Belt-sphere complements in low handle levels preserve the fundamental group

Put $Q=M\setminus {\bigcup }_j\mathrm{int}(S^{q-1}\times D^{n-q+1})$, using the actual framed attaching tubes. By [F1], $N\setminus {\bigcup }_j{B}_j$ is $Q$ with the pieces $(D^q\setminus \{0\})\times S^{n-q}$ glued along $S^{q-1}\times S^{n-q}$. Radially retract each punctured $D^q$ to its boundary, keeping that boundary fixed. Thus this complement deformation retracts to $Q$, with the corner collars included. Also $Q$ is homotopy equivalent to $M$ minus the attaching cores $S^{q-1}$. Their codimension is $n-q+1\ge 3$. By [F4] loops and disk nullhomotopies in $M$ can avoid these finitely many cores, relative to prescribed endpoints or boundary collars: their expected intersection dimensions are $q-n<0$ and $q+1-n\le -1$. Therefore ${\pi }_1(Q)\to {\pi }_1(M)$ is an isomorphism. These identifications commute with the inclusions into the trace $T$.

For $q\ge 3$, [F3] makes ${\pi }_1(M)\to {\pi }_1(T)$ an isomorphism. For $q=2$ it is a surjection, since attaching a $2$-handle only kills the attaching loop; the assumed injection makes it an isomorphism too. Reading the trace backwards attaches only handles of index $n+1-q\ge 3$, so [F2] and [F3] make ${\pi }_1(N)\to {\pi }_1(T)$ an isomorphism. Step 1.1 now identifies the complement-to-$N$ map as the comparison between two isomorphisms to ${\pi }_1(T)$, proving the first assertion. For a subfamily, every loop in its complement can be perturbed away from the remaining belts because they have codimension $q\ge 2$ and $1-q<0$. Hence the full-complement map onto the subfamily-complement fundamental group is surjective. Since its composite to ${\pi }_1(N)$ is an isomorphism, the subfamily map to ${\pi }_1(N)$ is also injective and surjective.

If $W$ is an h-cobordism, its remaining handles beyond $T=W_q$ have index at least $q+1\ge 3$, so [F3] gives ${\pi }_1(T)\cong {\pi }_1(W)$. The incoming inclusion $M\to W$ is an isomorphism by [F5], and therefore $M\to T$ is an isomorphism. If instead all $2$-handle attaching loops are nullhomotopic, their normal closures are zero and the incoming map is again an isomorphism. This proves the two advertised sufficient conditions, without assuming that a simply connected outgoing level alone controls an arbitrary codimension-two complement.

For the $q=2$ pair form the usual clean boundary annulus in its sheet and fixed corner collars. Its inner circle $\lambda$ misses all belts and the $2$-sphere $A$, and is homotopic in $N$ to the original Whitney circle. Step 2.1 makes the full belt-complement map injective, so $\lambda$ bounds a disk in that full complement. Smooth and embed the disk relative to its annulus collar using [F4] and [F6]. Perturb its interior in the belt complement to avoid $A$ and any additional prescribed $2$-spheres: each incidence dimension is $2+2-n<0$. Small relative perturbations preserve its compact embedded collar and embeddedness. Thus the resulting disk misses every belt in its interior and every additional $2$-sphere as requested.

The framing part is the exact $r=2$ construction of [F6]: opposite corner signs match the oriented endpoint values of its rank-one partial $E$ frame, tangent to $A$ on its arc and normal to the selected belt on the other. Trivialize the rank-$(n-2)$ disk-normal bundle by radial projection transport; the partial-frame loop lies in $V_1({\mathbb{R}}^{n-2})$, simply connected since $n\ge 5$. Extend and smooth $E$ relative to its collars, then frame its orthogonal complement $H$ over the disk. This gives an admissible frame with no arbitrary preassigned full boundary class. By [F7] a thin tube gives the desired isotopy of $A$. It avoids the other belts and additional spheres by step 3.2 and compact separation outside the designated collars, so no new intersection with them is created. The choices are finite or are the declared countable-choice approximation inputs.