# Can we know that a map is a submersion with using only its level sets

Let $$M$$ be a manifold of dimension $$2n$$, $$f:M\rightarrow\mathbb{R}^n$$ a smooth map.

We know that if $$f$$ is a submersion, then for any $$c\in f(M)$$, $$f^{-1}(c)$$ is a $$n$$-submanifold of $$M$$.

But what about the converse, if $$f$$ is surjective and if every level set of $$f$$ is a $$n$$-submanifold of $$M$$, does that imply that $$f$$ is a submersion?

• Small mistake: the submanifolds have codimension $n$, i.e dimension $\dim M - n$ (for $c$ in the image of $f$ otherwise the level set is empty). May 26 at 21:01
• @peek-a-boo: Here, $M$ has dimension $2n$, so $\dim M-n=n$. But that doesn't seem essential, so maybe drop the dimension of $M$, and then it has to be as you say. May 26 at 21:33
• @TobyBartels it seems the question was edited after I made the comment May 26 at 21:34
• @peek-a-boo: Ah, that explains it! May 26 at 21:34

No, the converse is false. To come up with a counterexample, let us take $$M$$ to be a vector space $$\Bbb{R}^m$$. Note that if $$f$$ were linear, then the converse is true (rank-nullity), so to get a counterexample, we need to use non-linear functions.
So, consider $$f:\Bbb{R}^2\to\Bbb{R}$$ defined as $$f(x,y)=x^3-y^3$$, which is clearly surjective, and has $$f'(x,y)= (3x^2, -3y^2)$$ which vanishes only at the origin; so $$f$$ is surjective but not a submersion. Now, for any $$c\in\Bbb{R}\setminus\{0\}$$, we have that $$(0,0)\notin f^{-1}(\{c\})$$ and thus at each $$p\in f^{-1}(\{c\})$$, we have $$f'(p)$$ being non-zero and thus $$f^{-1}(\{c\})$$ is an embedded 1-dimensional submanifold of $$\Bbb{R}^2$$. Also, the level set $$f^{-1}(\{0\})$$ is just the line $$\{(x,y)\in\Bbb{R}^2\,:\, x=y\}$$ which is clearly a 1-dimensional embedded submanifold. Hence, $$f$$ provides a counterexample to the converse.
• Easier: Take $f(x,y)=y^3$. May 27 at 1:24