# $S^{n + m}$ is not homeomorphic to $M \times N$

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.

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 $$\dim H^m(S^{n+m})$$?
• Can it be shown using de Rham cohomology instead Künneth formula? (and using that $M$ and $N$ are oriented) Jan 8, 2021 at 9:39
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 $$H^m(S^{m+n})=0$$.
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.