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What I tried: let $f:M\to N$ be a submersion, $S$ be a submanifold in $N$. Since by regular value theorem, for $q\in N$, $f^ {-1}(q)$ is a submanifold. Then $f^{-1}(S)$ is union of disjoint submanifolds, so $f^{-1}(S)$ is a submanifold.

I think the above proof is wrong, since it didn't use the condition $S$ is a submanifold in $N$. But I don't know where it is wrong.

Edited: Since the above proof is wrong, I was just wondering I haven't learned anything about transversality theorem, then how should I prove the title using regular value theorem?

  • why is a disjoint union of submanifolds a submanifold? – peek-a-boo Jun 17 '22 at 21:53
  • @peek-a-boo I think that similar as union of disjoint manifolds is manifolds? –  Jun 17 '22 at 21:57
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    Let $X$ be the $x$-axis, let $Y_+$ be the positive $y$-axis (${(x,y)\in\Bbb{R}^2,:, x=0,y>0}$). These are both embedded 1-dimensional submanifolds of $\Bbb{R}^2$. They are also disjoint. But their union is not a submanifold of $\Bbb{R}^2$. What you're probably thinking of is taking two manifolds $M_1,M_2$, and taking the disjoint union $M=M_1\sqcup M_2$, equipped with the disjoint union topology and corresponding smooth structure. But here we're talking about embedded submanifolds, so you're not free to mess around with the structures as you wish. – peek-a-boo Jun 17 '22 at 22:01
  • @peek-a-boo Oh, I see, thank you very much! I was just wondering I haven't learned anything about transversality theorem, then how should I prove the title using regular value theorem? –  Jun 17 '22 at 22:08

1 Answers1

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The nice thing about this question is that it automatically leads you to the definition of transversality. Fix a point $a\in f^{-1}(S)$, and consider the point $b:=f(a)\in S$. Since $S$ is a submanifold of $N$, there is an open neighborhood $V$ of $b$ in $N$ and a submersion $\pi:V\to \Bbb{R}^{\dim N - \dim S}$ such that $V\cap S=\pi^{-1}(\{0\})$. This is just saying submanifolds are locally level sets of submersions. Now, consider the restricted map $\pi\circ f: f^{-1}(V)\to \Bbb{R}^{\dim N - \dim S}$.

  • What is the $0$ level set of the restricted map $\pi\circ f$?
  • Is $0$ a regular value for $\pi\circ f$?
  • Bonus: rewrite an equivalent condition for $0$ being a regular value for $\pi\circ f$ in terms of only $f$ and the submanifold $S$. If you do this successfully, you'll be led to the definition of transversality.
peek-a-boo
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  • Thanks again!!! For (1) I think 0 level set is $f^{-1}(V)\cap f^{-1}(S)$. For (2) I think 0 should be a regular value, but I am not sure how to prove it? Then $f^{-1}(V)\cap f^{-1}(S)$ is a submanifold of M, but I am not sure how to conclude preimage of S is a submanifold from here? –  Jun 17 '22 at 22:48
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    @diiiiiklllllll What kind of map is $f$? What kind of map is $\pi$? So, what kind of map is $\pi\circ f$? Lastly, being a submanifold is a local condition (about every point, there is an open neighborhood such that blablabla... now look again at what this argument shows) – peek-a-boo Jun 17 '22 at 22:51
  • I think since both $\pi$ and $f$ are submersion, and composition of submersions is submersion then $\pi\circ f$ is a submersion, so need not to show that at $0$, $\pi\circ f$ has full rank? And I am not sure why $\pi$ is submersion? And to be a submanifold of M, need to find a chart $(U \phi ) $ of M, s.t. $\phi(U\cap \mathbb{R}^k)=\phi(U)\cap \mathbb{R}^k$, I just can not see why $f^{-1}(V)\cap f^{-1}(S)$ is a submanifold suggests such a chart? –  Jun 17 '22 at 23:29
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    @diiiiiklllllll $\pi\circ f$ being a submersion by definition means at each point it has full rank. $\pi$ is a submersion because I chose it to be (and I can do this because $S$ is a submanifold... review the equivalent ways of describing submanifolds). Lastly, I showed for every $a\in f^{-1}(S)$, there is an open set $U$ (namely $f^{-1}(V)$) such that $U\cap f^{-1}(S)$ is the zero level set of a submersion ($\pi\circ f$). This is equivalent to $f^{-1}(S)$ being a submanifold. You should review the equivalent ways of describing submanifolds. – peek-a-boo Jun 17 '22 at 23:35
  • Thanks, I just start learning differential geometry, and I haven't seen any equivalent definitions of submanifold, and don't understand why S is submanifold suggests such $\pi$ exists. Maybe need more time to think about that. –  Jun 17 '22 at 23:55
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    Specifically, I'm referring to the following: "Let $N$ be an $n$-dimensional smooth manifold, and $S\subset N$. Then, $S$ is a smooth $k$-dimensional embedded submanifold of $N$ if and only if for each point $p\in S$, there is an open set $U\subset N$ about $p$ and a smooth submersion $\pi:U\to\Bbb{R}^{n-k}$ such that $U\cap S=\pi^{-1}({0})$". The implication $\implies$ follows immediately from the slice-chart definition you quoted; just take $\pi=(\phi^{k+1},\dots, \phi^{k+n})$. For the converse, you use the regular value theorem to generate the slice charts. – peek-a-boo Jun 17 '22 at 23:56
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    ^ sorry I meant $\pi=(\phi^{k+1},\dots, \phi^n)$, not $\phi^{k+n}$, i.e $\pi$ is just the last $n-k$ component functions of $\phi$. – peek-a-boo Jun 18 '22 at 00:01