Choice-free homology and longitude computation for a supplied knot tube

Give $S^1$ its one-vertex, one-edge CW structure. By the oriented-knot convention [F1], the core circle has the specified positive orientation used for the section generator. The solid tube $S^1\times D^2$ retracts radially to the core circle, whose cellular boundary is zero, so $H_1(S^1\times D^2;\mathbb{Z})\cong \mathbb{Z}$. The product torus $T=S^1\times S^1$ has one $0$-cell, two $1$-cells and one $2$-cell; the two endpoint incidences cancel in degree $1$, and the attaching commutator of the $2$-cell has total signed incidence zero on each $1$-cell. Thus $H_1(T;\mathbb{Z})\cong {\mathbb{Z}}^2$, with basis the fibre class $[\mu ]$ and section class $[\sigma ]$. The projection to the first $S^1$ sends $[\mu ]$ to $0$ and $[\sigma ]$ to $1$.

Set $U=\mathrm{\Phi }(S^1\times \{\mathrm{\mid }z\mathrm{\mid }<3ϵ\mathrm{/}2\})$ and $V=S^3\setminus \mathrm{\Phi }(S^1\times D_{ϵ\mathrm{/}2}^2)$. They are open and cover $S^3$. Radial contraction in the disk coordinate retracts $U$ to the core circle and $U\cap V$ to $T$. The homotopy $r_t(r)=(1-t)r+tϵ$ for $ϵ\mathrm{/}2<r<ϵ$, fixed for $r\ge ϵ$, retracts $V$ onto $E$. The intersection map in the Mayer–Vietoris sequence is therefore the map $H_1(T)\to H_1(S^1)\oplus H_1(E)$ given by $x\mapsto (p_{\ast }x,-i_{\ast }x)$, where $p$ is product projection.

By exactness and [F3], this intersection map is an isomorphism. Its restriction to $\ker p_{\ast }=\mathbb{Z}[\mu ]$ is an isomorphism onto $0\oplus H_1(E)$: injectivity follows from the isomorphism, and any $(0,y)$ has a preimage $x$ with $p_{\ast }x=0$. Hence $H_1(E)\cong \mathbb{Z}$, generated up to sign by $i_{\ast }[\mu ]$.

Write $i_{\ast }[\sigma ]=c{i}_{\ast }[\mu ]$ for the unique integer $c$, and define $[\lambda ]=[\sigma ]-c[\mu ]$. Then $i_{\ast }[\lambda ]=0$ and its product projection is $1$. Every class in $\ker i_{\ast }$ has the form $a[\mu ]+b[\sigma ]$ and satisfies $a+bc=0$, so it is exactly $b[\lambda ]$; the projection condition forces $b=1$. The change-of-basis matrix from $([\mu ],[\sigma ])$ to $([\mu ],[\lambda ])$ has determinant $1$, so the latter pair is a basis. All choices were the supplied finite product data and integer operations; no choice axiom was used.