# Showing that a map between surfaces is a local isometry

Currently studying Differential Geometry of Curves and Surfaces.

We have: $$\sigma:(0,\infty) × \mathbb{ R} \to \mathbb{R}^3, \quad (u,v) \to \frac{1}{\sqrt{2}} (u \cos v,u\sin v ,u)$$

We need to show that $\phi$ is a local isometry from $\sigma$ to $\sigma '$ where $$\phi: (0, \infty) × \mathbb{R} \to \mathbb{R}^2, \quad (u,v)\to (u\cos(v/\sqrt{2}),u\sin(v/\sqrt{2}))$$ where $\sigma': \mathbb{R}^2 \to \mathbb{R}^3, (x,y) \to (x,y,0)$

The answers say you need to compare the first fundamental form of $\sigma$ with the first fundamental form of the composition $\sigma ' \circ \phi$

I was wondering why you have to do this? I get comparing the first fundamental forms but why do we choose to compose those two?

Because you want to prove that $\phi$ is a local isometry, and the definition of local isometry requires the 1st fundamental form of the domain to be equal to the pullback of the 1st fundamental form of the target. The target is $\sigma'$. Pullback means composition with $\phi$. There is not much to say here other than review the definition of local isometry.