Proving implicit function theorem with the fixed point theorem I am trying to solve an exercise in Kolmogorov's analysis text asking for a proof of the implicit theorem using the fixed point theorem. I am struggling with how to get started on this, though. It's clear that I need to define a mapping, $T$, demonstrate that $\rho(Tx, Ty) = \alpha \rho(x,y)$ for some $\alpha < 1$, and then conclude that because $T$ is a contraction mapping, it has a fixed point, $x$, such that $Tx = x$. I know of only one proof of the implicit function theorem, though, and it in no way uses any of these facts. (I proved it at a time when I didn't know what a contraction mapping was.) 
Any help on this would be greatly appreciated. 
 A: Consider a continuously differentiable map
$$\begin{array}{l|rcl}
f : & \mathbb R^n \times \mathbb R^m & \longrightarrow & \mathbb R^m \\
    & (x,y) & \longmapsto & f(x,y) \end{array}$$
a point $(a,b) \in \mathbb R^n \times \mathbb R^m$ such that $f(a,b) = 0$ and suppose that the partial Jacobian $J_{f,y}(a,b)$ is invertible. Those are the hypothesis of the implicit function theorem. With the given hypothesis, one can find a neighborhood $U_a$ of $a$ and $U_b$ of $b$ such that $J_{f,y}(x,y)$ is invertible for $(x,y) \in U_a \times U_b$.
For $x \in U_a$, the map $\varphi_x : y \mapsto y-J_{f,y}^{-1}(a,b) \circ f(x,y)$ is continuously differentiable in $U_b$ with $J_y(b) = 0$. Hence $\Vert J_y(y) \Vert <1$ for $y$ close enough to $b$. With that, you can apply the fixed point theorem to find $g(x)$ with $\varphi_x(g(x))= g(x)$ with means $f(x,g(x)) = 0$ as desired.
A: Hint:
For express $x_0$ over other variables, define $g(X) = \frac{f(X)}{\max_{[x_0-\epsilon,x_0+\epsilon]} f'}$ can be good candidate, since $\max f'$ is finite due to theorem hypothesis.
