the proof of implicit function theorem (Terence Tao) Implicit function theorem: Let $E$ be an open subset of $\mathbb{R}^n$, let $f : E \to \mathbb{R}$ be continuously differentiable, and let $y = ( y_1, ... , y_n)$ be a point in $E$ such that $f(y) = 0$ and $\frac{\partial{f}}{\partial{x_n}}(y) \not= 0$. Then there exists an open subset $U$ of $\mathbb{R}^{n-1}$ containing $(y_1, ... , y_{n-1})$, an open subset $V$ of $E$ containing $y$, and a function $g : U \to \mathbb{R}$ such that $g(y_1, ... ,y_{n-1}) = y_n$, and 
$$\{(x_1, ... ,x_n) \in V : f(x_1, ... ,x_n) = 0\} = \{(x_1, ... ,x_{n-1}, g(x_1, ... ,x_{n-1}) : (x_1, ... ,x_n) \in U\}.$$ 
Moreover, $g$ is differentiable at $(y_1, ... ,y_{n-1})$, and we have 
$$\frac{\partial{g}}{\partial{x_j}}(y_1, ... , y_{n-1}) = \frac{\partial{f}}{\partial{x_j}}(y)/ \frac{\partial{f}}{\partial{x_n}}(y)$$
for all $1 \le j \le n-1$. 


I don't understand some steps in the proof. I highlight with purple lines the step I am struggling with.


*

*The author let $F^{-1} (x) = (h_1(x), ..., h_n(x))$. Since $F$ is invertible, $x = F(h_1(x), ... ,h_n(x))$, but why $h_j(x) = x_j$? 

*We know that $W$ is open in $\mathbb{R}^n$, but how do we know that $U$ is open in $\mathbb{R}^{n-1}$? 

*Similarly, we know that $h_n$ is differentiable at $(y_1, ... ,y_{n-1}, 0)$, but how can we prove that $g$ is differentiable at $(y_1, ... ,y_{n-1})$? 
I really appreciate if you give some help! 
 A: *

*

$$(x_1, \ldots, x_j, \ldots, x_n) = x = F(F^{-1}(x)) = F((h_1(x), \ldots, h_j(x), \ldots, h_n(x))) = (h_1(x), \ldots, h_j(x), \ldots, f(h_1(x), \ldots, h_n(x)))$$
Therefore $x_j = h_j(x)$.


*Theorem 2.1.5 (c) states that if $f$ is a continuous function then the inverse image/preimage of an open set is also open. We can make a continuous function $f: \mathbb{R}^{n-1} \rightarrow \mathbb{R}^{n}$ that maps $(x_1, \ldots, x_{n-1})$ to $(x_1, \ldots, x_{n-1}, 0)$. By that theorem,
$$f^{-1}(W) = \{(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1} \vert f((x_1, \ldots, x_{n-1})) = (x_1, \ldots, x_{n-1}, 0) \in W\} = U$$
is open as $W$ is open.


*The derivative of $g$ evaluated at $(y_1, \ldots, y_{n-1})$ is the derivative of $h_n$ evaluated at $(y_1, \ldots, y_{n-1}, 0)$ which exists.
A: 1) From the definition of $F$, it fixes the first $n-1$ coordinates
2) $U$ is basically the intersection of the hyperplane $x_n=0$ and $W$, and this intersection is open in the hyperplane (which you identify as $\mathbb{R}^{n-1}$)
3) Composing a differentiable function with a linear projection gives you a differentiable function (at least, I think, it should be easy to prove using linearity and the definition of differentiability)
