The upper link-diagram region is free on strand meridians

Let $A_L^{\circ }=(({\mathbb{R}}^2\times (-1,\mathrm{\infty }))\cup \{\mathrm{\infty }\})\setminus L_D$. By [F1], ${\pi }_1(A_L^{\circ })\cong {\pi }_1(A_L)$. A small neighborhood $H$ of $\mathrm{\infty }$ in $A_L^{\circ }$ is the open upper half-ball together with its cone vertex $\mathrm{\infty }$, disjoint from $L_D$; other points of the level-sphere boundary have been deleted. The open cover $A_L^{\circ }=M\cup H$ has intersection $H\setminus \{\mathrm{\infty }\}$, which is homeomorphic to $(0,1)\times \mathrm{int}(S_{+}^2)$ and is contractible because the open hemisphere $\mathrm{int}(S_{+}^2)$ is an open disk. Since $H$ is also contractible, [F2] gives ${\pi }_1(M)\cong {\pi }_1(A_L^{\circ })$.

Each $D_j$ is the image of the open region $(s,t)\mapsto (\gamma (s),-1+t(\eta (s)+1))$ with $s\in I_j$ and $0<t<1$; this is an embedded open disk. At a projected double point, the underpassing branch has height $-1$ and is absent from $M$, while the overpassing branch has height greater than $-1$. Thus sheets over distinct intervals are disjoint. Each sheet is closed in $M$: its only limit points outside itself lie on the excluded plane $z=-1$ or on the excluded link. Hence it is a proper smooth embedded simply connected surface.

At a fixed vertical line, every point of $Z$ lies above all link points on that line. Since every link height is at most $0$, the straight homotopy $(x,y,z)\mapsto (x,y,(1-t)z+t)$ ($0\le t\le 1$) stays above those link points and contracts $Z$ onto the plane $z=1$. This plane is contained in $Z$, so $Z$ is contractible.

For $j=1,\dots ,n$, put $M_j=M\setminus {\bigcup }_{k>j}{D}_k$. Then $M_0=Z$ and $M_j=M_{j-1}\cup D_j$. At each stage $M_{j-1}$ is path-connected: it contains path-connected $Z$, and every point of an earlier sheet joins to $Z$ along a short normal segment. Orient $D_j$ by its interval and height coordinates. The ambient orientation orients its normal line, so that line bundle is trivial; a bundle metric gives the unique positive unit vector in each fibre. By [A1, F3], choose a tubular neighborhood disjoint from the other sheets. Radial contraction shows the whole neighborhood and each of its two side components are simply connected. Choose a small positive meridian circle around an interior point of the $j$th strand in a ball disjoint from all other sheets. It meets $D_j$ transversely once. Remove a short open arc about that intersection; the remaining compact arc lies in $M_{j-1}$ and joins the two sides of $D_j$. Shrink the tubular neighborhood so that it meets this arc only in two end balls. This gives the corridor required by [F4], and its complement hypothesis holds because $M_j\setminus D_j=M_{j-1}$ is path-connected.

The cut lemma [F4] gives ${\pi }_1(M_j)\cong {\pi }_1(M_{j-1})\ast \mathbb{Z}$. Its cyclic generator is represented by the short meridian arc crossing $D_j$ followed by the complementary arc from step 3.1. Joining this loop to the basepoint by its access path in $Z$ gives exactly $x_j$, with positive orientation fixed by [F5]. Since ${\pi }_1(M_0)={\pi }_1(Z)=1$, induction yields that ${\pi }_1(M)$ is free on $x_1,\dots ,x_n$. Step 1.1 identifies this with ${\pi }_1(A_L)$.