I'm currently studying manifolds and wanted to have a detailed insight on a part of some proof. This might be very easy, but I can't find the good words to express the correct idea.
My definition of manifold is the following (I'm not sure about the word manifold as the texts I'm reading are not in English, feel free to change it if needed) :
A subset $M \subset \mathbb{R}^n$ is a manifold of dimension $p$ of $\mathbb{R}^n$ if $\forall x \in M$, there exists $U$ an open neighbourhood of $x$ and a neighbourhood $V$ of $0$ in $\mathbb{R}^n$ and a diffeomorphism $f : U \rightarrow V$ such as $f(U \cap M) = V \cap (\mathbb{R}^p \times \{0\})$.
Here is the assertion : M is a manifold of $\mathbb{R}^n$ $\implies$ $\forall a \in M$, $\exists U \subset \mathbb{R}^n$ open, $a \in U$ and $g : U\rightarrow \mathbb{R}^{n-p}$ such as $U \cap M = g^{-1}(0)$ and $g$ is a submersion.
The proof is like the following :
We define $g$ such as $g = (f_{p+1}, \ldots, f_n)$ with $f_i$ coordinates functions of $f$.
Then $f(g^{-1}(0)) = \{ (y_1, \ldots, y_p, 0, \ldots, 0) \in V \}$ so it follows that $g^{-1}(0) = U \cap M$ (by the way, is this correct to say ?)
Then we have to prove that $g$ is a submersion, i.e. that $dg_x$ is surjective $\forall x \in U$.
Many texts do not mention this part as it is obvious to them, but I'd like to prove it with a few sentences, just to convince myself that it works easily. I tried to use the rank theorem that gives the equality : $n = rg(dg_x) + dim(Ker(dg_x))$. And then show that $dim(Ker(dg_x)) = p$. But I can't find the good words to have clear way of saying it.
It might sounds weird as many books don't even bother with these kinds of considerations but I feel like it does help to understand the proof in details.