$df_p:T_pM \to T_{f(p)}N$ has rank $k$ for every $p\in U$ , then $f^{-1}\{q\}$ is an embedded submanifold of $M$ of dimension $m-k$?

Question

Let $M,N$ be abstract smooth manifolds of dimension $m,n$ respectively , $f:M \to N$ be a smooth map such that for some open set $U $ in $M$ , $df_p:T_pM \to T_{f(p)}N$ has rank $k$ for every $p\in U$ , if $q\in f(M)$ be such that $ f^{-1}\{q\} \subseteq U$ then is it true that $f^{-1}\{q\}$ is an embedded submanifold of $M$ of dimension $m-k$ ?

Answer 1

This is a consequence of the constant rank theorem which states: If $f: M^m \to N^n$. Then by the constant rank theorem if $\textbf{rank}(Jac(Df(p))) = k$, then there exists charts $(U, \phi)=(U,x^1,...,x^m)$ and $(V, \psi)=(V, y^1,...,y^n)$ such that:

$$ \psi \circ f \circ \phi^{-1}: (x^1,...,x^m) \mapsto (x^1,...,x^k,0,...,0)$$

In your case we have $S = f^{-1}(q)$. Alter the chart $(V \cap f(M), \psi)$ such that $\psi(q) = \bf{0}$ and then:

$$S = (\psi \circ f \circ \phi^{-1})^{-1}(\textbf{0}) \Rightarrow S \cap U = \{p: x^1 = \cdots=x^k = 0\}$$

Therefore you can take the relative charts $(U \cap S, \tilde{\phi})$ where $\tilde{\phi}$ is gotten from $\phi = (x^1,...,x^m)$ by setting the last $k$ coordinates to zero. Hence $\{(U \cap S, \tilde{\phi} = (\tilde{x}^1,...,\tilde{x}^{m-k}))\}$ is an atlas on $S$ and so $\textbf{dim}(S) + k = \textbf{dim}(M)$, and we are done.