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Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite areas such that for each simply finite area there exist a diffeomorphic map that maps the area to a m-sphere? Bounded means the area $A<\infty$ with the boundary defined by the polynomials. To partition means to partition with k-dimension manifold($k=m-1$) as boundary(the intersection of the two subset) .

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    $\begingroup$ What does "partitioned" mean? A disjoint union? (This clearly can't happen.) An open cover? $\endgroup$ – Qiaochu Yuan Nov 15 '14 at 6:43
  • $\begingroup$ @QiaochuYuan, thank you, I will edit it to clarify the question. $\endgroup$ – XL _At_Here_There Nov 15 '14 at 6:53
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    $\begingroup$ I still don't understand the question. The $n$-sphere cannot even immerse into $\mathbb{R}^n$. Did you mean the $n-1$-sphere, or maybe the $n$-disk? $\endgroup$ – Qiaochu Yuan Nov 15 '14 at 7:17
  • $\begingroup$ @QiaochuYuan, thank you for your patience. Let me give a example in 1-dimension, $\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx $, [a,b] is partitioned into [a,c],[c,b]. $\endgroup$ – XL _At_Here_There Nov 15 '14 at 7:25
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    $\begingroup$ it is well-known that any semialgebraic set can be triangulated. The question you ask seems like a particular case of this fact. $\endgroup$ – Dima Pasechnik Nov 15 '14 at 9:41
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See for instance:

S. Łojasiewicz, Triangulation of semi-analytic sets, Ann. Scuola Norm. Sup. di Pisa, ser. 3, 18.4 (1964), pp. 449–474,

or start a search with the key-words "Triangulation" and "semi-algebraic", for more recent results, or more suitable to your needs.

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