Send each orientation-preserving <isometry> to its <isotopy> class. This is a <group homomorphism>
$$
\operatorname{Isom}^+(S_g)\longrightarrow\operatorname{Mod}(S_g).
$$
For $g\geq2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the <hyperbolic plane>: after composing with a <deck transformation>, its lift commutes with the surface <fundamental group>. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the <injection of a finite hyperbolic isometry group into a mapping class group>.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the <unit sphere> is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat <torus>, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 1 / a / Solution

If a finite group $G$ acts faithfully by orientation-preserving isometries on a closed hyperbolic surface $S$ of genus at least two, then $G\hookrightarrow\operatorname{Mod}(S)$. An isometry isotopic to the identity induces an inner automorphism of the surface group; its suitable lift commutes with every deck transformation and fixes their limit set, forcing the lift and the original isometry to be the identity.


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