This is exercise 1.2 in Svetlana Katok's Fuchsian Groups.
$\mathbb{H}$ is the upper half plane (with the hyperbolic metric), and $f:\mathbb{H}\rightarrow\mathbb{H}$ is a surjective $C^1$ map. I want to show $f$ is an isometry (in terms of the hyperbolic metric) if and only if it preserves the Riemannian norm on the tangent bundle of $\mathbb{H}$.
One direction I can do (isometry implies norm-preserving), but the other direction is giving me trouble. I've shown that if $f$ is norm-preserving, then it also preserves the length of curves, so that $$ d(f(z),f(w))\le d(z,w) $$
But I can't seem to show there's equality here. In particular, I can't show $f$ is injective. Am I missing something special about the upper half plane?
I think this works, based on comments by @TedShifrin.
We can characterize the geodesics as those curves $\gamma$ with $\nabla_{\dot{\gamma}_t}\dot{\gamma}_t\equiv0$. Since $f$ preserves the Riemannian norm, it preserves the Riemannian metric (via the parallelogram rule), and so it preserves the covariant derivative. That is, $$ \nabla_{df(\dot{\gamma}_t)}df(\dot{\gamma}_t)\equiv0 $$ So $f\circ\gamma$ is a geodesic.
I still don't understand why $f$ preserves the covariant derivative, but it seems to be true, and I've seen some messy formulas involving the Riemannian metric, claiming to prove it. Conceptually, I'm still unclear though.