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


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*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.
 A: 
  
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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.

  
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*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.

  
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*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.
