The Smooth H Cobordism Theorem

By [F1] the inclusion ${\iota }_0:M_0\hookrightarrow W$ is a homotopy equivalence, so by [F2] the induced map $H_k({\iota }_0;G):H_k(M_0;G)\to H_k(W;G)$ is an isomorphism for every $k\ge 0$, in particular surjective; the same applies to ${\iota }_1:M_1\hookrightarrow W$.

Fix $k\ge 0$ and read the exact sequence of [F3] for the pair $(W,M_0)$ around $H_k(W,M_0;G)$, namely $H_k(M_0;G)\stackrel{H_k({\iota }_0;G)}{\to }{H}_k(W;G)\to H_k(W,M_0;G)\stackrel{\delta }{\to }{H}_{k-1}(M_0;G)\stackrel{H_{k-1}({\iota }_0;G)}{\to }{H}_{k-1}(W;G)$. Surjectivity of $H_k({\iota }_0;G)$ together with exactness at $H_k(W;G)$ makes $H_k(W;G)\to H_k(W,M_0;G)$ the zero map; injectivity of $H_{k-1}({\iota }_0;G)$ together with exactness at $H_{k-1}(M_0;G)$ makes $\delta$ the zero map (for $k=0$ the group $H_{-1}(M_0;G)$ is zero, so $\delta$ is automatically zero). The image of the zero map $H_k(W;G)\to H_k(W,M_0;G)$ is $0$, so exactness at $H_k(W,M_0;G)$ gives $\ker \delta =0$, and since $\delta =0$ this says $H_k(W,M_0;G)=0$.

Interchanging the roles of $M_0$ and $M_1$, which is legitimate for an h-cobordism by [F1], the argument of step 2.1 gives $H_k(W,M_1;G)=0$ for every $k\ge 0$. By step 1.1 each inclusion induces an isomorphism in every degree, and specialising $G=\mathbb{Z}$ gives $H_k(W,M_0;\mathbb{Z})=H_k(W,M_1;\mathbb{Z})=0$.

Apply [F1] to the collared triad $(W;M_0,M_1)$, which is compact by the definition of an h-cobordism, to get an adapted excellent Morse function $f$ and an adapted complete downward gradient-like field $X$ agreeing with the product model on the fixed collars, so that by [F2] the faces are the level sets $f^{-1}(0)=M_0$, $f^{-1}(1)=M_1$ and the boundary collars contain no critical point.

Apply the rearrangement theorem [F3] to $(f,X)$ to obtain an adapted field for an excellent function whose critical levels are ordered by index, and then apply the self-indexing theorem [F4] to that excellent pair to put all critical points of index $k$ at one common level $c_k$ with $c_0<\cdots <c_{n+1}$; the modification is supported near the finitely many critical levels, so the boundary product model on the fixed collars is preserved, and compact interior field modifications preserve collar-extension completeness: extend the change by zero to the complete carrier and apply the compact-support construction of [F1]. Retain the original pair $(f,X)$ from step 1.1 for the statement’s first clause; the self-indexed function has its own adapted field. No assertion that the final field descends for the original $f$ is needed. The countable choice principle is inherited from [F1]–[F4] and [F8].

Apply [F6] to a small regular-endpoint band around each occupied critical level $c_k$ of the self-indexed function. Its critical points all have index $k$, and no other critical value lies in the band, so crossing it attaches disjoint $k$-handles, one per critical point. Concatenate these successive sublevel attachments in increasing order of $k$; [F4] supplies the resulting decomposition of the whole triad, and [F6] permits any order within each disjoint family. By [F7] this is a handle decomposition relative to $M_0$ with the stated handle list and level structure. The fixed boundary collars contain no critical point by [F2], so contain no handle.

Therefore every compact h-cobordism admits an adapted field and a Morse function whose critical levels are strictly ordered by index, together with the resulting handle decomposition relative to $M_0$ with all $k$-handles attached at one level before all $(k+1)$-handles and one handle per critical point. No simple connectivity of the faces or of $W$ and no dimension restriction beyond $\dim W=n+1$ was used, since the suppliers [F1], [F3], [F4], [F6] hold for arbitrary compact triads.

Product collars let us enlarge the lower stage slightly across each attaching seam and retract that enlargement onto the lower stage. Excision then identifies the relative homology of the next index stage with that of the disjoint union of its handle pairs; the retraction and excision preserve the relative core classes. By [F1]–[F2], $H_i(W_k,W_{k-1};\mathbb{Z})$ is zero unless $i=k$, and in that degree is free on these cores. At $k=0$, compressing the initial collar onto $M_0$ gives the same calculation, including disjoint $0$-handles.

Write ${\delta }_k:C_k\to H_{k-1}(W_{k-1})$ for the pair connector and $p_{k-1}:H_{k-1}(W_{k-1})\to C_{k-1}$ for the quotient map. By [F3], ${\mathrm{\partial }}_k=p_{k-1}{\delta }_k$. Exactness of the pair $(W_{k-1},W_{k-2})$ gives ${\delta }_{k-1}{p}_{k-1}=0$, so ${\mathrm{\partial }}_{k-1}{\mathrm{\partial }}_k=p_{k-2}{\delta }_{k-1}{p}_{k-1}{\delta }_k=0$ for $k\ge 2$. For $k=1$ this follows from ${\mathrm{\partial }}_0=0$.

The triple sequences and step 1.1 show inductively that $H_i(W_j,M_0)=0$ for $i>j$. Fix $k\ge 0$. The triple $(W_k,W_{k-1},M_0)$ therefore injects $H_k(W_k,M_0)$ into $C_k$. Its image is the kernel of the connector into $H_{k-1}(W_{k-1},M_0)$, which itself injects into $C_{k-1}$ by the triple $(W_{k-1},W_{k-2},M_0)$ when $k\ge 1$. Naturality and the connector factorization identify that composite with ${\mathrm{\partial }}_k$. For $k=0$ the target is zero. Hence $H_k(W_k,M_0)$ identifies with $\ker {\mathrm{\partial }}_k$.

The triple $(W_{k+1},W_k,M_0)$ identifies $H_k(W_{k+1},M_0)$ with the cokernel of $C_{k+1}\to H_k(W_k,M_0)$, because $H_k(W_{k+1},W_k)=0$. Under step 3.1 this map is ${\mathrm{\partial }}_{k+1}$. Adding stages of index $j\ge k+2$ changes neither $H_k$ nor its map, since both $H_{k+1}(W_j,W_{j-1})$ and $H_k(W_j,W_{j-1})$ vanish. The finite filtration therefore gives $H_k(W,M_0)\cong \ker {\mathrm{\partial }}_k\mathrm{/}\mathrm{im}{\mathrm{\partial }}_{k+1}$.

The pair connector sends an upper core class to its oriented attaching sphere. To read its coefficient at $e_j$, collapse the preceding stage and all other $k$-handle summands. In the outgoing piece $D^k\times S^{d-k-1}$ of $e_j$, $d=\dim W$, the resulting map to $D^k\mathrm{/}{S}^{k-1}$ is projection onto the $D^k$ factor. A regular value at its center has preimages exactly $A_i\cap B_j$; the local degrees are $I(A_i,B_j)$'s local signs by the normal-orientation convention. Summing gives the displayed formula. For $k=0$ the connector records the two signed endpoints, and for $k=d-1$ the same calculation is dual. This local argument applies to a relative triad as well as a closed manifold. The coordinate convention and the two-index assertion now follow from steps 1.1–4.1.

Temporarily stop the presentation at $W_{q+1}$. At the standard trivial embedding of hypothesis 2 introduce a $(q+1),(q+2)$ pair by [F1]. Use that framed isotopy and [F2] to move the new lower attachment to $\alpha$ in $N_{q+1}$, carrying the new upper attachment along. Restore all higher handles by transporting their attaching data. If a stage is disconnected, perform the construction on its affected connected component and leave the others fixed.

Because the entire image of $\alpha$ lies in the common part $N_q^{\circ }$, the new $(q+1)$-handle can instead be attached at $N_q$, before all old $(q+1)$-handles: their disjoint attaching-region quotient is the same whichever is attached first. Move $e$ to the end of its $q$-index family by [F3]. Apply the framed isotopy of hypothesis 1 to the new upper attachment in the resulting $N_q$, carrying all later attaching data by [F2]. Now $e$ and that new $(q+1)$-handle are consecutive and meet once.

Cancel this consecutive pair by [F4]. The deleted handles are $e$ and the newly introduced $(q+1)$-handle; the new $(q+2)$-handle survives, and every old higher handle is carried by the cancellation diffeomorphism. Thus the counts change exactly as asserted, relative to the incoming face. This is the two-isotopy and disjoint-reordering proof of the cited Lück Elimination Lemma, without an isotopy-to-slide decomposition premise.

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 $n\ge 5$.

By [F1] the reversed triad $(W;M_1,M_0)$ is again an h-cobordism with the same dimension, and its faces are the same closed simply connected $n$-manifolds in the opposite order while $W$ remains connected; hence the low-index elimination [F2] applies to the reversed triad and gives a presentation of $W$ relative to $M_1$ with no handles of index $0$ or $1$.

Negate the Morse function of the reversed presentation and dualise by [F3]: a handle of index $k$ in the presentation relative to $M_1$ becomes a handle of index $n+1-k$ in a presentation relative to $M_0$, with attaching and belt spheres interchanged. Absent indices $0$ and $1$ therefore become absent indices $n+1$ and $n$, so the dual presentation relative to $M_0$ has no handles of index $n+1$ or $n$.

To obtain the two exclusions simultaneously, first eliminate $0,1$ relative to $M_0$ by [F2]. Its reversed presentation has every index at most $n-1$. Apply the actual low-index procedure to that presentation relative to $M_1$: by [F4] it deletes $0,1$ and introduces only $3$, so its maximum stays at most $n-1$, because $n\ge 5$. Reverse back by [F3]. The resulting original indices lie between $2$ and $n-1$; both exclusions now hold in one presentation. No composition of two independently chosen presentations or face-fixing diffeomorphisms is being assumed.

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.

Isotope $f$ transverse to the finitely many belt spheres. By [F1] its signed intersection coordinates are $\pm 1$ at the distinguished belt and zero at every other belt. Unless the required configuration already holds, one belt therefore has an opposite-sign surplus pair.

Choose sheet arcs avoiding every other intersection by [F2]. Their circle is nullhomotopic because the outgoing level is simply connected. For $3\le q\le n-3$, both sheet dimensions are at least three and [F3] supplies the Whitney move; choose the disk and tube disjoint from all other spheres by their codimension-at-least-three dimension counts. For $q=2$, [F4] gives the full belt-complement injection from the explicit incoming assumption, fills the shifted circle in that complement and clears all other $2$-sphere attaching images. The helper supplies the clean admissibly framed disk and pair removal.

Each move lowers the finite intersection count by two, preserves the other intersections and transports any given normal frame. Iterating leaves one point at the distinguished belt and none at the others, because the signed sums remain $\pm 1$ and zero. This proves the stated generic range with its exact $q=2$ condition.

Fix a $(q+1)$-handle ${\phi }_j$. Choose a parallel copy $P_j$ of its attaching sphere just outside its closed attaching tube, using the extension over a neighborhood of the normal disk factor. It lies in $N_q^{\circ }$, the open complement of all closed upper attaching tubes. In $N_{q+1}$ it bounds the outgoing disk $D^{q+1}\times \{z\}$, $z\in S^{n-q-1}$, together with the short annular collar from its boundary to $P_j$. This is an embedded disk with interior in that handle's outgoing piece, disjoint from any sphere in the common part. Its normal disk coordinates give the standard bounding-disk framing. In $C_q$, $[P_j]={\mathrm{\partial }}_{q+1}[{\phi }_j]$ with compatible orientations, by [F1].

$N_q^{\circ }$ is connected: the upper attaching cores have codimension $n-q\ge 2$, so paths can be rerouted in finitely many product charts off the cores, and radial retraction in each punctured tube pushes them outside the closed smaller tubes. Apply [F2] within this common part to band-sum $f$ with $P_j$. The band interior can avoid both spheres and all other attaching tubes. The resulting $q$-sphere $g$ is embedded and remains in $N_q^{\circ }$. An oriented pair-of-pants bordism in a thin neighborhood of the band has boundary $g-f-P_j$, so in $C_q$ its fundamental chain gives $[g]=[f]+[P_j]$. Choose the orientation of the added parallel copy, rather than reverse $f$, to obtain either sign.

In $N_{q+1}$, slide the added lobe back along the band and across the embedded disk of step 1.1. The disk interior is in the new handle's outgoing region, whereas $f$ and the band are in the common part. A thin product neighborhood of their union therefore gives the usual embedded isotopy from the band-sum to $f$, fixed outside that neighborhood. If $f$ has a normal frame, use the bounding-disk frame for $P_j$ and the matching band frame; this isotopy transports the full frame and gives a framed isotopy as well. Thus no implication from a vanishing homology class to embedding triviality is used.

Repeat with fresh disjoint parallel copies $\mathrm{\mid }{x}_j\mathrm{\mid }$ times, using the sign of $x_j$, and then over the finite upper-handle list. At every iteration the new sphere is in the common part and is isotopic one level higher to the preceding sphere. The additive computation gives $[g]=[f]+{\sum }_j{x}_j{\mathrm{\partial }}_{q+1}[{\phi }_j]$. In particular, a standard trivial framed starting sphere yields a sphere framed-isotopic to it in $N_{q+1}$. This proves the assertion, including zero coefficients and an empty upper-handle list.