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Let $f:R^{k+n} \rightarrow R^n$ be of class $C^1$; suppose that $f(a)=0$ and that $Df(a)$ has rank n. Show that if $c$ is a point of $R^n$ sufficiently close to $0$, then the equation $f(x)=c$ has solution.

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it follows from Implicit function theorem? Or is there anything else you have in mind? –  Sanchez Nov 8 '12 at 18:47
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Notation is the main impediment to a solution here.

Let $\Pi$ be a permutation matrix such that the first $n$ columns of $Df(a)\Pi$ are linearly independent, and write a point of $\mathbb{R}^{k+n}$ as $(x,y)$, with $x \in \mathbb{R}^n$. Let $(\hat{x}^T, \hat{y}^T)^T = \Pi^{-1} a $, and define $\phi(x) = f(\Pi (x^T,\hat{y}^T)^T)$. Note that $\phi$ is $C^1$ and $\phi(\hat{x}) = 0$.

Then we have that $D\phi(\hat{x})$ is invertible (since it is the first $n$ columns of $Df(a)\Pi$, ie, square matrix with $n$ linearly independent columns), we can apply the inverse function theorem to get a neighbourhood $U_{\hat{x}}$ of $\hat{x}$, a neighbourhood $V_{0}$ of $0$ (in $\mathbb{R}^n$) and a function $\eta:V_{0} \to U_{\hat{x}}$ such that $\phi(\eta(c)) = c$ for all $c \in V_{0} $. Since this implies $f(\Pi (\eta(c)^T,\hat{y}^T)^T) = c$, we see that the equation $f(x) = c$ has a solution for $c \in V_{0} $.

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I do not understand the notation $(x^T,y^T)^T$. –  Cleiton Ricardo Nov 8 '12 at 20:57
    
If anyone can help me, thank you –  Cleiton Ricardo Nov 8 '12 at 21:17
    
It is the vector $\pmatrix{ x_1 \\ \vdots \\ x_n \\ y_1 \\ \vdots \\ y_k}$. I was just trying to be notationally consistent. You could write it horizontally, but then I would need to write $\Pi^{-1}(a)$, $Df(a) \circ \Pi$, etc., this was I can use standard matrix notation. –  copper.hat Nov 8 '12 at 22:15
    
thank you very much copper.hat, helped me a lot. –  Cleiton Ricardo Nov 9 '12 at 1:45
    
You are very welcome, glad to be of help. –  copper.hat Nov 9 '12 at 2:49
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