A crossing-ball cover of a knot complement

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.