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I don't have a clue for question (b) at all. Can I get some help?

Let $A,B \subset S^n$ be disjoint closed subsets. The Smooth Urysohn Theorem guarantees that there is a smooth $\phi: S^n \to [0,1]$ with $\phi|_A \equiv 0$ and $\phi|_B \equiv 1$.

(a) Prove that there is a compact, oriented, codimension-1 submanifold $X$ in $S^n$ that is disjoint from $A$ and $B$ and separates them in the sense that any curve from $A$ to $B$ must intersect $X$.

(b) For any $X$ as in (a), suppose $p$ is in $S^n \subset \mathbb{R}^{n+1}$ with $p, -p \notin X$. Let $S_p$ be the unit $(n-1)$-sphere centered at 0 in the boundary of the half-space in $\mathbb{R}^{n+1}$ for which $p$ is the outward normal. (Note that this definition orients $S_p$.) Let $\pi_p: S^n - \{\pm p\} \to S_p$ be the obvious projection, and let $\delta(p) = \deg(\pi_p|X)$. Prove that $X$ can be constructed as in (a) so that $| \delta{p}|<10$ for all $p$. What are the possible values of $\delta$ in your construction? Prove your answer.

WishingFish
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    Small nitpicking: you want $y \in (0,1)$, because if $y=0$ or $y=1$, then $X=\Phi^{-1}(Y)$ will contain either $A$ or $B$. – Najib Idrissi Jul 28 '13 at 07:57
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    The codimension and orientability are given by the fact that $X$ is globally defined by $\phi$ (as a level set of a regular value). The compactness, by continuity of $\phi$ and the compactness of the surrounding space ($S^n$). – Daniel Fischer Jul 28 '13 at 08:19
  • Exactly @nik, I should correct it. Thanks so much! – WishingFish Jul 28 '13 at 20:03
  • Hi @DanielFischer, I get your points - thanks. I added it in my question - hope I got it correctly. However, I couldn't follow your argument proving orientability. Could you give me some (more) hint? Thanks. – WishingFish Jul 28 '13 at 20:26
  • When you have a $1$-codimensional submanifold $S$ of an orientable manifold $M$, you can locally define orientations on $S$ by specifying a unit normal vector (orthogonal to $T_x S$ in $T_x M$). The local orientations fit together if ... (something that is given here). – Daniel Fischer Jul 28 '13 at 20:35
  • While you are editing the question, consider the title. "A [tag name] problem" title adds zero information to the tag. – 40 votes Jul 28 '13 at 20:43
  • Hey I've been thinking for a name, @40votes~ – WishingFish Jul 28 '13 at 20:45
  • Besides, I've seen titles called, a problem from GP... – WishingFish Jul 28 '13 at 20:46
  • @WishingFish That's a well-known effect: seeing litter around, people add their own more freely. (A severe comparison, I know; don't take it literally.) "Existence of a separating codimension 1 submanifold" is one of options for the title. Second, hiding the actual question (b) at the bottom of a long post makes it hard to read. There's no reason why thanks to a user and thanks to another user should precede the actual question. – 40 votes Jul 28 '13 at 20:53
  • Back to math: in (a), the gradient of $\phi$ gives you a normal vector, and thus orientation of the submanifold. – 40 votes Jul 28 '13 at 20:57
  • Here's a historical reason for hiding actual question (b) at the end - (a) was actually a question... Thanks for pointing out though, I modified.. And if you guys didn't answer/hint in the comment, I would have a more normal way(upvote and/or accept answer) to thank... And about the title, from the beginning, especailly at this point, is more focused on (b), which is not easy to verbalize it... – WishingFish Jul 28 '13 at 21:13

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