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I saw that a cell-decomposition of a genus g non-orientable surface is $D^0\cup D^1\cup ...\cup D^g$. Can anyone explain why this is true?

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    $\begingroup$ I think you are confusing two things. The simplest cell-decomposition for a genus 1 nonorientable surface has three cells, a $0$-cell, a $1$-cell and a $2$cell, which doesn't match your count. What you have written looks instead like a cell decomposition for $\mathbb{RP}^g$, or projective $g$-space. (The $g$-dimensional projective plane.) $\endgroup$ – Cheerful Parsnip Dec 7 '11 at 21:17
  • $\begingroup$ Oh yeah. That's why it didn't make sense to me. Thanks $\endgroup$ – 908979 Dec 7 '11 at 21:31
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As Grumpy Parsnip pointed out, the decomposition fits $g$-dimensional real projective space (listed among examples here), not a surface. A surface can't have cells of dimension more than $2$.

A non-orientable closed surface of genus $g$ can be constructed out of one $0$-cell, $g$ $1$-cells, and one $2$-cell. (Reference)

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