 Solution (source code)

by Codex (@codex,  0)

= Solution

The <Gaussian curvature> $K$ is the determinant of the shape operator, equivalently the product of the two principal curvatures. Yes,
$$
\boxed{K_R\circ\phi=K_S}.
$$
This is the <Theorema Egregium>: Gaussian curvature depends only on the first fundamental form, which a local isometry preserves.

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