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Let $f:X\to Y$ be a homotopy equivalence between closed smooth manifolds. Is there a closed manifold $Z$ such that $X$ and $Y$ are (strong) deformation retracts of $Z$? (There is the mapping cylinder, but it isn't a manifold.)

EDIT. As Ryan Budney pointed out, there are counterexamples in dimension 3. Probably, the right question must include $\dim X=\dim Y>4$ (or maybe even higher dimension).

EDIT. Not a very good question. The dimension is a homotopy invariant of a closed manifold, so $Z$ must have the same dimension, which can't be. Why nobody pointed it out?

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1 Answer 1

Generally the answer is no. For example, the 3-dimensional lens spaces $L_{7,2}$ and $L_{7,1}$ are homotopy-equivalent but not diffeomorphic. Let $f : L_{7,2} \to L_{7,1}$ be the homotopy equivalence. If you had your manifold $Z$, that would mean the two lens spaces are h-cobordant. But h-cobordant lens spaces are diffeomorphic, see Agol's response in this thread:

4-dimensional h-cobordisms

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Ryan, are you maybe answering a different (more interesting) question, where $Z$ is allowed to be a manifold with boundary? – Dylan Thurston May 17 '14 at 14:33

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