Inverse Function Theorem in Immersions. 
Let $\varphi: U \to \mathbb{R}^{n}$ of class $C^{k}$ ($k\geq 1$) in the open $U \subset \mathbb{R}^{m}$. If $a \in U$ is such that $\varphi'(a): \mathbb{R}^{m} \to \mathbb{R}^{n}$ is injective, then there exists a decomposition in direct sum $\mathbb{R}^{n} = \mathbb{R}^{m}\oplus\mathbb{R}^{n-m}$ and an open $V$, with $a \in V$, such that $\varphi(V)$ is the graph of an aplication $f: W \to \mathbb{R}^{n-m}$, of class $C^{k}$ in the open $W \subset \mathbb{R}^{m}$.

We can write $C^{k} \ni \varphi: U \to \mathbb{R}^{m+(n-m)}$. As a consequence of the Inverse Function Theorem applied to immersions and by hypothesis about $\varphi$, there is a diffeomorphism ($C^{k}$) $h: Z \to X \times Y$ where $Z \ni \varphi(a)$ is open in $\mathbb{R}^{m+(n-m)}$ and $X \times Y \ni (a,0)$ is open in  $\mathbb{R}^{m}\times \mathbb{R}^{m+(n-m)}$, such that $(h\circ \varphi)(x) = (x,0)$ for all $x \in X$ and $h$ is strongly differentiable in $\varphi(a)$. 
Seems intuitive that $``h$ is $f"$ and $``X$ is $W"$, but I couldn't develop more than this. A detail that confused me is: the dimmensions of $X \times Y$ and $\mathbb{R}^{n-m}$ are not equals. Maybe I'm using this result incorrectly. Can anybody help me? Thanks for the advance!
 A: If $d\phi(a)$ is injective then the matrix $[d\phi(a)]$ has rank $m$, hence an $(m\times m)$- submatrix with nonzero determinant. Assume this is the submatrix $J:=[d\phi(a)]_{1\leq i\leq m, \>1\leq k\leq m}$. Let
$$\eqalign{\pi':\quad&(y_1,\ldots, y_m,y_{m+1},\ldots, y_n)\mapsto (y_1,\ldots, y_m)\cr
\pi'':\quad&(y_1,\ldots, y_m,y_{m+1},\ldots, y_n)\mapsto (y_{m+1},\ldots, y_n)\cr}$$
be the projections associated to the factorization ${\mathbb R}^n={\mathbb R}^m\times{\mathbb R}^{n-m}$. We write $y=(y',y'')$ for the points of ${\mathbb R}^n$, in particular  $\phi(a)=:b=(b',b'')$.
Consider now the map
$$\psi:=\pi'\circ\phi:\quad U\mapsto{\mathbb R}^m\ .$$
Since $[d\psi(a)]=J$ is nonsingular the inverse function theorem tells us that $\psi$ maps an open neighborhood $U_a$ of $a$ diffeomorphically onto an open neighborhood $V_{b'}$ of $b'$. The composition
$$f:=\pi''\circ\phi\circ \psi^{-1}:\quad V_{b'}\to{\mathbb R}^{n-m},\qquad y'\mapsto \bigl(f_1(y'),\ldots,f_{n-m}(y')\bigr)$$
is then the function you are looking for. Check for yourself that it "lifts" each point $y'\in V_{b'}$ "vertically" up to the proper "height".
