Is the map from the circular half cone to the $xy$ plane a local isometry?

Question

This is a text book exercise. And I think that this map is not a local isometry. But, I don't know how to show this question. Please help me explaining this question. Thanks a lot.

I posted its answer:

Answer 1

Consider $$\pi : (C = \{ (x,y,z)|\ x^2+y^2=z^2 >0 \},g_C) \rightarrow ({\bf R}^2,g),\ (x,y,z)\mapsto (x,y)$$ where $g_C,\ g$ are metrics induced from ${\bf R}^3$.

If $c$ is a curve and $\pi$ is an local isometry then $$ g_C(c'(t),c'(t)) = g(\frac{d}{dt}\pi\circ c(t),\frac{d}{dt}\pi\circ c(t))$$

Hence $$ \int_1^2 g_C(c'(t),c'(t))^{1/2}\ dt = \int_1^2 g(\frac{d}{dt}\pi\circ c(t),\frac{d}{dt}\pi\circ c(t))^{1/2}\ dt$$

Hence local isometry preserves length.

Note that $c(t)=(t,0,t)$, $t\in [1,2]$, is a geodesic of length $\sqrt{2}$ on cone.

But $\pi\circ c(t)=(t,0,0) $ is a geodesic of length $1$.

Hence $\pi$ is not local isometry.