$S^{n + m}$ 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 $\dim H^m(S^{n+m})$?
