Show that if $c$ is a point of $\Bbb{R}^n$ sufficiently close to $0$, then the equation $f(x)=c$ has a solution. 
Let $f:\Bbb{R}^{k+n}\xrightarrow{}\Bbb{R}^n$ 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 $\Bbb{R}^n$ sufficiently close to $0$, then the equation $f(x)=c$ has a solution.

My attempt:
Since, $Df(a)$ has full rank we know that the $\det Df(a) \ne 0$, as we all $f\in C^1$ and $f(a)=0$. So we can apply the implicit function theorem giving us an open neighbourhood $B$ of $a$. Such that $c\in B(0,\epsilon)$ then $f(x)=c$ must have a solution.
How can I improve my answer?
 A: As $f$ is a map from ${\mathbb R}^{n+k}$ to ${\mathbb R}^n$ so is the linear map $Df(a)$. If $k>0$ this linear map does not have a determinant. In fact, the matrix $J:=[Df(a)]$ is an $n\times(n+k)$-matrix. Assuming that $Df(a)$ has rank $n$ means that $J$ has a nonvanishing subdeterminant of size $n$, and you may assume that this is the determinant of the submatrix
$$\left[{\partial f_i\over\partial x_j}\right]_{1\leq i\leq n, \ 1\leq j\leq n}\ .$$
We now write the points of ${\mathbb R}^{n+k}$ in the form $(x_1,\ldots, x_n,x')$ with $x'\in{\mathbb R}^k$. By the implicit function theorem the map
$$\hat f:\quad (x_1,\ldots, x_n)\mapsto f(x_1,\ldots, x_n,a')$$ maps a neighborhood $U$ of $(a_1,\ldots, a_n)\in{\mathbb R}^n$ to a full neighborhood of $0\in{\mathbb R}^n$. It follows that for small enough $|c|$ there is a point $(\xi_1,\dots,\xi_n)\in U$ such that $$f(\xi_1,\ldots,\xi_n, a')=\hat f(\xi_1,\dots,\xi_n)=c\ .$$
(In fact there is a full $k$-dimensional set of solutions to $f(x)=c$ in the neighborhood of $a$. In order to simplify matters we have held $a'$ constant here.)
