Let's assume that $f:\mathbb{R}^n\to\mathbb{R}^n$ with \begin{align*} f(y_1,y_2,\cdots ,y_n):=\begin{pmatrix}f_1(y_1,y_2,\cdots ,y_n)\\f_2(y_1,y_2,\cdots ,y_n)\\\vdots\\f_n(y_1,y_2,\cdots ,y_n)\end{pmatrix} \end{align*} is continuously differentiable and there exists a point $c\in\mathbb{R}^n$ such that $Df(c)\neq 0$.
Show that $f$ is locally invertible by applying the implicit function theorem.
My approach:
First, we define $F: \mathbb{R}^{2n}\to\mathbb{R}^n$ by \begin{align*} F(x_1,x_2,\cdots ,x_n,y_1,y_2,\cdots ,y_n):=\begin{pmatrix}f_1(y_1,y_2,\cdots ,y_n)-x_1\\f_2(y_1,y_2,\cdots ,y_n)-x_2\\\vdots\\f_n(y_1,y_2,\cdots ,y_n)-x_n\end{pmatrix}. \end{align*} $f$ is by assumption continuously differentiable and $h:\mathbb{R}^n\to\mathbb{R}^n$, with \begin{align*} h(x_1,\cdots, x_n):=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix} \end{align*} is obviously continuously differentiable. The same holds for $\varphi_1,\varphi_2:\mathbb{R}^{2n}\to\mathbb{R}^n$ with \begin{align*} \varphi_1(x_1,\cdots, x_n,y_1,\cdots,y_n):=\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}\text{ and } \varphi_2(x_1,\cdots, x_n,y_1,\cdots,y_n):=\begin{pmatrix}y_1\\\vdots\\y_n\end{pmatrix}. \end{align*} So applying the rules of continuously differentiable functions shows that $F=f\circ \varphi_2-h\circ \varphi_1$ is continuously differentiable. We define the point $a\in\mathbb{R}^{2n}$ by $$ a:=(f_1(c_1,\cdots,c_n),\cdots, f_n(c_1,\cdots,c_n), c_1,\cdots, c_n).$$ So $F(a)=0$. If we take a closer look at $DF$ we see that: \begin{align*} DF(x_1,\cdots, x_n,y_1,\cdots,y_n)=\begin{pmatrix} -1&0&\cdots&0&0&D_{y_1}f_{1}&D_{y_2}f_{1}&\cdots &D_{y_n}f_{1}\\ 0&-1&\cdots&0&0&D_{y_1}f_{2}&D_{y_2}f_{2}&\cdots &D_{y_n}f_{2}\\ &&\cdots&0&0&\cdots&\cdots&\cdots &\cdots\\ 0&0&\cdots&0&0&\cdots&\cdots&\cdots &\cdots\\ 0&0&\cdots&-1&0&\cdots&\cdots&\cdots &\cdots\\ 0&0&\cdots&0&-1&D_{y_n}f_{n}&D_{y_n}f_{n}&\cdots &D_{y_n}f_{n}\\ \end{pmatrix}= (-I_n\mid Df). \end{align*} By assumption $Df(c)$ has rank $n$, so we can apply the implicit function theorem.
The theorem guarantees the existence of a continuously differentiable function $g:U\to V'$, where $U\subseteq\mathbb{R}^n$ and $V'\subseteq\mathbb{R}^n$ are open neighborhoods such that $(a_1,\cdots,a_n)\in U$ and $(a_{n+1},\cdots, a_{2n})\in V'$. For each $x\in U$ it follows that $F(x,g(x))=0$. On the other hand if for two $x,y$ with $x\in U$, $y\in V'$ it holds that $F(x,y)=0$ then $g(x)=y$. Now we define $V:=g(U)$ and shrink the codomain $V'$ of $g$ to its image, so $g:U\to V$. If we assume for $u_1, u_2\in U$ that $g(u_1)=g(u_2)$ then $$F(u_1,g(u_1))=F(u_2,g(u_2))=0\implies f(g(u_1))-u_1 f(g(u_2))-u_2\implies u_1=u_2.$$ So $g:U\to V$ is injective and by definition surjective (hence bijective).
Next, we restrict the domain of $f:\mathbb{R}^n\to\mathbb{R}^n$ and get $f_{\mid V}:V\to \mathbb{R}^n$. If we take some $y\in V$ and look at its image $f(y)=x$ then we know that there exists a $u\in U$ such that $g(u)=y$ and $F(u,g(u))=f(g(u))-u=0\implies f(y)=u$. So it must be $x=u$ otherwise $f$ wasn't a well defined function. Hence the image of $f_{\mid V}$ is a subset of $U$ and the function $f_{\mid V}:V\to U$ is indeed well defined. If we plug $(x_1,\cdots,x_n)\in U$ and $g(x_1,\cdots,x_n)=\begin{pmatrix}y_1\\\vdots\\y_n\end{pmatrix}$ into the equation \begin{align*} F(x_1,x_2,\cdots ,x_n,y_1,y_2,\cdots ,y_n)=\begin{pmatrix}f_1(y_1,y_2,\cdots ,y_n)-x_1\\f_2(y_1,y_2,\cdots ,y_n)-x_2\\\vdots\\f_n(y_1,y_2,\cdots ,y_n)-x_n\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix} \end{align*} we get \begin{align*} \begin{pmatrix}f_1(g_1(x),g_2(x),\cdots ,g_n(x))-x_1\\f_2(g_1(x),g_2(x),\cdots ,g_n(x))-x_2\\\vdots\\f_n(g_1(x),g_2(x),\cdots ,g_n(x))-x_n\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}\implies \begin{pmatrix}f_1(g_1(x),g_2(x),\cdots ,g_n(x))\\f_2(g_1(x),g_2(x),\cdots ,g_n(x))\\\vdots\\f_n(g_1(x),g_2(x),\cdots ,g_n(x))\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix} \end{align*} which shows that $f_{\mid V'}\circ g =id_U$. On the other hand we know that for all $(x,y)\in U\times V$ it holds \begin{align*} \begin{pmatrix}f_1(y_1,y_2,\cdots ,y_n)-x_1\\f_2(y_1,y_2,\cdots ,y_n)-x_2\\\vdots\\f_n(y_1,y_2,\cdots ,y_n)-x_n\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}\implies \begin{pmatrix}f_1(y_1,y_2,\cdots ,y_n)\\f_2(y_1,y_2,\cdots ,y_n)\\\vdots\\f_n(y_1,y_2,\cdots ,y_n)\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}. \end{align*} and $g(x)=y$. So if we consider $g\circ f$ we get \begin{align*} g(f_1(y_1,y_2,\cdots ,y_n),f_2(y_1,y_2,\cdots ,y_n),\cdots,f_n(y_1,y_2,\cdots ,y_n))=\begin{pmatrix}y_1\\y_2\\\vdots\\y_n\end{pmatrix}=id_V. \end{align*} So $g$ is the local inverse function of $f$. If we apply the formula of the derivative of the implicit function $g$ we get \begin{align*} Dg(x_1,\cdots, x_n)= (-1)\begin{pmatrix} D_{y_1}f_{1}&D_{y_2}f_{1}&\cdots &D_{y_n}f_{1}\\ D_{y_1}f_{2}&D_{y_2}f_{2}&\cdots &D_{y_n}f_{2}\\ \cdots&\cdots&\cdots &\cdots\\ \cdots&\cdots&\cdots &\cdots\\ \cdots&\cdots&\cdots &\cdots\\ D_{y_n}f_{n}&D_{y_n}f_{n}&\cdots &D_{y_n}f_{n}\\ \end{pmatrix}\begin{pmatrix} -1&0&\cdots&0&0\\ 0&-1&\cdots&0&0\\ 0&0&\cdots&0&0\\ 0&0&\cdots&0&0\\ 0&0&\cdots&-1&0\\ 0&0&\cdots&0&-1\\ \end{pmatrix}= Df. \end{align*} $Dg$ has full rank and we know that images of open sets are also open if the function is conntinuously differntiable and has a non-vanishing derivative. So $g(U)=V$ is also open. We are done.
Is this correct? Do you have any suggestions?