Let's consider the continuously differentiable functions $f:\mathbb{R}^n\to\mathbb{R}^n$ and $F: \mathbb{R}^{2n}\to\mathbb{R}^n$ where \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*} and assume that we can apply the implicit function theorem to equation $F(x,y)=0$ at point $a:=(a_1:=f(a_{n+1}),\cdots,a_{n}:=f(a_{2n}),a_{n+1},\cdots,a_{2n},)$ and can solve for $y$.
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 $(a_{n+1},\cdots, a_{2n})\in V'$.
Now we define $V:=g(U)$ and set $g:U\to V$. Note that $(a_{n+1},\cdots, a_{2n})\in V$. 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).
I am not sure about the following part:
Next we restrict $f$ 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. By using the implicit function theorem we immediately see $f\circ g= id_U$ and $g\circ f =id_V$ hence $f$ is the invere of $g$ (and vice versa).
By assumption the submatrix of the partial derivatives w.r.t $y$ has full rank so $Dg$ also has full rank. We know that a function with a derivative of full rank maps open sets to open sets so $V$ is open.
In our lecture the step where we proved that $g$ and $f$ are the inverse functions of each other and map between the corresponding open sets was done differently. In fact we first adjusted the domain of $f$ ... So I am a bit unsure if my approach is valid. Maybe someone can give feedback?