Definition 1 : A subset $M ⊂ \Bbb R^n$ is a $m$-dimensional submanifold of $\Bbb R^n$ if for all $x$ in $M$, there exists open neighborhoods $U$ and $W$ of $x$ and $0$ in $\Bbb R^n$ respectively, and a diffeomorphism $f : U \rightarrow W$ such that $f(U ∩M) = W ∩ (\Bbb R^m × \{0\}^{n-m})$.
Definition 2 : A subset $M ⊂ \Bbb R^n$ is a $m$-dimensional submanifold of $\Bbb R^n$ if for all $x$ in $M$, there exists an open neighborhood $U$ of $x$ in $\Bbb R^n$, an open set $V$ in $\Bbb R^m$, and an injective immersion $\phi : V \rightarrow U$ such that $\phi(V) = U ∩ M$ and $\phi$ is an homeomorphism between $V$ and $U ∩ M$.
I can prove $1 \Rightarrow 2$. Indeed, by definition, for all $x$ in $M$, there exists open neighborhoods $U$ and $W$ of $x$ and $0$ in $\Bbb R^n$ and a diffeomorphism $f : U \rightarrow W$ such that $f(U ∩M) = W ∩ (\Bbb R^m × \{0\}^{n-m})$. Define $V = \{(x^1, ..., x^m) \in \Bbb R^m \mid (x^1, ..., x^m, 0, ..., 0) \in W\}$ and $\phi : V \rightarrow U$ by $\phi(x) = f^{-1}(x, 0)$.
However, I'm struggling with the other direction even though I have this theorem :
Theorem (Local normal form for immersions) : Let $U$ in $\Bbb R^n$, $V$ in $\Bbb R^m$ be open set and let $\phi : V \rightarrow U$ be an injective immersion. Then for all $x$ in $V$, there exists open neighborhoods $V'$ and $U'$ of $x$ and $\phi(x)$ in $V \subset \Bbb R^m$ and $U \subset \Bbb R^n$ respectively, with $\phi(V') \subseteq U'$, and a diffeomorphism $F : U' \rightarrow F(U')$ such that $F \circ \phi(x^1, ...., x^m) = (x^1, ..., x^m, 0, ..., 0)$ on $V'$.
The function we are looking for as $f$ in Definition 1 is $F$. However, I cannot find good neighborhoods (because of the inclusion $\phi(V') \subseteq U'$) to get the result.
I have Jacques Lafontaine's book "An Introduction to Differential Manifolds" and in this book, he just says that we can restrict ourself so that it's good. Well, it's not enough of an argument for me.
Any help here ?
By local immersion theorem, we can find a diffeomorphism $\psi$ from open $V$ containing $\phi(0) = x$ to $\mathbf{R^n},\ \ni$ $$(\psi\circ\phi)(x_1, x_2,\dots,x_m) = (x_1,x_2,\dots,x_m,0,0,\dots,0)$$ Then, $$\psi(U\cap M) = \psi(\phi(V)) = \psi(U) \cap (\mathbf{R^m}\times\{0\}^{n-m})$$ Since, $\phi$ is a homeomorphism.