Homology and preferred longitude of a knot exterior

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.