The double branched cover of the trivial two-arc tangle is a solid torus

Regard $D^2$ as the unit disk in $\mathbb{C}$. On the Riemann sphere define $q(w)=(a_2{w}^2+a_1)\mathrm{/}(w^2+1)$, with its usual values at the poles and at infinity. This is a degree-two map branched precisely over $a_1,a_2$: $q(w)-a_1=(a_2-a_1)w^2\mathrm{/}(1+w^2)$ near $w=0$, and $q(1\mathrm{/}v)-a_2=(a_1-a_2)v^2\mathrm{/}(1+v^2)$ near $v=0$. The nonzero factors have local square roots, giving exactly the square charts of [F2]. Thus $S=q^{-1}(D^2)$ is a compact oriented surface with smooth boundary and a simple branched map to $D^2$. Its punctured restriction is connected: lifting a meridian about either branch point interchanges the two sheets. By [F1] it is the prescribed disk cover, since the two puncture meridians generate its fundamental group. This constructs the extension rather than assuming it from a link-only definition.

Triangulate the disk with both branch points as vertices. A fan triangulation of an $m$-gon has $V=m$, $E=2m-3$, $F=m-2$ and Euler characteristic $1$; inserting interior vertices preserves this count. Each open cell lifts twice except the two branch vertices, which lift once. Consequently $\chi (S)=2\cdot 1-2=0$. The boundary monodromy is $1+1=0$, so there are two boundary circles. Capping gives a closed connected orientable surface with Euler characteristic $2$, hence a sphere by [F3]. To apply [F4] with smooth circles in the standard sphere, use the explicit model of step 1.1: the closed exterior disk $\widehat{\mathbb{C}}\setminus \mathrm{int}{D}^2$ contains no branch value and is simply connected, so its inverse image consists of two disks by [F1]. Their boundaries are disjoint smooth circles, because $q$ is unramified over $\mathrm{\partial }{D}^2$. Thus $S$ is exactly the closed region between these circles in the standard Riemann sphere, and [F4] identifies it as an annulus.

Take $q{\mathrm{\mid }}_S\times {\mathrm{i}\mathrm{d}}_{[0,1]}$. Its local models along the two arcs, including their boundary endpoints, are $(z,t)\mapsto (z^2,t)$ by step 1.1. Away from the arcs it is the connected product cover with the required meridians. Projection of the punctured product to its disk factor induces an isomorphism on fundamental groups, so [F1] identifies this restriction with the specified cover of $B\setminus A$. Transporting the constructed extension along that covering isomorphism proves the required extension. Its total space is $S\times [0,1]\cong S^1\times [0,1{]}^2$, a solid torus after the usual rounding of product corners.