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Let $M$ be a closed manifold such that $M\times \mathbb{S}^1$ is a torus. Is it true that $M$ is homeomorphic to a torus?

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  • $\begingroup$ You used the 4-manifolds tag, but did not place any dimension restriction in the question. $\endgroup$ Commented Nov 14, 2022 at 11:08
  • $\begingroup$ @MichaelAlbanese I just thought that that an example might come from 4-dimensional topoogy. $\endgroup$ Commented Nov 14, 2022 at 11:29

1 Answer 1

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Suppose $M\times S^1$ is homeomorphic to $T^{n+1}$. Then $\pi_1(M\times S^1) \cong \pi_1(T^{n+1})$, so $\pi_1(M)\oplus\mathbb{Z} \cong \mathbb{Z}^{n+1}$, and hence $\pi_1(M) \cong \mathbb{Z}^n$. Moreover, as $T^{n+1}$ is aspherical, so is $M$. Since $\pi_1(M) \cong \pi_1(T^n)$ and both are aspherical, we see that $M$ is homotopy equivalent to $T^n$ (aspherical spaces are uniquely determined up to homotopy by their fundamental group). Finally, by combining the results here, a closed manifold of any dimension homotopy equivalent to a torus is in fact homeomorphic to a torus, so $M$ is homeomorphic to $T^n$.

If you ask the same question at the level of diffeomorphism, then I suspect it is not true. As this answer explains, there are exotic tori in every dimension $n \geq 5$. If $M$ is an exotic torus, I don't know if $M\times S^1$ can be diffeomorphic to the standard torus.

Added Later: I asked about the diffeomorphism analogue here.

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  • $\begingroup$ There are manifolds which are homotopy equivelant to Tori but not homeo to it in all dimensions at least 5 see FAKE TORI, THE ANNULUS CONJECTURE, AND THE CONJECTURES OF KIRBY* BY W. C. HSIANG AND J. L. SHANESON. (I am not questioning your answer but I think the word aspherical needs to be inserted into some sentance.) $\endgroup$
    – Nick L
    Commented Nov 14, 2022 at 11:40
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    $\begingroup$ @NickL No, the Hsiang-Shaneson result says that there exist PL-exotic tori. They are homeomorphic, just not PL-homeomorphic. $\endgroup$
    – mme
    Commented Nov 14, 2022 at 11:41
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    $\begingroup$ The results on tori are also collected in the answers to this question: mathoverflow.net/q/403202/40804 $\endgroup$
    – mme
    Commented Nov 14, 2022 at 11:41
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    $\begingroup$ Oh, my mistake. Thanks! $\endgroup$
    – Nick L
    Commented Nov 14, 2022 at 11:44

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