It is well-known the regular value theorem holds. Does this generalization also hold? I can not think of a counter-example.
$f:M\to N$ be a smooth map of two smooth manifolds of arbitrary dimension $m,n$. For a level set $S:=f^{-1}(q)\neq\emptyset$, if $\text{rank}(f)_p\equiv r$ for all $p\in S$, then S is a smooth sub-manifold of dimension $m-r$.
The regular value theorem is just a special case of it. But do we really need the surjectivity?
Best Answer
It does not hold: consider $S^1=\{(x,y)\in\mathbb{R}^2;x^2+y^2=1\}$ and the map $p_1:S^1\to\mathbb{R}$ defined by $p_1(x,y)=y$. Then $p_1^{-1}(1)=\{(0,1)\}$ and $p_1$ has rank $0$ on $(0,1)$ (if you parametrize by $\theta\mapsto(\cos(\theta,\sin(\theta))$, then$(0,1)$ has coordinate $\frac{\pi}{2}$ and $$\frac{\partial}{\partial\theta}\hat{p_1(\theta)}|_{\frac{\pi}{2}}=\frac{\partial}{\partial \theta}\sin(\theta)|_{\frac{\pi}{2}}=0,$$ where $\hat{p_1}$ denotes $p_1$ read in these coordinates). But $\{(0,1)\}$ is not a manifold of dimension $1-0=1$.
What fails here is that the rank of $p_1$ is not constant in a neighborhood of $p_1^{-1}(1)$, and so we can't apply the constant rank theorem which would give local coordinates of the expected form around your points. That's why we require the map being a submersion at a point: by openess of being of maximal rank, you know that you will stay of maximal rank in a neighborhood of your preimage, and so on.