The Implicit Function Theorem I am working with: Let $f \in C^1$ from an open set in $\mathbb{R}^m \times \mathbb{R}^n$ to $\mathbb{R}^m$. Let $(y_0, x_0)$ be a point in this open set such that $f(y_0, x_0) = 0$, and \begin{align*} L = \left( \frac{\partial f_i}{\partial y_j} (y_0, x_0) \right)_{i,j = 1, \dots, m} \end{align*} is nonsingular. Then there exist open sets $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ such that $(y_0, x_0) \in V \times U$, and to every $x \in U$, there exists a unique $y = \phi (x)$ in $V$ such that $f(\phi (x), x) = 0$, and $\phi \in C^1$ on $U$.
The proof begins by defining a function $F: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^m \times \mathbb{R}^n$ as $F(y,x) = (f(y,x), x)$. Then the Inverse Function Theorem is applied to $F$. We have a neighborhoods $U' \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$ so that $F$ has a $C^1$ inverse on $W = F(V \times U')$, $U = \{x \in U': (0,x) \in W\}$
My question is: how can we show that there exists a $C^1$ function $\Phi: W \rightarrow \mathbb{R}^m$ so that $F^{-1}(y,x)) = (\Phi (y,x),x))$ on $W$? My approach has been to find the derivative of $F$ and then try to invert it by the Inverse Function Theorem, but I don't know how this shows $\Phi$ exists.
Furthermore, if we let $\phi(x) = \Phi(0,x)$, how can we show that $\phi$ then satisfies the conclusion of the theorem?
Forget about all the open neighbourhoods. If you have an invertible mapping $F:\Bbb{R}^m\times\Bbb{R}^n\to \Bbb{R}^m\times\Bbb{R}^n$ then we can consider the following two functions: \begin{align} \alpha:= \pi_{\Bbb{R}^m}\circ F^{-1}\quad \text{and}\quad \beta:= \pi_{\Bbb{R}^n}\circ F^{-1}, \end{align} where the $\pi$'s are the standard projections. In other words, we can express $F^{-1}$ in the form $F^{-1}(u,v)=(\alpha(u,v),\beta(u,v))$. In your case $\Phi$ is $\alpha$ and $\beta=\pi_{\Bbb{R}^n}$ because of the additional assumption that $F(y,x)$ has the special form $(f(y,x),x)$.