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I'm thinking about whether there exists a smooth surjective map $f: M^m \twoheadrightarrow N^n$ where both $M$ and $N$ are smooth manifolds, and $\dim M = m < n = \dim N$. (We might assume that $M$ and $N$ are boundaryless, although I don't think this is necessary.)

Intuition tells me the answer is no, since smooth manifolds and smooth maps between them ought to behave nicely. Things like space-filling curves are obviously excluded from this class of nice objects.

However, I find it hard to justify such a claim. In particular, I tried to approach from the measure of $M$ and $N$ but failed. Any ideas? Thanks.

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2 Answers 2

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Sard's theorem tells us that the image has measure zero.

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  • $\begingroup$ Looks like you're scooped :) $\endgroup$
    – 4ae1e1
    Apr 29, 2014 at 4:18
  • $\begingroup$ @KevinSayHi, if that is significant to you, you can check the times at which answers are posted. $\endgroup$ Apr 29, 2014 at 4:20
  • $\begingroup$ Yeah I've already checked, you're later by one minute (less than one actually) so I accepted the other one. $\endgroup$
    – 4ae1e1
    Apr 29, 2014 at 4:21
  • $\begingroup$ @4ae1e1 no, he answered earlier actually :) $\endgroup$ Oct 1, 2021 at 15:54
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You're looking for Sard's theorem. If $m<n$, the image of $f$ consists solely of critical values, and thus must have measure zero in $N$.

Note that the proof of Sard's theorem is extremely difficult, so you shouldn't feel too bad about not discovering it!

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  • $\begingroup$ I do know Sard's theorem and have seen a proof of it (and I thought about applying it...), but the version I know is from $\mathbb{R}^m$ to $\mathbb{R}^n$, so to apply it to $M$ and $N$, I need to go to the ambient space of $N$. But the measure of $N$ could be zero in the ambient space; that's why it wasn't clear to me how to proceed. However, obviously Sard's theorem also applies to general manifolds. Then it's crystal clear. Thank you. $\endgroup$
    – 4ae1e1
    Apr 29, 2014 at 4:18

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