# 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 $$k \leq \dim M$$.

• 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.

• Oh darn, thanks for that example... So that basically means I need to have injectivity, right? – Lattice Jun 6 '19 at 13:10
• I clearified the point about continuity, I was hoping that $f$ can be continuous instead of smooth – Lattice Jun 6 '19 at 13:13
• 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. – S. Dewar Mar 11 at 15:27

• 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.