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

I can only see that since $Df(a)$ has rank n, hence by implicit function thm, some n of the variables are actually functions of the remaining k variables, but I can't see how to use it.

Thanks in advance.

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1 Answer 1

The Implicit Function Theorem allows you to write a fixed level set as a graph. I recommend trying to apply the Inverse Function Theorem. Write $\mathbb R^{n+k}=\mathbb R^n\times\mathbb R^k$ and write points as $(x,y)$. Assume the $n\times n$ matrix $\left[\dfrac{\partial f}{\partial x}(a)\right]$ is nonsingular, and consider $g\colon \mathbb R^{n+k}\to\mathbb R^{n+k}$ defined by $g(x,y)=\big(f(x,y),y\big)$. Note that $Dg(a)$ is invertible, so $g$ has a local $C^1$ inverse $G$. Use $G$ and $c$ somehow to get an $(x,y)$ near $a$ with $f(x,y)=c$.

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can you please clarify a little more? I am new to these things, and I can't understand how to proceed after that. –  mathmansujo Aug 11 '13 at 19:08
    
I've added mire steps. Now you have to show me what you do with it. –  Ted Shifrin Aug 11 '13 at 19:23

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