite vertex, edge, segment and crossing lists, so the rational-coordinate, incidence, cyclic-order and starting-arc checks are finite. Each pairwise segment test uses terminating rational determinant, intersection and endpoint-order operations. A failed validation returns invalid-input. If validation succeeds, the one-component tracing follows the finite cyclic list of edges and paired half-edges from $a_0$, returns to $a_0$ after finitely many steps, and labels each maximal under-arc once. The crossing pass then visits the finite crossing list once, reads each source and target from its verified cyclic order, appends exactly the corresponding positive-meridian relator, and returns the finite presentation. Thus the procedure terminates with one generator per under-arc and one relator per crossing, as claimed. If $c=0$, the explicit punctured-product model in [F1] gives its fundamental group as infinite cyclic on the positive meridian (the sphere factor is simply connected), while [F6]–[F8] identify the returned one-generator empty-relator presentation with the same free cyclic group. If $c>0$, [F1] and [F5] give the free meridian basis for the meridian region.

Assume $c>0$ and order the crossings $v_1,\dots ,v_c$. Let $b_{\ast }$ be the common collar basepoint in $U$, and let $X_0=U$ and $X_i=U\cup V_1\cup \cdots \cup V_i$. For each $i$, the relative-open cover $X_i=X_{i-1}\cup V_i$ has intersection $A_i=U\cap V_i$, the annulus specified in [F1]. Choose $q_i$ on its core and use the supplied collar access path ${\gamma }_i$ in $U$ from $b_{\ast }$ to $q_i$. Concatenation and reparameterization make $[\alpha ]\mapsto [{\bar{\gamma }}_i\ast \alpha \ast {\gamma }_i]$ a well-defined homomorphism from loops based at $b_{\ast }$ to loops based at $q_i$. Its inverse uses ${\bar{\gamma }}_i$, since the inserted path-reversal pairs contract by the piecewise retracing homotopy. The same concatenation formula commutes with inclusion. Apply [F3] at $q_i$: since $V_i$ is contractible, [F4] identifies ${\pi }_1(X_i,q_i)$ with ${\pi }_1(X_{i-1},q_i)$ modulo the normal closure of the annular-core image. Compute that image in ${\pi }_1(U,b_{\ast })$, before attaching any crossing piece. In the cap model of [F1], push its boundary to the top plate at $z=0$; deleting the two exit disks makes this plate a disk with two holes. Let ${\mu }_s,{\mu }_t$ be the exit meridians, and let $\eta$ run from their basepoints across the plate between the exits, below the over-arc. The oriented knot enters one exit and leaves the other, so their positive meridians have opposite planar orientations. Cutting the two-holed plate along access arcs reads its outer boundary, up to orientation and basing conjugacy, as ${\mu }_s\eta {\mu }_t^{-1}{\eta }^{-1}$. All these paths lie in $A\subset U$. For the common collar paths ${\alpha }_s,{\alpha }_t$, the loop ${\alpha }_s\eta {\alpha }_t^{-1}$ represents $m_o$ in $U$: in the over-normal $(y,z)$ plane it runs from the high collar down the source side, across below the over-arc, and back up the target side, winding once positively. Its straightening to the over-meridian stays above the plate and avoids both exit tubes and the over-tube. This is the same local path calculation as [F2], performed entirely in the upper region, without using an attached cap. Therefore the outer-boundary image is, up to conjugacy and inversion, $m_s{m}_o{m}_t^{-1}{m}_o^{-1}=m_o(m_o^{-1}{m}_s{m}_o{m}_t^{-1})m_o^{-1}$. Its normal closure is exactly that of the constructor's relator, whether or not the understrand orientation follows the source-to-target direction. This computes the attaching word in the free group, rather than inferring it from an equation in the final complement. Distinct $V_i$ are disjoint, so these are precisely the intersections at every stage. Repeating over the finite crossing list kills exactly the normal closure of the constructor's relators.

By [F6] and [F7], the quotient in step 2.1 is the returned presentation group; [F8] identifies the quotient map with the homomorphism sending each symbol to its positive meridian. It is an isomorphism onto the group of the covered exterior by the pushout construction.

For $c=0$, step 1.1 already identifies the returned cyclic presentation with the Euclidean complement group on the positive meridian. For $c>0$, step 3.1 identifies the presentation with the covered exterior group, and the compact-tube retraction identifies that group with ${\pi }_1({\mathbb{R}}^3\setminus K)$. In both cases [F9] then gives the isomorphism to ${\pi }_1(S^3\setminus K)$, with exactly the stated positive-meridian assignment.

By [F2], replacing the common basing path conjugates every generator image by the same element. Simultaneous conjugation is an inner automorphism of the target group, so the resulting presentation group remains isomorphic to ${\pi }_1(S^3\setminus K)$.

Changing the designated start arc reindexes the same finite list of under-arcs. The induced bijection of generator symbols carries each crossing's over/under labels, and therefore each constructor relator, to the corresponding relator in the new indexing. It extends to an isomorphism of the free groups and carries one relator normal closure onto the other; [F7] and [F8] give an isomorphism of the presented quotients.

The eight stated routes give 24 segments, 16 degree-two ports and four crossing vertices. Here is an exact reproducible intersection test for the displayed integers. For segments $a+tu,b+sv$, with $u=a^{\mathrm{\prime }}-a,v=b^{\mathrm{\prime }}-b$ and $\mathrm{\Delta }=\det (u,v)\mathrm{\ne }0$, their intersection parameters are $t=\det (b-a,v)\mathrm{/}\mathrm{\Delta },s=\det (b-a,u)\mathrm{/}\mathrm{\Delta }\mathrm{.}$ Retain the intersection just when $0\le t,s\le 1$. If $\mathrm{\Delta }=0$, the test $\det (b-a,u)=0$ detects collinearity, and comparison of the two closed endpoint intervals in either nonconstant coordinate detects overlap. Applying these formulas to the table gives 276 unordered segment pairs: precisely 40 meet, each solely at its common declared endpoint, and the other 236 are disjoint. The 40 consist of one pair at each of the 16 ports and six pairs at each of the four crossing centers. No pair overlaps in positive length. At each center the $Oi,C,Oo$ and $Ui,C,Uo$ triples are collinear with the center strictly between the ports. The determinants of the full oriented over/under vectors are respectively $-69696,3168,3168,-69696$, so those are transverse opposite pairs with exactly the stated cyclic orders. The cyclic visit list uses every branch once, and following the paired branches gives that single list, not several circles. Hence every validity condition in [F1] holds. Finitely many nonincident compact segments have positive distance, so sufficiently small disjoint bend disks introduce no new crossing; over/under height separation in the crossing disks realizes the smooth knot.

The geometric figure-eight admission can be checked without inferring knot type from a group presentation. Draw the downward three-strand braid with successive crossings ${\sigma }_1,{\sigma }_2^{-1},{\sigma }_1,{\sigma }_2^{-1}$, where a positive generator has its left strand over, and close corresponding endpoints outside the braid rectangle without additional crossings. Number its successive crossings $1,2,4,3$. Following its first top strand through the closure gives exactly $1O,2U,3O,1U,4O,3U,2O,4U$. In right/up plane coordinates its positive-generator over and under tangents point respectively down-right and down-left, giving negative determinant; the negative generator reverses that sign. Thus the four labeled determinants are $-,+,+,-$. Together with the visit list these determine precisely the cyclic orders in the statement. For an additional check, the six face boundary cycles, listing each departure port and then taking the next counterclockwise port after traversing its edge, are $(1Oo,2Oo,4Oi),(1Ui,3Uo,2Ui),(1Oi,4Oo,3Oo),(1Uo,4Uo),(2Uo,3Ui,4Ui),(2Oi,3Oi)\mathrm{.}$ These are the same two bigons and four triangular faces of that closed braid drawing; the table merely subdivides its eight edges at their ports. To see that this comparison concerns embeddings, take small crossing disks and narrow edge rectangles in each drawing. The identical cyclic orders identify these neighborhoods respecting over/under pairs. The six face boundaries are simple polygonal circles. Their disk coordinates can be obtained by the elementary finite polygon construction: use the vertical-slab convex-cell triangulation of [F6] and peel boundary triangles to identify the filled polygon with a disk. At each peeling step a triangle with one boundary edge and interior opposite vertex leaves a disk; if all vertices of the remaining triangulation are on its boundary, the dual is a finite tree and a leaf triangle has two boundary edges. Reversing these elementary triangle attachments gives disk coordinates with the prescribed boundary parametrization. Matching the resulting boundary circle coordinates extends over the disk by $(r,z)\mapsto (r,h(z))$. Applying this to each paired face extends the neighborhood identification over every face; the exterior is treated as a disk on the sphere. After rounding, this is the same regular spherical diagram, and choosing the same outer face gives the planar diagram. The closed braid just described is the standard two-clasp four-crossing figure-eight diagram (equivalently its usual alternating closed-three-braid description). This verifies admission by the full embedded diagram, not solely its under-arc relators.

Cutting the oriented visit circle at its four $U$ visits gives the four claimed under-arcs. The intervening $O$ visits put $O1$ on $x_1$, $O3$ on $x_2$, $O4$ on $x_3$, and $O2$ on $x_4$. For a direct side check, let $o_r=rOo-rOi$. The values of $\det (o_r,rUi-C_r)$ are $15840,-2376,-2376,15840$; those of $\det (o_r,rUo-C_r)$ are $-53856,792,792,-53856$. Positive values are the left side. This proves every entry of the crossing table, including which underpass follows the positive normal. By [F2] and [F3], $x_o^{-1}{x}_R{x}_o{x}_L^{-1}=1$ is equivalently $x_o{x}_L=x_R{x}_o$. Substitution of the four rows gives exactly $R_1,R_2,R_3,R_4$ as displayed. By [F4] these are the actual complement group and positive meridians. Starting at $x_1$ the underpasses are $R_2,R_1,R_3,R_4$ and the corresponding crossing signs sum to $1-1+1-1=0$.