4

Let $f: M \to N$ a submersion between two manifolds, and let $S\subset N$ a subset of N. Proof that $S$ is a regular submanifold of $N$ if and only if $f^{-1}(S)$ is a regular submanifold of M.

So far I have half of the problem. Using the transversality theorem, we see that if $S$ is a submanifold, then using that $f$ is a submersion, we see that $S$ is transverse to $f$, so the preimage of $S$ is a submanifold of M. Now, for the second part I tried to use coordinate charts and the constant rank theorem, but nothing seems to work.

Any help will be appreciated.

alexp9
  • 879
  • FYI: In mathematics, "proof" is a noun, not a verb. The verb form is "prove." A proof is what you create when you prove something. (Proof can be used as a verb in other contexts, but not in math.) – Jack Lee Oct 04 '17 at 17:44
  • 1
    (However, the word "prove," confusingly, is pronounced as if it were spelled proov.) – Akiva Weinberger Oct 04 '17 at 17:49
  • 1
    @JackLee: I'm fine with your proofing the yeast. I'll be by for the bread in a few hours. – Ted Shifrin Oct 04 '17 at 17:50
  • 1
    Another comment: Your title is misleading. The statement you're trying to prove as about the preimage of a submanifold under a submersion. The image of a submanifold under a submersion need not be a submanifold. – Jack Lee Oct 04 '17 at 17:50
  • 2
    @TedShifrin: If you can get here in a few hours, I'll make bread for you!! [Grammatical aside: hats off for "your proofing." I hardly know anyone who worries about fused participles any more, including me! You must have proofed your comment before sending it. :-) ] – Jack Lee Oct 04 '17 at 17:55
  • 1
    @JackLee (Well, seriously, I do need to visit Seattle. I haven't been since around 1981 and you're not my only [old] friends there!!) – Ted Shifrin Oct 04 '17 at 17:56
  • Thank you all for the comments. I'll improve my redaction next time. In fact, the image of a submanifold doesn't need to be a submanifold. – alexp9 Oct 04 '17 at 17:57
  • Evidently I didn't proof my own comment ("as about" -> "is about"). Discovered that 15 seconds too late to edit. – Jack Lee Oct 04 '17 at 17:57
  • Definitely edit the title, @Rhcpy99. And work out my hint and let me know if you get it. (And if not, follow up with a question.) – Ted Shifrin Oct 04 '17 at 18:01

1 Answers1

4

HINT: First choose local coordinates on $M$ and $N$ so that $f(x_1,\dots,x_m) = (x_1,\dots,x_n)$ ($n\le m$). (Fix points corresponding to the origin and choose appropriate open sets, but I'm not going to mess with those.) Suppose the $(m-k)$-dimensional submanifold $Z=f^{-1}(S)$ given in these coordinates by $g(x)=0$, $g\colon \Bbb R^m\to\Bbb R^k$, with $0$ a regular value of $g$. Note that $g^{-1}(0)$ contains the fibers of the projection $f$.

Now define $h\colon\Bbb R^n\to\Bbb R^k$ by $h(x_1,\dots,x_n) = g(x_1,\dots,x_n,0,\dots,0)$. Then check that $h^{-1}(0) = S$ and that $0$ is a regular value of $h$. (The crucial observation is that at points of $Z$, we have $\dfrac{\partial g}{\partial x_i} = 0$ for $i=n+1,\dots,m$.)

Ted Shifrin
  • 115,160
  • 1
    You're welcome. Make sure you accept the answer when you're satisfied, so that the question will go off the unanswered list. :) – Ted Shifrin Oct 05 '17 at 14:39