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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1$\begingroup$ What does "partitioned" mean? A disjoint union? (This clearly can't happen.) An open cover? $\endgroup$– Qiaochu YuanCommented Nov 15, 2014 at 6:43
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$\begingroup$ @QiaochuYuan, thank you, I will edit it to clarify the question. $\endgroup$– XL _At_Here_ThereCommented Nov 15, 2014 at 6:53
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1$\begingroup$ 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 YuanCommented Nov 15, 2014 at 7:17
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$\begingroup$ @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_ThereCommented Nov 15, 2014 at 7:25
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1$\begingroup$ 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 PasechnikCommented Nov 15, 2014 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.
answered Nov 15, 2014 at 9:42
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$\begingroup$ Thank you, I will read the article, and search with the keywords. $\endgroup$ Commented Nov 15, 2014 at 9:48
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1$\begingroup$ Actually there is a lot of material online. Chapter 3 of these notes: perso.univ-rennes1.fr/michel.coste/polyens/SAG.pdf $\endgroup$ Commented Nov 15, 2014 at 10:25
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$\begingroup$ Yes, I have just found a lot of material online. Thank you again. $\endgroup$ Commented Nov 15, 2014 at 10:36