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Suppose I have a manifold which has a CW structure with cells $e^0 \cup e^1 \cup e^2$, where $e^i$ represents an $i$-cell. If I took the direct product of this manifold with another manifold which has cell structure say $e^0 \cup e^2$, is it acceptable to say that I get a manifold with a CW structure with cells $(e^0 \cup e^2) \times (e^0 \cup e^1 \cup e^2)=e^0 \cup e^1 \cup e^2 \cup e^2 \cup e^3 \cup e^4$?

Is the resulting manifold at least homotopy equivalent to a manifold with such a cell decomposition?

Stefan Hamcke
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Traxter
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    You may be interested in the answer to this question: http://math.stackexchange.com/questions/100170/cartesian-product-of-two-cw-complexes – fixedp Apr 30 '14 at 12:23
  • @fixedp Thanks for the link. This at least tells me that I'm getting cells of the right dimension, however that answer expresses the resulting CW complex in terms of products of cells.

    I suppose my question now is whether the product of two cells, say $e^p \times e^q$, is homotopic (or any other relation?) to a cell $e^{p+q}$?

    This seems to make sense if I look at it in lower dimensions, e.g. $e^1 \times e^2 \cong D^1 \times D^2$ would give a cylinder, which is homeomorphic to $D^3$, or a 3-cell.

     – Traxter Apr 30 '14 at 12:35
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    Yeah, if $e_\alpha$ is an $m$-cell and $e_\beta$ is an $n$-cell, then $e_\alpha \times e_\beta$ is an $(m+n)$-cell. – fixedp Apr 30 '14 at 12:40
  • Excellent! Thank you very much for your very fast help. – Traxter Apr 30 '14 at 12:48
  • @fixedp: "is" an $(m+n)$-cell only if you're using cubical cells. But, yes, in general, a product of disks is homeomorphic to a disk. – Ted Shifrin Apr 30 '14 at 13:15
  • Oops you are correct. I did mean homeomorphism. – fixedp Apr 30 '14 at 13:24

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