A punctured-sphere cover from a labeled link diagram

The height condition says that the link meets the level sphere $\mathrm{\Sigma }$ exactly at the $n$ underpassing points. Hence $A_L\cup B_L=X$ and $A_L\cap B_L={\mathrm{\Sigma }}_L$. The one-point compactification of the plane ${\mathbb{R}}^2\times \{-1\}$ is a sphere, so removing those $n$ points gives the stated punctured sphere. The punctured sphere is path connected: remove one puncture to obtain a plane, in which a polygonal path detours around the remaining finite points.

Because the crossing set is finite, choose pairwise disjoint coordinate balls about its points, then shrink the product tubes so each crossing ball meets precisely the two prescribed local branches and contains no other tube. Shrinking a product tube radially is an isotopy of its boundary and leaves the exterior homeomorphic. Use the unit-sphere stereographic coordinates $(u_1,u_2,u_3,u_4)=(2x,2y,2z,x^2+y^2+z^2-1)\mathrm{/}(1+x^2+y^2+z^2)$. Here $\mathrm{\Sigma }$ is $h\cdot u=c$, with $h=(0,0,1,-1)\mathrm{/}\sqrt{2}$, $c=-1\mathrm{/}\sqrt{2}$. Write $q=ch+\sin (3\pi \mathrm{/}4)w$, $w\perp h$, $\mathrm{\mid }w\mathrm{\mid }=1$. Its explicit unit-speed normal collar is $\cos (3\pi \mathrm{/}4-t)h+\sin (3\pi \mathrm{/}4-t)w$, for $\mathrm{\mid }t\mathrm{\mid }<\rho =\pi \mathrm{/}8$; it is injective since its angle stays in $(0,\pi )$, and positive $t$ is the $A$ side. In particular no general tubular-neighborhood or choice theorem is required. For $q\in {\mathrm{\Sigma }}_L$, set $ϵ(q)=\min \{\rho \mathrm{/}4,d(q,L)\mathrm{/}4\}$. This is positive and continuous. The points ${\exp }_q(t{\nu }_q)$ with $\mathrm{\mid }t\mathrm{\mid }<ϵ(q)$ avoid $L$, since their distance from $q$ is less than $d(q,L)$. The variable-width collar $C=\{{\exp }_q(t{\nu }_q):q\in {\mathrm{\Sigma }}_L,\mathrm{\mid }t\mathrm{\mid }<ϵ(q)\}$ is embedded and open in $X$. Set $U=A_L\cup C_{<0}$ and $V=B_L\cup C_{>0}$, where $C_{<0}$ and $C_{>0}$ are its negative and positive normal halves. Then $U,V$ are open, cover $X$, and $U\cap V=C$. Clamping the normal coordinate to $0$ retracts $U$ to $A_L$ and $V$ to $B_L$, while the collar itself retracts to ${\mathrm{\Sigma }}_L$. For the final assertion, choose a continuous function $\chi :[0,1]\to [0,1]$ equal to $1$ on $[0,1\mathrm{/}4]$ and $0$ on $[1\mathrm{/}2,1]$. On the positive collar of $A_L$, the homotopy $t\mapsto t+sϵ(q)\chi (t\mathrm{/}ϵ(q))\mathrm{/}4$ ($0\le s\le 1$) pushes every boundary point into $A_L^{\circ }$, fixes points outside the collar, and stays below $ϵ(q)$; it stays in $X$ by the distance bound. It maps $A_L^{\circ }$ to itself, so the inclusion $A_L^{\circ }\hookrightarrow A_L$ is a homotopy equivalence.

Every finite point of $B_L$ has height at most $-1$. For $0\le t<1$, move $(x,y,z)$ vertically to $(x,y,z-t\mathrm{/}(1-t))$, and send $\mathrm{\infty }$ to itself; at $t=1$ send every point to $\mathrm{\infty }$. This is continuous in the one-point compactification, avoids $L_D$ because the link lies in $z\ge -1$, and contracts $B_L$ to $\mathrm{\infty }$.

Choose disjoint small coordinate disks about the punctures on $\mathrm{\Sigma }$. Within each punctured disk, the radial homotopy $r\mapsto (1-t)r+t\max \{r,\epsilon \}$ pushes its punctured cap onto its boundary collar, fixing points outside the cap. Thus ${\mathrm{\Sigma }}_L$ retracts to the compact sphere with $n$ open disks removed. Base the $n$ boundary circles along any specified accesses. Filling all $n$ boundary circles by disks gives $S^2$, which is simply connected: two slightly enlarged hemispheres are contractible with path-connected annular overlap, so [F3] gives the trivial group. Filling all but the $j$th boundary circle gives a disk, also simply connected. The radial retraction extends by the identity on the newly attached disks. Consequently attaching disks to the original noncompact punctured sphere gives spaces homotopy equivalent to these filled compact models, rather than literally the compact models. The van Kampen quotient in [F3] proves that all the based puncture circles normally generate, and that the $n-1$ circles other than $c_j$ still normally generate. For $n=1$ the omitted list is empty and ${\mathrm{\Sigma }}_L\cong {\mathbb{R}}^2$ is contractible, as required. Independent access changes conjugate each relator, leaving its normal closure unchanged.

The lower piece is contractible by step 1.3. For the upper piece, a finite point outside the link has a small relative ball disjoint from the link. Move within that ball to a point whose plane projection misses the whole projected diagram; such projections are dense because a finite regular immersed compact curve has locally finitely many one-dimensional graph branches and empty interior. The vertical path from that point to height $1$ now avoids the link. The height-$1$ plane is connected. The point at infinity first moves into this finite upper piece along the explicit positive collar in step 1.2. Thus the upper piece too is path connected. This uses explicit paths rather than a general-position theorem for arbitrary arcs. The retractions of step 1.2 now show that $U,V,U\cap V$ are path connected as well: any point is joined to its retraction image by that retraction path.

In a product tube $S^1\times D^2$, the punctured tube $S^1\times (D^2\setminus \{0\})$ retracts radially onto $S^1\times \mathrm{\partial }{D}^2$. Applying this on each of the finitely many disjoint tubes and fixing the exterior gives a deformation retraction of $X$ onto the chosen exterior. Thus the induced map on fundamental groups is an isomorphism.