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I have two compact manifolds $M$ and $N$ and a smooth map $f:M\to N$ with constant rank $k \leq \dim M$.

  • Is it true that $f(M)$ is a submanifold in $N$?
  • If not, what other properties do I need?
  • Would it suffice that $f$ is continuous instead of smooth?

I've been trying to see how this could work using the constant rank level set theorem by constructing a sort of "inverse" $g : N \to M$, but that approach seems to fail, as $f$ can be non-injective.

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  • $\begingroup$ Oh darn, thanks for that example... So that basically means I need to have injectivity, right? $\endgroup$ – Lattice Jun 6 '19 at 13:10
  • $\begingroup$ I clearified the point about continuity, I was hoping that $f$ can be continuous instead of smooth $\endgroup$ – Lattice Jun 6 '19 at 13:13
  • $\begingroup$ With some extra assumptions you can reduce the requirement that the map is differentiable to just Lipschitz continuous; see "On the inverse function theorem" by F. H. Clarke. $\endgroup$ – S. Dewar Mar 11 at 15:27
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  • Is it true that $f(M)$ is a submanifold in $N$?

No. When $k=\dim M$ then $f$ is also known as immersion. And the simpliest counterexample is a smooth immersion $S^1\to S^2$ with a single self-intersection, i.e. the $\infty$ shape in $S^2$ which is not even a topological manifold.

  • If not, what other properties do I need?

When $f$ is an injective immersion then (since $M$ is compact) it is an embedding. And in that situation $f(M)$ is a submanifold of $N$ diffeomorphic to $M$.

The case when $f$ is not injective (but still immersion) we've already discussed earlier and I'm not sure about $k<\dim M$ case unfortunately.

Note that in some situations $f(M)$ is a submanifold even when $f$ is not injective. E.g. when $N=\{*\}$ is the trivial $0$-manifold or the double winding map $S^1\to S^1$, $z\mapsto z^2$. So the property you are looking for is probably far from trivial.

  • Would it suffice that $f$ is continuous instead of smooth?

Of course not. First of all smooth functions are continuous so previous counterexample applies. Secondly if $f$ is not smooth (or at least differentiable) then "constant rank" condition is meaningless. And in this situation everything can happen. For example every Peano space (i.e. a compact, connected, locally connected, second-countable space) is an image of $S^1$ via a space filling curve by the Hahn–Mazurkiewicz theorem.

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