I am reading a proof about the implicit function theorem that uses the inverse function theorem. There they make a statement that I do not understand.
The inverse function theorem statement is:
Assume that U is an open subset of $\mathbb{R}^{m+1}$ and let $f:U \rightarrow\mathbb{R}$ be a function with continuous partial derivatives. Assume that $(\bar{x},\bar{y})=(\bar{x}_1,\bar{x}_2,\ldots,\bar{x}_m,y)$ is a point in U such that $f(\bar{x},\bar{y})=0.$ Assume also that $\frac{\partial f}{\partial y}(\bar{x},\bar{y})\ne 0$. Then there exists an open set $U_0$ about $\bar{x}$ and a differentiable function $g: U_0\rightarrow \mathbb{R}$ such that $g(\bar{x})=\bar{y}$ and $f(x,g(x))=0$ for all $x \in U_0$. The derivative of $g$ is given by $\frac{\partial g}{\partial x_i}=-\frac{\frac{\partial f}{\partial x_i}(\bar{x},\bar{y})}{\frac{\partial f}{\partial y}(\bar{x},\bar{y})}$.
In the proof they define the function $F: \mathbb{R}^{m+1}\rightarrow \mathbb{R}^{m+1}$ given by $F(x_1,x_2,\ldots,x_m,y)=(x_1,x_2,\ldots,x_m,f(x_1,x_2,\ldots,x_m,y))$.
They calculate the Jacobian of $F$ and show that the determinant in $(\bar{x},\bar{y})$ is nonzero. And then they can use the inverse function theorem.
Then they make the statement: It is easy to see that the inverse function of $F$ must be of the form $G(x_1,x_2,\ldots,x_m,z)=(x_1,x_2,\ldots,x_m,h(x_1,x_2,\ldots,x_m,z))$, where $h$ is a differentiable funciton.
But why is this easy to see? How do we know that the function must be of this form?