Let $M\subseteq\mathbb{R}^n$ be a $m$-dimensional smooth submanifold. We can prove that for all $p\in M$ there're an open neighborhood $U\subseteq\mathbb{R}^n$ of $p$ and a submersion $\varphi :U\to \mathbb{R}^{n-m}$ such that $\varphi^{-1}(\{0\})=U\cap M$.
My question is: can we prove that for all $p\in M$ there're an open neighborhood $U\subseteq\mathbb{R}^n$ of $p$ and a submersion $\varphi :U\to \mathbb{R}^{n-m}$ such that $\varphi^{-1}(\{0\})=\color{red}{ M}$?
I tried to answer the above question using the partition of unity but I failed. I also don't know any counterexample.
Thank you for your attention!
If $M$ is globally defined by as the preimage of $0\in \mathbb{R}^{n-m}$ under a submersion $\phi$ then the normal bundle of $M$ has a global frame defined by $(\operatorname{grad} \phi_1, \ldots, \operatorname{grad} \phi_{n-m})$, so it's trivial. This imposes some conditions on the manifold. In particular it has to be orientable ( since the ambient $\mathbb{R}^n$ is). I am not sure whether the triviality of the cotangent bundle implies the existence of a globally defined equation, but I am inclined to think so.
As a concrete example, take a non-orientable manifold like the Klein bottle imbedded in $\mathbb{R}^4$.
$\bf{Added:}$ The condition is also sufficient. As @Kajelad: indicated, we should consider a tubular neighborhood $U$ of $M$, which is diffeomorphic to the total space of the normal bundle. If the bundle is trivial, we can define $\phi= (\phi_1, \ldots, \phi_{n-m})$ ( the pullbacks of the coordinates in a trivialization) to define $M$ locally on $U$ as $M= M\times \{0\} = \{ \phi=0\}$.