Zero- and one-handles are eliminated in a simply connected h-cobordism

By [F1] choose an ordered presentation without $0$-handles. Write $T_j=W_j$ for the trace through index $j$ and $N_j={\mathrm{\partial }}_1{T}_j$. The initial outgoing level is $M_0$. Deleting the finitely many $1$-handle attaching disks leaves it connected: paths may be rerouted around each disk through its connected boundary sphere, since $n\ge 5$.

In the outgoing piece of a selected $1$-handle $e$, take $D^1\times \{x\}$ with $x\in S^{n-1}$. Join its endpoints by an embedded arc in the punctured $M_0$ and smooth the two corners; the arc can be chosen by smoothing a path and applying the relative curve embedding case of [F3]. The resulting circle $\gamma \subset N_1$ meets the belt of $e$ once at the midpoint of that outgoing core arc, and misses the other $1$-handle belts. Perturb the old $2$-handle core circles to miss $\gamma$: $1+1<n$. Shrink their normal tubes, so $\gamma$ and a small neighborhood lie in the common unchanged part of $N_1$ and $N_2$. These attaching isotopies transport the later handles.

$\gamma$ is nullhomotopic in $N_2$. Indeed $W$ is simply connected by its incoming homotopy equivalence. The remaining forward handles after $T_2$ have indices at least three, so ${\pi }_1(T_2)\to {\pi }_1(W)$ is an isomorphism. The reversed presentation of $T_2$ relative to $N_2$ has indices $n-1,n$, both at least four, so ${\pi }_1(N_2)\to {\pi }_1(T_2)$ is an isomorphism as well. These applications concern the correct forward and reverse traces; no claim that index-three cells preserve ${\pi }_2$ is needed.

A continuous filling disk can be made smooth relative to $\gamma$: insert the smooth boundary loop on a radial collar, extend that collar slightly past the disk boundary, and apply relative approximation on this extended source. Then use [F3] to embed the disk relative to its boundary, since $5\le n$. Its normal bundle is trivial: in a finite Euclidean embedding, radial transport of its smooth orthogonal projection produces a frame over the contractible disk. Append the disk's outward boundary-normal line to that frame. For the tube near the disk boundary, extend its embedding slightly in the outward boundary-normal direction. The local inverse-function tube construction in [F4], followed by compactness of the disk and separation of distinct zero-section points, gives one sufficiently small embedded tube along the full disk, including its boundary. This does not require the closed boundaryless-submanifold hypothesis of the global tubular theorem. That tube gives the standard framed circle $S^1\times D^{n-1}$ along $\gamma$ in $N_2$. In tubular disk coordinates radial contraction into a small interior disk exhibits its standard triviality by a framed isotopy. Shrink its circle tube to remain in the common part of $N_1,N_2$, and read that same framed circle as an attachment in $N_1$. No disk avoidance of full-dimensional higher attaching regions is asserted or needed.

This framed circle satisfies both hypotheses of [F5], so trade $e$ for a $3$-handle. Repeating finitely removes all $1$-handles. The procedure removes only $0$- and $1$-handles and introduces only $3$-handles, carrying all other data. Index ordering can be restored by [F1]. Hence every remaining index is at least two, and an existing upper bound of at least three is preserved. The sole disk-embedding restriction is $n\ge 5$.