A homology cobordism need not be an h-cobordism

(The $1$-handlebody.) Let $Y$ be the $0$-handle with the two $1$-handles attached orientation-compatibly. It is compact, connected and oriented, and has the homotopy type of the wedge of two circles by [F1], so ${\pi }_1(Y)$ is freely generated by $a,b$. Its boundary is connected: each $1$-handle deletes two $5$-disks from the connected boundary and joins their boundary spheres by the connected cylinder $[-1,1]\times S^4$. The dual decomposition of $(Y,\mathrm{\partial }Y)$ has only indices $5,6$ by [F4]. A relative CW pair with cells only in dimensions at least $3$ induces a fundamental-group isomorphism: cellular approximation moves loops into the boundary and homotopies of loops into the boundary because their domains have dimensions $1,2$. Hence ${\pi }_1(\mathrm{\partial }Y)\to {\pi }_1(Y)$ is an isomorphism and the words $w_1,w_2$ can be represented by loops in $\mathrm{\partial }Y$.

(Embedded framed attaching loops.) Apply [F2] first to smooth representatives and then to their disjoint union $S^1\bigsqcup S^1\to \mathrm{\partial }Y$. The target has dimension $5\ge 2\cdot 1+1$, so a homotopic embedding gives two disjoint embedded circles representing the words (up to basepoint conjugacy, which leaves their normal closure unchanged). Each oriented circle in the oriented $5$-manifold has an oriented rank-four normal bundle. Pull that bundle back to $[0,1]$ by cutting the circle at one point: finite successive local trivializations give a smooth oriented frame over the interval, arranged near the endpoints to be the pullbacks of one fixed seam trivialization up to constant matrices. The endpoint gluing is then a constant $A\in {\mathrm{G}\mathrm{L}}^{+}(4,\mathbb{R})$. By [F3] join the identity to $A^{-1}$ by a smooth path, made constant near its endpoints, and multiply the interval frame by that path. The adjusted frames agree under endpoint gluing and are smooth across the seam, giving a global framing. Tubular neighbourhoods supply disjoint attaching regions $S^1\times D^4$, so attach two orientation-compatible $2$-handles to obtain a compact oriented smooth $6$-manifold $X$.

(Boundary connectedness and group.) Removing the two circle cores from $\mathrm{\partial }Y$ leaves it path connected: a path between two points can be perturbed relative to endpoints to be transverse to those circles, and $1+1<5$ forces the perturbed path to miss them. Radially pushing the punctured normal $4$-disks outward shows that deleting the interiors of small tubular neighbourhoods also leaves a connected complement. Each $2$-handle replaces $S^1\times D^4$ by the connected $D^2\times S^3$, glued along the connected $S^1\times S^3$; hence $N=\mathrm{\partial }X$ is connected. It is a closed smooth $5$-manifold. The dual decomposition of $(X,N)$ has indices $4,5,6$, so the same relative cellular-approximation argument as in step 1.1 gives ${\pi }_1(N)\cong {\pi }_1(X)$.

(Acyclicity.) By [F1], $X$ has one $0$-cell, two $1$-cells and two $2$-cells with attaching words $w_1,w_2$. Traversing a letter contributes its signed exponent to the incidence degree on the corresponding $1$-cell; the exponent vectors are $(2,-3)$ for $w_1$ and $(-3,5)$ for $w_2$, since $2-5=-3$ for the exponent of $a$ in $w_2$. Thus the cellular chain complex is $0\to {\mathbb{Z}}^2\stackrel{d_2}{\to }{\mathbb{Z}}^2\stackrel{0}{\to }\mathbb{Z}\to 0$, with $d_2=\left(\begin{array}{cc}2& -3\\ -3& 5\end{array}\right)$. Its determinant is $10-9=1$ and its integral inverse is $\left(\begin{array}{cc}5& 3\\ 3& 2\end{array}\right)$. Consequently $H_0(X;\mathbb{Z})=\mathbb{Z}$ and ${\tilde{H}}_j(X;\mathbb{Z})=0$ for every $j$.

(A nontrivial fundamental group.) Van Kampen gives ${\pi }_1(X)=⟨a,b\mid a^2{b}^{-3}=1,a^2(b^{-1}a{)}^{-5}=1⟩$. Map $a$ to $(12)(34)$ and $b$ to $(135)$ in the permutation group on five letters, composing right to left. Then $a^2=b^3=1$, and direct composition gives $b^{-1}a=(12534)$, of order $5$, so both relators map to the identity. This defines a homomorphism from ${\pi }_1(X)$ whose value on $a$ is nonidentity, proving ${\pi }_1(X)\mathrm{\ne }1$; step 3.1 also gives ${\pi }_1(N)\mathrm{\ne }1$.

(Puncturing and the open cover.) Choose one interior coordinate disk $j:B^6\hookrightarrow \mathrm{int}X$, let $D=j(B^6)$ and $S=\mathrm{\partial }D\cong S^5$, and put $W=X\setminus \mathrm{int}D$. It is a compact oriented smooth manifold with boundary $N\bigsqcup S$. Set $D_0=j(\{\mathrm{\mid }x\mathrm{\mid }\le 1\mathrm{/}2\})$, $U=X\setminus D_0$, $V=\mathrm{int}D$. These are open in $X$, cover it, and overlap in $j(\{1\mathrm{/}2<\mathrm{\mid }x\mathrm{\mid }<1\})\cong S^5\times (1\mathrm{/}2,1)$. The overlap retracts onto its radius-$3\mathrm{/}4$ sphere, and $U$ strongly deformation retracts onto $W$ by pushing radius $r$ to $(1-t)r+t$ on $1\mathrm{/}2<r\le 1$ and fixing $W$. The radius-$3\mathrm{/}4$ sphere inclusion and $S$ inclusion in $U$ are homotopic by interpolating radii. The open ball $V$ contracts radially to its centre. The reduced degree-zero Mayer--Vietoris sequence, with connected overlap, connected $V$ and connected $X$, gives ${\tilde{H}}_0(U)=0$; since $U$ is a manifold it is locally path connected and therefore path connected, as is $W$.

(The sphere end is a homology equivalence.) For $j\ge 1$, the Mayer--Vietoris map $H_j(U\cap V)\to H_j(U)\oplus H_j(V)$ is an isomorphism because both adjacent positive-degree groups of $X$ vanish by step 3.2 and $H_j(V)=0$. Via the radial homotopies in step 4.2 this is exactly the inclusion-induced map $H_j(S;\mathbb{Z})\to H_j(W;\mathbb{Z})$. In degree zero it is an isomorphism since $S,W$ are nonempty and connected. The long exact sequence of $(W,S)$ therefore gives $H_j(W,S;\mathbb{Z})=0$ for all $j\ge 0$.

(The other end is a homology equivalence.) The relative singular chain complex is free over $\mathbb{Z}$ (its basis is the singular simplices not lying wholly in $S$). Its homology vanishes by step 5.1, so the universal coefficient sequence in [F7] has zero Hom and Ext terms and gives $H^p(W,S;\mathbb{Z})=0$ for every $p$. Duality with $\mathrm{\partial }W=N\bigsqcup S$ then gives $H_j(W,N;\mathbb{Z})=H^{6-j}(W,S;\mathbb{Z})=0$ for $0\le j\le 6$; for $j>6$, apply the Morse/handle correspondence of [F4] to the compact triad $(W;N,S)$ and the CW comparison of [F1]: its relative cells have indices at most $6$, so those higher relative homology groups also vanish. Thus the long exact sequence of $(W,N)$ makes its inclusion an integral homology isomorphism in every degree.

(Failure of the h-cobordism condition.) In the cover of step 4.2, $V$ is contractible and $U\cap V\simeq S^5$ is simply connected; van Kampen gives ${\pi }_1(U)\cong {\pi }_1(X)$. The radial deformation retraction identifies ${\pi }_1(W)\cong {\pi }_1(U)\mathrm{\ne }1$ by step 4.1. But ${\pi }_1(S)={\pi }_1(S^5)=1$ by [F5], so $S\hookrightarrow W$ cannot be a homotopy equivalence by [F8]. Both end inclusions are integral homology isomorphisms by steps 5.1 and 6.1, yet one fails the defining homotopy-equivalence condition. Hence $W$ is a homology cobordism which is not an h-cobordism, refuting the claim.