Knot exteriors and peripheral systems

Use an orientation-preserving identification of the standard three-sphere with the unit sphere in ${\mathbb{R}}^4$. Write the knot as $\gamma :S^1\to S^3\subset {\mathbb{R}}^4$ and set $\tau (s)={\gamma }^{\mathrm{\prime }}(s)\mathrm{/}\parallel {\gamma }^{\mathrm{\prime }}(s)\parallel$. Since $\gamma (s)$ is unit and $\gamma$ is an immersion, $\gamma (s)$ and $\tau (s)$ are orthonormal. Put $P_s=\{v\in {\mathbb{R}}^4:⟨v,\gamma (s)⟩=⟨v,\tau (s)⟩=0\}$, oriented by the rule in the statement. Near any $s_0$, project two fixed vectors spanning $P_{s_0}$ onto $P_s$ by $Q_s(v)=v-⟨v,\gamma (s)⟩\gamma (s)-⟨v,\tau (s)⟩\tau (s)$. They remain independent on a small arc; Gram–Schmidt and, if necessary, a sign change give a smooth oriented orthonormal frame $e_1(s),e_2(s)$. Equip the quotient set in [F2] with the charts $[v]\mapsto (s,⟨v,e_1(s)⟩,⟨v,e_2(s)⟩)$ for $v\in T_{\gamma (s)}{S}^3$. These are well-defined bijections on each fibre, with inverse $(s,a,b)\mapsto [a{e}_1(s)+b{e}_2(s)]$; their transition matrices consist of smooth inner products of the local frames. They give the standard quotient smooth structure: in any slice chart, changing the tangent frame to $(\tau ,e_1,e_2)$ is a smooth invertible matrix, and the induced change on the two transverse quotient coordinates is smooth with smooth inverse. Thus $P$ and ${\nu }_K$ are smooth rank-two bundles here, and $v\mapsto v-⟨v,\tau (s)⟩\tau (s)$ on $T_{\gamma (s)}{S}^3$ induces a smooth orientation-preserving isomorphism ${\nu }_K\to P$. This constructs the needed smooth structure directly, without an additional choice assumption.

Cut $S^1$ at one point and fix an oriented local frame on an arc about the cut. Compactness of the remaining interval gives a finite subdivision whose closed subintervals each lie in a bundle-trivializing chart. Inductively extend a frame from one subinterval to the next: on a small overlap, the incoming and chart frames differ by a smooth map into ${\mathrm{G}\mathrm{L}}^{+}(2,\mathbb{R})$. Shrink the overlap so that this map lies in a neighbourhood of its value at the junction where straight interpolation remains in ${\mathrm{G}\mathrm{L}}^{+}$. Then join that constant value to the identity by a path in ${\mathrm{G}\mathrm{L}}^{+}$ and use a smooth cutoff to extend the transition across the new subinterval, making the adjusted frame agree on a collar of the junction and be a fixed chart frame away from it. This can be done at each of the finitely many junctions, with the first and last collars expressed in the fixed cut-arc frame by constant matrices. The required paths exist because Gram–Schmidt writes every matrix in ${\mathrm{G}\mathrm{L}}^{+}(2,\mathbb{R})$ as a rotation times an upper-triangular matrix with positive diagonal; rotations are path-connected and the positive-diagonal triangular matrices contract linearly to the identity. We obtain a smooth oriented interval frame with constant transition matrices at its endpoint collars.

Under the endpoint identification, the two endpoint frame germs differ by a constant matrix $A\in {\mathrm{G}\mathrm{L}}^{+}(2,\mathbb{R})$ by step 1.2. Choose a smooth path $B(t)$ from the identity to $A^{-1}$ that is constant on collars of $0$ and $1$, and multiply the interval frame by $B(t)$. The endpoint values and germs now agree under the identification, because the original germs differ by the constant $A$ and $B$ is constant on their collars. The adjusted frame descends smoothly to an oriented frame on $S^1$, giving the required bundle isomorphism.

By [F1], $K$ is continuous and $K(S^1)$ is compact because $S^1$ is compact. Since $S^3$ is Hausdorff, $K(S^1)$ is closed; [F1] also identifies it as a smooth embedded submanifold. Thus the closed smooth embedded hypothesis of [F4] holds, and [A1] permits its application. Take a tubular map $\mathrm{\Phi }:\mathrm{\Omega }\to U\subset S^3$ defined on an open neighbourhood of the zero section of $\nu (K)$.

Use the trivialization [F3] and its product Euclidean norm as the chosen smooth fibre norm in [F2], so $\mathrm{\Omega }$ is open in $S^1\times {\mathbb{R}}^2$. For each $s\in S^1$, openness gives an arc $I_s$ about $s$ and a radius $r_s>0$ with $I_s\times D_{r_s}^2\subset \mathrm{\Omega }$. Compactness of $S^1$ gives a finite subcover by such arcs; taking $ϵ$ smaller than half the least of their radii yields $S^1\times D_{2ϵ}^2\subset \mathrm{\Omega }$. The closed disk subbundle $S^1\times D_{ϵ}^2$ therefore lies in $\mathrm{\Omega }$.

Put $N(K)=\mathrm{\Phi }(S^1\times D_{ϵ}^2)$. The restricted map is a smooth embedding, so $N(K)$ is a compact embedded solid torus with boundary $T_K=\mathrm{\Phi }(S^1\times \mathrm{\partial }{D}_{ϵ}^2)\cong S^1\times S^1$. Since $S^3$ is compact and $E_K=S^3\setminus \mathrm{int}N(K)$ is closed, $E_K$ is compact. The annulus $S^1\times \{z:ϵ\mathrm{/}2<\mathrm{\mid }z\mathrm{\mid }<2ϵ\}$ maps to a collar of $T_K$; in these coordinates $E_K$ is the side $\mathrm{\mid }z\mathrm{\mid }\ge ϵ$. Thus $E_K$ is a smooth three-manifold with boundary exactly $T_K$, oriented by restriction of the orientation of $S^3$ as in [F2].

On $S^3\setminus K$, define $H_t$ to be the identity outside $N(K)$ and, in the punctured disk fibres with $0<\mathrm{\mid }z\mathrm{\mid }\le ϵ$, set $H_t(\mathrm{\Phi }(s,z))=\mathrm{\Phi }(s,((1-t)\mathrm{\mid }z\mathrm{\mid }+tϵ)z\mathrm{/}\mathrm{\mid }z\mathrm{\mid })$ for $0\le t\le 1$. The formula is continuous for $z\mathrm{\ne }0$ and is the identity at $\mathrm{\mid }z\mathrm{\mid }=ϵ$, so it glues continuously to the identity outside $N(K)$. It fixes every point of $E_K$, stays in $S^3\setminus K$, and at $t=1$ sends every point of the punctured tube to $T_K\subset E_K$. Hence $H_t$ is a strong deformation retraction of $S^3\setminus K$ onto $E_K$.

In product coordinates on $T_K=S^1\times S^1$, the meridian $\mu$ is the second factor loop and the section $\sigma$ is the first factor loop. Give the torus its standard product CW structure: take the quotient of a square by identifying opposite edges, with one vertex, the two oriented factor loops as its $1$-cells, and the open square as its $2$-cell attached by the commutator $\mu \sigma {\mu }^{-1}{\sigma }^{-1}$.

The cellular chain groups in dimensions $2,1,0$ are $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$. The differential $d_1$ is zero because each $1$-cell starts and ends at the only vertex. By [F7], each coefficient of $d_2$ is the degree of the attaching map after collapsing the other $1$-cell. The commutator traverses each factor once in each orientation, so both degrees are zero. Thus by [F5] and [F6], $H_1^{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}(T_K;\mathbb{Z})={\mathbb{Z}}^2$ with the factor loops as basis.

By [F4], the same two classes form a basis of singular $H_1(T_K;\mathbb{Z})$.

Under countable choice [A1], [F5] supplies a product tubular map and a closed product disk tube with the stated norm and radius data for every smooth oriented knot.

Write $D_r^2=\{z\in {\mathbb{R}}^2:\mathrm{\mid }z\mathrm{\mid }\le r\}$ in the product coordinates and set $U=\mathrm{\Phi }(S^1\times \{\mathrm{\mid }z\mathrm{\mid }<3ϵ\mathrm{/}2\})$ and $V=S^3\setminus \mathrm{\Phi }(S^1\times D_{ϵ\mathrm{/}2}^2)$. The set $U$ is open because its source is open in the tubular domain and $\mathrm{\Phi }$ maps that domain diffeomorphically onto an open set. The removed disk image is compact, hence closed in Hausdorff $S^3$, so $V$ is open. They cover $S^3$, and $U\cap V=\mathrm{\Phi }(S^1\times \{ϵ\mathrm{/}2<\mathrm{\mid }z\mathrm{\mid }<3ϵ\mathrm{/}2\})$.

Radial homotopies retract $U$ onto $N(K)$ by replacing each fibre radius $r$ with $(1-t)r+t\min (r,ϵ)$, fixing the zero vector; retract $V$ onto $E_K$ by replacing $r$ with $(1-t)r+t\max (r,ϵ)$ for $ϵ\mathrm{/}2<r\le 2ϵ$ and fixing points outside the closed radius-$2ϵ$ product tube; and retract $U\cap V$ onto $T_K$ by replacing $r$ with $(1-t)r+tϵ$. Each formula remains within its domain, agrees with the identity at the relevant boundary, and fixes its target. Thus these are deformation retractions.

On the overlap, composing its inclusion into $U$ with the first retraction gives radius $\min (r,ϵ)$, while retracting first to $T_K$ and including into $N(K)$ gives radius $ϵ$; linear interpolation stays in $N(K)$. For the inclusion into $V$, the corresponding radii are $\max (r,ϵ)$ and $ϵ$, and their interpolation stays in $E_K$. Hence, after the deformation-retraction identifications, the two maps in Mayer–Vietoris are the maps induced by $T_K\hookrightarrow N(K)$ and $T_K\hookrightarrow E_K$. By [F7], homotopic maps induce the same homomorphism on homology.

By [F9], the Mayer–Vietoris sequence for the open cover of step 1.2 has the segment $H_2(S^3;\mathbb{Z})\to H_1(U\cap V;\mathbb{Z})\to H_1(U;\mathbb{Z})\oplus H_1(V;\mathbb{Z})\to H_1(S^3;\mathbb{Z})\mathrm{.}$ The outer groups vanish by [F6], so the middle map is an isomorphism. Under the identifications of step 3.1, it is induced by the two boundary inclusions.

By [F8], $H_1(T_K;\mathbb{Z})\cong {\mathbb{Z}}^2$ with basis $([\mu ],[\sigma ])$. The product tube $N(K)\cong S^1\times D^2$ retracts to its core $S^1$ by contraction in each disk fibre, so homotopy invariance [F7] and [F6] give $H_1(N(K);\mathbb{Z})\cong \mathbb{Z}$. Put $H=H_1(E_K;\mathbb{Z})$. Steps 2.1, 3.1, 4.1 give ${\mathbb{Z}}^2\cong \mathbb{Z}\oplus H$. The isomorphism embeds $H$ as a subgroup of ${\mathbb{Z}}^2$. Every subgroup of ${\mathbb{Z}}^2$ is free of rank at most two: projection to one coordinate has cyclic image; if nonzero, choose a preimage of its positive generator to split it off, and the kernel is a subgroup of $\mathbb{Z}$; if zero, the subgroup already lies in $\mathbb{Z}$. Since the complementary $\mathbb{Z}$ has rank one and the total rank is two, $H$ has rank one. Therefore $H\cong \mathbb{Z}$.

Choose $c$ to be the first component of the Mayer–Vietoris image of $[\sigma ]$; it is a generator of $H_1(N(K);\mathbb{Z})$ because the product section projects with degree one to the core of the product tube [F5, F8]. Choose a generator $h$ of $H_1(E_K;\mathbb{Z})$, which exists by step 5.1. The meridian maps to zero in the $N(K)$ component because it bounds its fibre disk. In the bases $([\mu ],[\sigma ])$ and $(c,h)$, the Mayer–Vietoris isomorphism of step 4.1 has first column $(0,a)$ for some integer $a$ and second column whose first coordinate is $1$. Its determinant is therefore $\pm a$; since the map is an isomorphism over $\mathbb{Z}$, $a=\pm 1$. Thus $[\mu ]$ generates $H_1(E_K;\mathbb{Z})$.

Let $k\in \mathbb{Z}$ be the unique integer with $i_{\ast }[\sigma ]=k[\mu ]$, and put $[\lambda ]=[\sigma ]-k[\mu ]$. Then $i_{\ast }[\lambda ]=0$, $p_{\ast }[\lambda ]=1$, and the change from the basis $([\mu ],[\sigma ])$ to $([\mu ],[\lambda ])$ has determinant $1$. In the new basis, $i_{\ast }(a[\mu ]+b[\lambda ])=a[\mu ]$, so the kernel is exactly $\mathbb{Z}[\lambda ]$, a primitive rank-one direct summand. Every degree-one class is $[\sigma ]+a[\mu ]$ and maps to zero exactly when $a=-k$, proving uniqueness. The curve $t\mapsto (e^{2\pi it},e^{-2\pi ikt})$ is embedded in $S^1\times S^1$ because its first coordinate is injective on $S^1$; it represents $[\lambda ]$ and is a preferred longitude.