String-link exterior homology and end-meridian maps

Let $P=D^2\setminus {\bigcup }_i\mathrm{int}{D}_i$ be either endpoint surface. The supplied product tubes parametrize the pairwise disjoint endpoint disks by ${\varphi }_i:D^2\to D_i$, with ${\varphi }_i(0)=p_i$. Put $Q=D^2\setminus \{p_1,\dots ,p_r\}$. On $D_i\setminus \{p_i\}$ write a point as ${\varphi }_i(ru)$ with $0<r\le 1$, $\mathrm{\mid }u\mathrm{\mid }=1$, and set $R_t({\varphi }_i(ru))={\varphi }_i(((1-t)r+t)u)\mathrm{.}$ Put $R_t=\mathrm{i}\mathrm{d}$ outside the interiors of these disks. At $r=1$ the formulas agree, so finite pasting gives a deformation retraction $Q\to P$. This uses the supplied disk coordinates, not a triangulation of an arbitrary smooth planar region. A collar of the outer unit circle misses all $p_i$. In its radial coordinate a continuous inward push, equal to the identity off that collar, gives a homotopy inverse to $\mathrm{int}{D}^2\setminus \{p_i\}\hookrightarrow Q$; throughout the homotopy no puncture is crossed. By [F8] the open punctured disk has a finite circle-wedge model and a positive meridian free basis. Transfer it across these displayed equivalences, changing basepoint along the recorded collar path when needed. Small loops about $p_i$ become the positive endpoint meridians: the annulus ${\varphi }_i(ru)$ carries their orientation to the tube boundary. The accesses are the jointly transported stem paths of that finite spine; their images need not be arbitrary preselected accesses. If $a_i$ and $a_i^{\mathrm{\prime }}$ are compatible and arbitrary accesses to the same meridian circle, then $c_i=a_i^{\mathrm{\prime }}{a}_i^{-1}$ is a based loop and $a_i^{\mathrm{\prime }}{\mu }_i(a_i^{\mathrm{\prime }}{)}^{-1}=c_i(a_i{\mu }_i{a}_i^{-1})c_i^{-1}$. Thus independent basings conjugate generators, without asserting they remain a free basis. Conjugation leaves their homology classes unchanged by [F9]. The homotopy type and abstract free rank are unchanged.

For a closed link, enlarge each solid torus tube by a small radial collar and apply [F1] to the pair $(S^3,E_L)$; the collar deformation retracts the enlarged-tube pair to $(S^1\times D^2,S^1\times S^1)$. The product torus has a CW structure with one vertex, two edges, and one face attached by a commutator, so [F7] gives $H_1(S^1\times S^1)\cong {\mathbb{Z}}^2$ with meridian and section basis. The solid torus retracts to its core circle, so [F3, F6] give $H_2(S^1\times D^2)=0$ and $H_1(S^1\times D^2)\cong \mathbb{Z}$, with the meridian mapping to zero and the section to a generator. The pair sequence [F2] therefore identifies $H_2(S^1\times D^2,S^1\times S^1)$ with the kernel generated by the meridian. Excision gives $H_2(S^3,E_L)\cong {\mathbb{Z}}^r$. By [F2] and [F3], the boundary homomorphism $H_2(S^3,E_L)\to H_1(E_L)$ is an isomorphism. Each relative generator is a normal disk and its boundary is the positive meridian by [F4]. Hence the positive meridians form the ordered basis of $H_1(E_L)$.

Let $D_i^{+}$ be disjoint disks slightly larger than the removed endpoint tube disks. The closed complement of their interiors lies in the interior of $P$ relative to $D^2$. By [F1], $H_{\ast }(D^2,P)$ is the direct sum of the relative groups of the $r$ disk pairs $(D_i^{+},D_i^{+}\setminus \mathrm{int}{D}_i)$. Radial collars identify each pair with $(D^2,S^1)$. By [F2], [F3], and [F6], the relative group of this disk pair is $\mathbb{Z}$ in degree $2$ and zero in every other degree. Its boundary map sends the relative disk to the positive endpoint meridian. Applying [F2] to $(D^2,P)$, with $D^2$ contractible by radial contraction, therefore gives $H_1(P)\cong {\mathbb{Z}}^r$ with that basis and $H_2(P)=0$.

For each string component, enlarge its chosen product tube by a small radial collar. The complement of these enlarged tubes has closure in the interior of $X_{\tau }$; excising it by [F1] reduces $(D^2\times [0,1],X_{\tau })$ to the disjoint union of tube pairs. In each pair, the radial collar deformation retracts the pair to $([0,1]\times D^2,[0,1]\times S^1)$: the endpoint cap annuli retract to their boundary circles and the lateral shell retracts to the original tube boundary. Contracting the interval factor identifies this pair with $(D^2,S^1)$. By [F2], [F3], and [F6], each pair has relative homology $\mathbb{Z}$ in degree $2$ and zero in every other degree. Therefore $H_2(D^2\times [0,1],X_{\tau })\cong {\mathbb{Z}}^r$ and its relative homology vanishes in degrees $1$ and $3$.

The ambient cylinder is contractible, so its positive-degree homology vanishes by [F6]. Its pair long exact sequence from [F2], together with step 3.1, gives $H_2(X_{\tau })=0$ from the segment $H_3(B,X_{\tau })\to H_2(X_{\tau })\to H_2(B)$, and gives an isomorphism $\mathrm{\partial }:H_2(B,X_{\tau })\to H_1(X_{\tau })$ from the segment $H_2(B)\to H_2(B,X_{\tau })\stackrel{\mathrm{\partial }}{\to }{H}_1(X_{\tau })\to H_1(B)$. Thus $H_1(X_{\tau })\cong {\mathbb{Z}}^r$.

Put $B=D^2\times [0,1]$ and let $j_{\pm }:(D^2,P_{\pm })\to (B,X_{\tau })$ be the endpoint pair inclusions. Represent the positive $i$th endpoint disk by an oriented singular disk chain $d_{i,\pm }$. Under the excision and interval contraction in step 3.1, its image is the positive relative disk generator in the $i$th tube summand: the slice $\{0\}\times D^2$ or $\{1\}\times D^2$ projects to $D^2$ preserving the normal-disk orientation. Thus the classes $[j_{\pm \mathrm{\#}}{d}_{i,\pm }]$ form the ordered basis of $H_2(B,X_{\tau })$. By the lift-and-boundary recipe [F10], ${\delta }_B([j_{\pm \mathrm{\#}}{d}_{i,\pm }])=[\mathrm{\partial }{j}_{\pm \mathrm{\#}}{d}_{i,\pm }]=[j_{\pm \mathrm{\#}}\mathrm{\partial }{d}_{i,\pm }]$: the equality holds on chains because composing a singular simplex with $j_{\pm }$ commutes with every face restriction in its boundary. These are exactly the images of the positive endpoint meridians. The connecting map is an isomorphism by step 4.1, and the endpoint meridians form the basis by step 2.1. Each end map on $H_1$ is therefore an isomorphism. Since both endpoint and exterior $H_2$ vanish, each end map on $H_2$ is surjective.

The product torus $T_i$ has $H_1(T_i)\cong {\mathbb{Z}}^2$ with basis the positive meridian and the chosen degree-one section. Write the image of the section in the basis from step 1.2 and let $k_i$ be its $i$th coordinate. Subtracting $k_i$ times the meridian gives the unique degree-one class with zero $i$th coordinate. Its coordinates in the product basis are $(-k_i,1)$, so it is primitive; the curve $t\mapsto (e^{2\pi it},e^{-2\pi i{k}_{i}t})$ in product coordinates is an embedded representative.