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Let $F:M\to N$ be a submersion. I want to show that for any submanifold $S\subset N$, $F^{-1}(S)$ is a submanifold of $M$.

We have not yet covered transversality theorems. However, we have regular value theorem that states the preimage of regular value under submersion is a submanifold. Can we somehow use regular value theorem to prove this? Any hints would be appreciated! Thanks.

jack
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1 Answers1

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HINT: This is a local question, so choose $p\in F^{-1}(S)$ and let $q=F(p)$. In a neighborhood $U$ of $q$ you can represent $S$ as the preimage of a regular value of a map $g\colon U\to\Bbb R^\ell$.

Ted Shifrin
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  • Sorry but I don't see how can we use this to show $F^{-1}(S)$ is a submanifold... Why do we need to show $S$ is preimage of a regular value? Since we have assumed $S$ is a submanifold of $N$. Could you elaborate more on it? – jack Mar 06 '17 at 19:36
  • Well, if $S=g^{-1}(0)$, what is $F^{-1}(S)$? – Ted Shifrin Mar 06 '17 at 20:18
  • How can we say globally $S=g^{-1}(0)$? According to your statement, shouldn't it only hold when $S\subset U$? – jack Mar 07 '17 at 02:47
  • I said "if." I wanted you to see the big picture here. Yes, officially we're talking about $S\cap U$ and $F^{-1}(S\cap U)$. Now what's the answer to the question? – Ted Shifrin Mar 07 '17 at 03:25
  • @TedShifrin this might be a stupid question, but how do we know there is a map $g$ such that $S$ is the preimage of a regular value of $g$? – Mikkel Rev Apr 03 '20 at 00:49
  • @MikkelRev Locally there is a chart in which $S$ is a linear slice. – Ted Shifrin Apr 03 '20 at 04:33