Every submanifold of $\mathbb R^n$ is locally a level set

Question

Is it true a very submanifold $M$ of $\mathbb R^n$ is locally a level set? Given a chart $\phi$ about $p \in M$, how can we construct a smooth function $f$ s.t. $f^{-1}(0)= M \cap U$ for some open neighbourhood $U$ of $p$?

Answer 1

By assumption, we have a parametrization $\Psi:V\subset\mathbb{R}^k\to M\subset\mathbb{R}^n$, with $\Psi(0)=p.$ Think of $\Psi$ as a map into $\mathbb{R}^n$. The differential matrix $d\Psi_0$ has degree $k$, and without loss of generality, we assume the first $k$ rows of $d\Psi_0$ are linearly independent. We define $$\chi:V\times\mathbb{R}^{n-k}\to\mathbb{R}^n$$by$$w=(x_1,\ldots,x_k,y_1,\ldots,y_{n-k})\mapsto(\Psi_{1}(x),\ldots,\Psi_k(x),\Psi_{k+1}(x)+y_1,\ldots,\Psi_n(x)+y_{n-k}).$$The differential matrix of $\chi$ at $0$ is$$d\chi_0=\left(\begin{array}{cc}\mathrm{something}\;\mathrm{invertible}&0\\*&I\end{array}\right),$$and by the inverse function theorem, $\chi$ is a diffeomorphism on some neighborhood of $0$. Let $\pi:\mathbb{R}^n\to\mathbb{R}^{n-k}$ denote the projection on the right $n-k$ coordinates. The composition $\pi\circ\chi^{-1}$ is the function you are looking for.