PROPOSITION:
Let $U \subset R^{d}$, and $V \subset R^{n}$
Consider a function $\alpha: U \rightarrow V$. Assume a($\textbf{0}$)= $\textbf{p}$, and that Da($\textbf{x}$) has rank d for all $\textbf{x} \in U$. Therefore, there exists a matrix $B \in Mat_{n,n-d}$ such that Da($\textbf{0}$)B is non degenerate.
Consider the mapping: $\beta : U \times R^{n-d} \rightarrow R^{n}$
$(\textbf{x},\textbf{y}) \rightarrow \alpha(\textbf{x}) + B\textbf{y} $
Then $D\beta(\textbf{0})= D\alpha(\textbf{0})B$.
I tried testing this proposition, but my solution isn't working out as shown below:
Let $\alpha: U \rightarrow R^{3}$
where $U= ((u,v): \sqrt{(u-3)^2 + (v-3)^2}<3)$
$\alpha(u,v)= (u,v, (u-3)^2+ (v-3)^2)$
Therefore, $D\alpha(u,v)= \begin{pmatrix} 1 & 0 & 2(u-3)\\ 0& 1 & 2(v-3) \end{pmatrix}$
This matrix has rank 2 for $(u,v) \in U$.
Therefore $B \in Mat_{3,1}$, and $y \in R$
Let $B= \begin{pmatrix} B_{1}\\ B_{2}\\ B_{3} \end{pmatrix}$
Therefore $\beta(\textbf{x},y)= \beta(u,v,y)= \alpha(u,v) + By $
so
$$D\beta(u,v,y)=\begin{pmatrix} 1& 0 & 2(u-3)\\ 0& 1& 2(v-3)\\ 0& 0& 0\\ \end{pmatrix} + \begin{pmatrix} 0& 0& 0\\ 0& 0& 0\\ B_{1}& B_{2}& B_{3}\\ \end{pmatrix}. $$
Therefore, $D\alpha(0,0)B = \begin{pmatrix} B_{1}- 6B_{3}\\ B_{2}- 6B_{3}\\ \end{pmatrix}$
But $D\beta(0,0)= \begin{pmatrix} 1& 0 &-6\\ 0& 1& -6\\ B_{1}& B_{2} & B_{3}\\ \end{pmatrix}$