[**Edit:** I have added some details and a more explicit example by Milnor.] I will present a couple of examples verifying the conditions required in the question.$\newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\ZZ}{\mathbb{Z}}$ ### $H$-cobordant manifolds ### We will make fundamental use of the following fact: if $M$ and $N$ are $h$-cobordant, closed, smooth manifolds of dimension greater than 3, then $M\times\RR$ is diffeomorphic to $N\times\RR$. [*Aside:* A similar statement holds for topological manifolds; simply replace diffeomorphism with homeomorphism.] Here are a couple of quick references for this fact, both making use of the $s$-cobordism theorem: - In a related mathoverflow question, Oscar Randal-Williams provides a fairly direct argument in a <a href="http://mathoverflow.net/questions/95129/manifolds-with-homeomorphic-interiors/95136#95136">comment to one of the answers</a>. - The article "<a href="http://arxiv.org/abs/1212.5618">*Contact manifolds with symplectomorphic symplectizations*</a>" states the result for smooth manifolds as corollary 2.5. [The proof described there applies also to topological manifolds.] [*Further aside*: Incidentally, the converse also holds — although we will not make use of it. In another mathoverflow question, Igor Belegradek remarks in <a href="http://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere/64036#64036">his answer and the ensuing comments</a> that if $M\times\RR$ is homeomorphic to $N\times\RR$, then $M$ and $N$ are $h$-cobordant.] So it suffices to find closed smooth manifolds $M$ and $N$ which are $h$-cobordant, yet not homeomorphic. As required by the question, we also want $M$ to be null-cobordant; then $N$ will be null-cobordant as well. As a minor bonus, since $M$ and $N$ are smoothly $h$-cobordant, $M\times\RR$ is actually *diffeomorphic* to $N\times\RR$. <!-- Obviously, we could also use closed *topological* manifolds $M$, $N$ which are $h$-cobordant but not homeomorphic. Then we would obtain only a homeomorphism between $M\times\RR$ and $N\times\RR$. --> ### An example by Farrel and Hsiang ### One such example is given by Farrel and Hsiang in their article "<a href="http://projecteuclid.org/euclid.bams/1183529046">*$H$-cobordant manifolds are not necessarily homeomorphic*</a>". There, the manifold $M$ is a product $M=L\times\TT^n$, where $L$ is any 3-dimensional lens space such that $\pi_1(L)\simeq\ZZ/p^2$ for some prime $p$, and $\TT^n=(S^1)^{\times n}$ is a torus of dimension $n\geq 3$. Note that $M$ is null-cobordant since the torus $\TT^n$ is null-cobordant. Unfortunately, the methods used by Farrel and Hsiang do not yield an explicit description of the manifold $N$. ### Milnor's examples ### A more explicit example is given in the celebrated article "<a href="http://www.maths.ed.ac.uk/~aar/papers/milnhaup.pdf">*Two complexes which are homeomorphic but combinatorially distinct*</a>" by John Milnor. Milnor essentially proves that we can take $M=L(7,1)\times S^{2n}$ and $N=L(7,2)\times S^{2n}$ for any $n>0$, where $L(p,q)$ denotes the lens space of type $(p,q)$. This is *only partially* stated as theorem 4 in that article. By the way, observe that $M$ and $N$ are null-cobordant since spheres are null-cobordant. I will briefly describe how to conclude from Milnor's article that the preceding $M$ and $N$ meet our requirements. In section 2, Milnor sketches a proof that $M$ and $N$ are $h$-cobordant. At the end of section 4, in the proof of corollary 2, he uses the fact that $L(7,1)$ and $L(7,2)$ have distinct Reidemeister torsions, and concludes that $M$ and $N$ also have distinct Reidemeister torsions, and thus are *not diffeomorphic*. But more is true: since the Reidemeister torsions of $M$ and $N$ differ, $M$ and $N$ are *not homeomorphic*. This follows from the topological invariance (i.e. invariance under homeomorphisms) of Reidemeister torsion, which is itself a simple consequence of the topological invariance of Whitehead torsion — proved by Chapman after the publication of Milnor's article. In general, the results in sections 2 and 4 of Milnor's article actually prove that we can take $M=K\times S^n$ and $N=L\times S^n$, where: - $K$, $L$ are parallelizable, smooth, closed $k$-manifolds; - $K$, $L$ are homotopy equivalent; - $n$ is even and $n\geq k$; - the Reidemeister torsions of $K$ and $L$ are defined (for some complex-valued representation of their common fundamental group), and have different absolute values. For instance, we can take $K$ and $L$ to be 3-dimensional lens spaces, and not just $L(7,1)$ and $L(7,2)$. The homotopical classification and Reidemeister torsions of lens spaces are well-known: see http://www.map.mpim-bonn.mpg.de/Lens_spaces. <!-- Note that we cannot take $M$ and $N$ to be three dimensional lens spaces. If $M\times\RR$ is homeomorphic to $N\times\RR$, then $M$ is $h$-cobordant to $N$ by Belegradek's remark. On the other hand, $h$-cobordant lens spaces of dimension 3 are homeomorphic. -->