Existence of Solutions for f(x)=c given f(a)=0 and Df(a) with Rank n

[JG89]

Homework Statement

Let $$f: \mathbb{R}^{k+n} \rightarrow \mathbb{R}^n$$ 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 $$\mathbb{R}^n$$ sufficiently close to 0, then the equation f(x) = c has a solution.

Homework Equations

The Attempt at a Solution

I'm pretty sure I have to use the Implicit Function Theorem here. Write f in the form f(x,y) for x in $$\mathbb{R}^k$$ and y in $$\mathbb{R}^n$$. Also write $$a = (a_1, a_2)$$. So $$f(a) = f(a_1, a_2) = 0$$.

Let f_x denote the partial derivative of f with respect to x, and f_y the partial derivative with respect to y. Then Df(a) = [f_x (a) f_y (a)]

Where f_x is a n by k matrix and f_y is an n by n matrix. Since Df(a) has rank n, then f_y (a) has rank n, and so its inverse exists. So by the inverse function theorem, there is a neighborhood B of $$a_1$$ and a function $$g: B \rightarrow \mathbb{R}^n$$ such that $$g(a_1) = a_2$$, f(x, g(x)) = 0 for all x in B, and g is of class $$C^1$$.

This is about as far as I can get. I know that the key is that the rank of Df(a) is n, because if this weren't so then the theorem wouldn't be true (I could find a counter-example).

Any help?

EDIT: Ok. f cannot be constant in any neighborhood of a, since Df(a) has rank n. So by continuity of f, there must be points in some neighborhood of a, such that the image of these points are very close to 0. But that still doesn't prove what I'm looking to prove...

 

[mighty2000]

One approach you can take is to consider the function g(x) = f(x,0) for x in \mathbb{R}^k . Since Df(a) has rank n, we know that f_y (a) is invertible and therefore g is of class C^1. Also, g(a_1) = f(a_1,0) = 0. Now, let c be a point in \mathbb{R}^n sufficiently close to 0. Since f(a) = 0, we know that c is also sufficiently close to f(a_1,0) = g(a_1). By the continuity of g, there exists a neighborhood B of a_1 such that g(B) contains c. Therefore, for any x in B, we have g(x) = f(x,0) = c, which means that the equation f(x) = c has a solution.