If every point in $f^{-1}(0)$ is regular, is $f^{-1}(0)$ a manifold?

Question

The area of differential manifold is foreign to me. I want to find a convenient way to prove that a set is a manifold.

Here is a smooth map $f:\mathbb{R}^n\rightarrow \mathbb{R^m}$, where $m<n$. Now we know that $rank(\frac{df}{dx})$ reaches its maximum $r<n$, i.e., $r=\max\{rank(\frac{df}{dx}):x\in\mathbb{R}^n\}$, at each point $x\in f^{-1}(0)$.

Can I conclude that $f^{-1}(0)$ is a manifold of dimension $n-r$ ? How to prove it? If there is some theorem to employ, please try to give me a reference or textbook.

Thank you very much!

Answer 1

Your question is extremely unclear. But, suppose $X$ and $Y$ are smooth submanifolds of their respective ambient euclidean spaces and $f$ is transversal to $\{y\} \subset Y$ (i.e. its a regular value) then $f^{-1}(y)$ is a submanifold in $X$. Also:

$$ \dim f^{-1}(y) = \dim X - \text{codim } \{y\}$$

Even if the map is not transversal, it can be perturbed to become one. Cf. Thom's Transversality Theorem.

Answer 2

Just have a look at Lee's Introduction to Smooth Manifolds, 2nd edition. To be precise, look at Theorem 5.12 (Constant-Rank Level Set Theorem), Which I quote for convenience:

Theorem 5.12 (Constant-Rank Level Set Theorem). Let $M$ and $N$ be smooth manifolds, and let $\Phi \colon M \to N$ be a smooth map with constant rank $r$. Each level set of $\Phi$ is a properly embedded submanifold of codimension $r$ in M.