Diffeomorphisms to $S^n$ Is $S^4$ diffeomorfhic to $S^2\times S^2$?
Moreover. Is $S^n$ diffeomorphic to some cross product of manifolds $X\times Y$ for $n\geq2$?
Is there a elemental topological invariant to let me see this?
Any suggestions are welcome!
thanks
 A: The second homotopy group, or the second homology group of $S^2\times S^2$ is isomorphic to $\mathbb Z\times\mathbb Z$, which that of $S^4$ is zero, so the two spaces are not homeomorphic.
Suppose that $S^n\cong X\times Y$, and fix an homeo. Pick points $x\in X$ and $y\in Y$, and let $r$ be the point in $S^k$ corresponding to $(x,y)$. The Künneth formula for relative homology tells us then that $$H_p(S^n,S^n\setminus r)\cong H_p(X\times Y,X\times Y\setminus\{(x,y)\}\cong\bigoplus_{p'+p''=p}H_{p'}(X,X\setminus x)\otimes H_{p''}(Y,Y\setminus y),$$ taking coefficients in any field, for all $p\geq0$.
If $X$ and $Y$ are manifolds of dimensions $n'$ and $n''$, then this implies that $H_{n'-1}(S^n,S^n\setminus r)\neq0$ and $H_{n''-1}(S^n,S^n\setminus r)\neq0$, because in the direct sum decomposition above there are non-zero terms. Moreover, we know thhat $H_{n-1}(S^n,S^n\setminus r)\neq0$.
Since the sphere is a manifold, this is only possible if the set $\{n,n',n''\}$ has only two elements. Since $n=n'+n''$, one of $n'$ or $n''$ must be zero. Suppose $n'=0$. Since $X$ and $Y$ must be connected, because $S^n$ is, we see that $X$ is point.
You can probably push this to $X$ and $Y$ any space, as in https://mathoverflow.net/questions/60375/is-r3-the-square-of-some-topological-space
