Let $f: \Omega \to \mathbb{R}$ with $\Omega \subset \mathbb{R^n}$ open be smooth.
Define: $g: \mathbb{R^n} \to \mathbb{R^{n+1}}$ by $g(x) = (x,f(x))$ does the map $g^{-1}|_{M\cap U}$ exist and is it continuous? where $M = \text{graph}(f)$ and $U = \mathbb{R^{n+1}}$
i tried using hte inverse funciton theorem but the jacobian is not square so I cannot find a determinant. How would I go about doing this?
Well, the inverse map is given by $g^{-1}|_M(x,f(x)) = x = \pi((x,f(x)))$ where $\pi \colon \mathbb{R}^{n+1} \rightarrow \mathbb{R}^n$ is the projection onto the first $n$ coordinates. That is, $g^{-1}|_{M} = \pi|_M$. Clearly this is a continuous map.