Let $f:\mathbb{R}^{k+n}\rightarrow \mathbb{R}^n$ be 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.
Write $f=f(x,y)$ with $x\in \mathbb{R}^k$ and $y\in \mathbb{R}^n$
As rank of $Df(a)=n$, we see that $\det\frac{\partial f}{\partial y}(a)\neq 0$.. write $a=(a_1,a_2)$ where $a_2\in \mathbb{R}^n$
Then by implicit function thoerem, there exists $g:\mathbb{R}^k\rightarrow \mathbb{R}^n$ such that $f(a_1,g(a_1))=0$
We have $f(m,g(m))=0$ in a nbd around $a_1$... I do not know how to use this to solve $f(x)=c$..
May be i should start with the function $h(x)=f(x)-c$
then there is a nbd around $a_1$ such that $h(m,g(g(m)))=f(m,g(m))-c=0$
Something like this works but i am not sure...
I dont know why $c$ has to be sufficienlty close to $0$
Since $Df(a)$ has rank $n$ there are indices $i_j$ ($j = 1,...,n$) such that the corresponding columns of $Df(a)$ are linearly independent. Let $I$ be all the other indices.
Let $L:\mathbb{R}^n \to \mathbb{R}^{n+k}$ be the map defined by $L(e_j) = e_{i_j}$, and define $\phi(x,c) = f(L(x)+\sum_{l\in I} a_l e_l) -c$.
Let $x_0 = (a_{i_1},...,a_{i_n})$, and note that $\phi(x_0, 0) = 0$. Also, ${\partial \phi(x_0,0) \over \partial x}$ is invertible.
The implicit function theorem tells that there is a $C^1$ function $\xi$ defined on a neighbourhood of $0$ such that $\xi(0) = x_0$ and $\phi(\xi(c),c) = 0$ for $c$ in this neighbourhood.
Expanding shows that $f(L(\xi(c))+\sum_{l\in I} a_l e_l) = c$ as required.