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I've seen referenced here that if $M$ and $N$ are closed topological $n$-manifolds and $f: \mathbb{R}\times M \to \mathbb{R}\times N$ is a homeomorphism, then $M$ and $N$ are h-cobordant.

I know that the first step of the proof is to consider the region between $f(0\times M)$ and $\varepsilon \times N$ in $\mathbb{R}\times N$. I see how this gives a cobordism from $M$ to $N$, but I fail to see how this region is an h-cobordism.

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  • $\begingroup$ This can't be true. Let $M$ be a twice-punctured copy of $\mathbb R^2$, and let $N$ be a once-punctured torus. Then $M \times \mathbb R$ and $N \times \mathbb R$ are homeomorphic -- they're both open genus 2 handlebodies. I imagine you want some additional assumptions like compactness. $\endgroup$
    – Ryan Budney
    Commented Jan 24 at 17:01
  • $\begingroup$ Ah! You're totally right, let's assume that the manifolds are closed. $\endgroup$
    – nick5435
    Commented Jan 24 at 17:08
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    $\begingroup$ See Igor Belegradek's comments to this question and the answer I wrote following them. Although the discussion there is in the smooth category, the arguments work exactly the same in the topological case. $\endgroup$
    – Michael Albanese
    Commented Jan 24 at 17:23
  • $\begingroup$ Thank you! This is fantastic! $\endgroup$
    – nick5435
    Commented Jan 24 at 17:28

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