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Let $M$ and $N$ be two oriented manifolds such that $m = \dim M$ and $n = \dim N$. Using the de Rham cohomology I have to show that $S^{n + m}$ is not homeomorphic to $M \times N$. I have no idea how to proceed, but maybe the next statement could be helpful: if $M$ is an oriented and compact manifold such that $m = \dim M$, then $\dim H^m(M) \geq 1$. I think I have to use just geometry and topology statements, not analysis theorems like Fubini.

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I assume that $M,N$ are manifolds of dimensions $m,n>0$ (this part is obviously necessary, otherwise we would get trivial counterexamples as $S^n\simeq \{0\}\times S^n)$. If $M\times N\cong S^{n+m}$ then $M,N$ are compact and orientable (why?). Now notice that $\dim H^m(M)>0$ and $\dim H^0(N)>0$, so by the Künneth formula we get $\dim H^m(M\times N)\geq \dim H^m(M)\otimes_{\mathbb{R}}H^0(N)>0$, but what is $\dim H^m(S^{n+m})$?

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  • $\begingroup$ Can it be shown using de Rham cohomology instead Künneth formula? (and using that $M$ and $N$ are oriented) $\endgroup$
    – joseabp91
    Jan 8 '21 at 9:39
  • $\begingroup$ @joseabp91 One can prove the Künneth formular using de Rham comology (it is basically an application of the 5-lemma and the Mayer-Vietoris-sequence, at least in the case when the manifold is compact). But if you want a proof without the Künneth formula, look for example at this $\endgroup$
    – leoli1
    Jan 8 '21 at 9:42
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The Künneth formula is the general mechanism at play here, but this particular case can be proved ad hoc without much effort. Take a volume form $\omega\in\Omega^m(M)$ and pull it back along the projection $\pi\colon M\times N\rightarrow M$ to a form $\pi^{\ast}\omega\in\Omega^m(M\times N)$. Pick a point $p\in N$ and consider the integral $\int_{M\times\{p\}}\pi^{\ast}\omega$. Can you relate this integral to $\int_M\omega$? What could you say about this integral if $H^m(M\times N)=0$? This will lead to a contradiction with $H^m(S^{m+n})=0$.

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I think an idea would be to use the Kunneth Formula applied to the cohomology groups of $H^k(M\times N)$ and the fact that we know the cohomology groups of the sphere,i.e they are trivial except for the $0-$degree and the top degree, and that this are homeomorphic invariant to get a contradiction.

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