Consider the “pacman” region M given by $\textbf{x}(u,v)=(u\cos(v),u\sin(v),0), 0 \leq u \leq R, 0 \leq v \leq V$ , with $V < 2\pi$. Let $c = V/2\pi$. Let M* be given by the parametrization $\textbf{x*}(u,v)=(cu\cos(v/c),cu\sin(v/c),\sqrt{1-c^2}u)$
Compute that $E = E^{*}$ , $F=F^{*}$ , $G=G^{*}$, and conclude that the mapping $\textbf{f}=\textbf{x*} \circ \textbf{x}^{-1}: M\rightarrow M*$ is a local isometry. Describe this mapping in concrete geometric terms.
I had no problem computing E,F,G etc. and showing that but I'm struggling with the idea of $\textbf{x}^{-1}$ and that final mapping/function.
[edit] After a comment instructed me to draw the images of $\textbf{x}$ and $\textbf{x*}$ I see that M and M* are an incomplete circle (hence pacman) and an incomplete cone respectively. However, I'm still not quite sure how to approach the idea of the inverse and thus the mapping from one surface to the other.