Question:
Let $f: \mathbb{R}^2 \to \mathbb{R}^3$ be a $C^1$ map. Assume $f(0) = (0,0,0)$ and $$\frac{\partial f}{\partial x}(0) \times \frac{\partial f}{\partial y}(0) =(0,0,1)$$ Show there exists neighborhoods $\mathcal O$ and $\Omega$ of $0 \in \mathbb{R}^2$ and a $C^1$ maps $u: \Omega \to \mathbb{R}$ such that the image of $\mathcal O$ under $f$ in $\mathbb{R}^3$ is the graph of $u$ over $\Omega$.
Hint: Let $\Pi: \mathbb{R}^3 \to \mathbb{R}^2$ be $\Pi(x,y,z) = (x,y)$ and consider $$\phi(x,y) = \Pi(f(x,y)) \qquad \phi: \mathbb{R}^2 \to \mathbb{R}^2$$ Show that $D\phi(0): \mathbb{R}^2 \to \mathbb{R}^2$ is invertible, and apply the inverse function theorem. Then let $u$ be the $z$-component of $f \circ \phi^{-1}$
I've gone through all the steps except for the part where it says "let $u$ be the $z$-component of $f \circ \phi^{-1}$. I see why $u$ would satisfy all the necessary conditions, except for the main one being that $f(\mathcal O)$ is the graph of $u$ over $\Omega$.
I've done everything the hint mentioned except I am not quite sure why $u$ would have the property that the graph of $u$ over $\Omega$ is the image of $\mathcal O$ under $f$. I'm assuming it has something to do with the fact that we used a projection function in creating $\phi$ but I'm still uncertain. Thanks in advance.