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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

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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

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ok let me see, thanks for answering. – Juan Carlos Castro Feb 5 '13 at 7:04

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