I am reading "Analysis on Manifolds" by James R. Munkres.
There is the following exercise (exercise 6) on p.79 in this book.
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 answer is here:
For $\mathbf{x}\in\mathbb{R}^{k+n}$ and $\mathbf{y}\in\mathbb{R}^n$, we define $g(\mathbf{x},\mathbf{y}):=f(\mathbf{x})-\mathbf{y}$.
Then, $g:\mathbb{R}^{k+2n}\to\mathbb{R}^n$ is of class $C^1$.
$g(\mathbf{a},\mathbf{0})=\mathbf{0}$.
$Dg(\mathbf{a},\mathbf{0})=\begin{bmatrix}\frac{\partial f}{\partial(x_1,\dots,x_{k+n})}(\mathbf{a})&-I_n\end{bmatrix}$.
Since $Df(\mathbf{a})$ has rank $n$, $$\det\frac{\partial g}{\partial(x_{i_1},\dots,x_{i_n})}(\mathbf{a},\mathbf{0})=\det\frac{\partial f}{\partial(x_{i_1},\dots,x_{i_n})}(\mathbf{a})\ne0$$ for some $\{i_1,\dots,i_n\}\subset\{1,\dots,k+n\}$.
So, by the implicit function theorem, if $\mathbf{c}$ is a point of $\mathbb{R}^n$ sufficiently close to $\mathbf{0}$, then the equation $g(\mathbf{x},\mathbf{c})=0$ has a solution.
So $f(\mathbf{x})=\mathbf{c}$ has a solution.
Prof. Theodore Shifrin, thank you very much for your comment.
My answer is here:
We write $f$ in the form $f(\mathbf{x},\mathbf{y})$ for $x\in\mathbb{R}^k$ and $y\in\mathbb{R}^n$.
Without loss of generality, we assume that $\det\frac{\partial f}{\partial (y_1,\dots,y_n)}(a_1,\dots,a_k,b_1,\dots,b_n)\ne 0$
We define $g:\mathbb{R}^{2n}\to\mathbb{R}^n$ by $$g(w_1,\dots,w_n,y_1,\dots,y_n):=f(a_1,\dots,a_k,y_1,\dots,y_n)-\begin{bmatrix}w_1\\\vdots\\w_n\end{bmatrix}.$$
$$g(0,\dots,0,b_1,\dots,b_n)=\begin{bmatrix}0\\\vdots\\0\end{bmatrix}.$$ $$Dg(0,\dots,0,b_1,\dots,b_n)=\begin{bmatrix}-I_n&\frac{\partial f}{\partial (y_1,\dots,y_n)}(a_1,\dots,a_k,b_1,\dots,b_n)\end{bmatrix}.$$ $$\det\frac{\partial g}{\partial (y_1,\dots,y_n)}(0,\dots,0,b_1,\dots,b_n)=\det\frac{\partial f}{\partial (y_1,\dots,y_n)}(a_1,\dots,a_k,b_1,\dots,b_n)\ne 0.$$ By the implicit function theorem, there is a neighborhood $B$ of $(0,\dots,0)$ in $\mathbb{R}^n$ and a (unique $C^1$) function $h:B\to\mathbb{R}^n$ such that $h(0,\dots,0)=(b_1,\dots,b_n)$ and $$g(w_1,\dots,w_n,h(w_1,\dots,w_n))=\mathbf{0}$$ for all $\mathbf{w}\in B$.
If $\mathbf{c}\in B$, then $g(c_1,\dots,c_n,h(c_1,\dots,c_n))=f(a_1,\dots,a_k,h(c_1,\dots,c_n))-\begin{bmatrix}c_1\\\vdots\\c_n\end{bmatrix}=\mathbf{0}$.
So, the equation $f(\mathbf{x})=\mathbf{c}$ has a solution.
Prof. Theodore Shifrin, thank you very much for your huge hint.
My answer is here:
We mimic the proof of the Implicit function theorem on pp.74-75.
From the 6th line from the bottom on p.74, to the 4th line from the top on p.75, we modify as follows:
Let $B$ be a (connected) neighborhood of $\mathbf{a}$ in $\mathbb{R}^k$, chosen small enough that $B\times\mathbf{c}$ is contained in $W$. See Figure 9.1. We prove existence of the function $g:B\to\mathbb{R}^n$. If $\mathbf{x}\in B$, then $(\mathbf{x},\mathbf{c})\in W$, so we have: $$G(\mathbf{x},\mathbf{c})=(\mathbf{x},h(\mathbf{x},\mathbf{c})),$$ $$(\mathbf{x},\mathbf{c})=F(\mathbf{x},h(\mathbf{x},\mathbf{c}))=(\mathbf{x},f(\mathbf{x},h(\mathbf{x},\mathbf{c}))),$$ $$\mathbf{c}=f(\mathbf{x},h(\mathbf{x},\mathbf{c})).$$ We set $g(\mathbf{x})=h(\mathbf{x},\mathbf{c})$ for $\mathbf{x}\in B$; then $g$ satisfies the equation $f(\mathbf{x},g(\mathbf{x}))=\mathbf{c}$, as desired. (Q.E.D.)
