I am reading "Analysis on Manifolds" by James R. Munkres.
Is my solution to the following exercise ok?
This exercise is in the section 9 "THE IMPLICIT FUNCTION THEOREM".
But I did not use the implicit function theorem to solve exercise 6 at all.
So I worry about whether my solution is ok.
On p.79,
Exercise 6.
Let $f:\mathbb{R}^{k+n}\to\mathbb{R}^n$ be of class $C^1$; suppose that $f(\mathbf{a})=\mathbf{0}$ and that $Df(\mathbf{a})$ has rank $n$. Show that if $\mathbf{c}$ is a point of $\mathbb{R}^n$ sufficiently close to $\mathbf{0}$, then the equation $f(\mathbf{x})=\mathbf{c}$ has a solution.
My solution is here:
Without loss of generality, we can assume that $$\det\begin{bmatrix}\frac{\partial f_1}{\partial x_1}(a_1,\dots,a_n,a_{n+1}\dots a_{k+n})&\dots&\frac{\partial f_1}{\partial x_n\\}(a_1,\dots,a_n,a_{n+1},\dots,a_{k+n})\\&\dots&\\\frac{\partial f_n}{\partial x_1}(a_1,\dots,a_n,a_{n+1}\dots a_{k+n})&\dots&\frac{\partial f_n}{\partial x_n\\}(a_1,\dots,a_n,a_{n+1},\dots,a_{k+n})\end{bmatrix}\neq 0.$$
$f_1(a_1,\dots,a_n,a_{n+1},\dots,a_{k+n}) = 0$,
$f_2(a_1,\dots,a_n,a_{n+1},\dots,a_{k+n}) = 0$,
$\cdots$,
$f_n(a_1,\dots,a_n,a_{n+1},\dots,a_{k+n}) = 0$.
$g_1(x_1,\dots,x_n):=f_1(x_1,\dots,x_n,a_{n+1},\dots,a_{k+n})$,
$g_2(x_1,\dots,x_n):=f_2(x_1,\dots,x_n,a_{n+1},\dots,a_{k+n})$,
$\cdots$,
$g_n(x_1,\dots,x_n):=f_n(x_1,\dots,x_n,a_{n+1},\dots,a_{k+n})$.
$g:=(g_1,\dots,g_n)$.
Then, $g:\mathbb{R}^n\to\mathbb{R}^n$ is of class $C^1$ and $\det Dg(a_1,\dots,a_n)\neq 0$.
By the inverse function theorem, there is a neighborhood $U$ of the point $(a_1,\dots,a_n)$ such that $g$ carries $U$ in a one-to-one fashion onto an open set $V$ of $\mathbb{R}^n$.
$g(a_1,\dots,a_n)=\mathbf{0}\in V$.
So, if $\mathbf{c}=(c_1,\dots,c_n)$ is a point of $\mathbb{R}^n$ sufficiently close to $\mathbf{0}$, then the equation $g(x_1,\dots,x_n)=\mathbf{c}$ has a solution $(x_1,\dots,x_n)=(b_1,\dots,b_n)$.
So,
$f_1(b_1,\dots,b_n,a_{n+1},\dots,a_{k+n}) = c_1$,
$f_2(b_1,\dots,b_n,a_{n+1},\dots,a_{k+n}) = c_2$,
$\cdots$,
$f_n(b_1,\dots,b_n,a_{n+1},\dots,a_{k+n}) = c_n$.
So,
the equation $f(\mathbf{x})=\mathbf{c}$ has a solution.