We know that the Cartesian product of two manifolds is a manifold, but is the converse true? Let us assume that we have $A$ and $B$ two second countable Hausdorff topological spaces, and $M = A \times B$. If we assume that $M$ is a $n$-manifold, with $n \geq 0$ finite, do we obtain that $A$ and $B$ are $k$- and $l$-manifolds with $k+l = n$?
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1Do you want any relationship between the structure of $M$ and that of $A$ and $B$? For example, $M=\mathbb{R}^2$ is a $2$-dismensional smooth manifold, and it is also the product of $A={(x,y)\in\mathbb{R}^2:\ y^2=x^3}$ and $B=\mathbb{R}$. These two can be made smooth manifolds, but the structure of $A$ will no be induced by that of $M$. Maybe you want topological manifolds. – YAlexandrov Mar 17 '18 at 11:25
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You are right, I was thinking about topological manifolds. Thanks for the remark. – Nicolas Boutry Mar 17 '18 at 11:33
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The dogbone space is not a manifold but its product with $\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$. This was proved in this paper by Bing.
Michael Albanese
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Do you know if there is any sufficient condition so that "$A$ is not a manifold" implies that "$A \times B$ is not a manifold? For example if $B = ]0,1[$ and $A$ is a cubical set? – Nicolas Boutry Mar 17 '18 at 12:55
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