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Let $S$ be a compact Riemann surface of genus $g$. The mapping class group $\text{Mod}(S)$, constitued by all homotopy classes of orientation-preserving diffeomorphisms, acts on $\mathcal{T}(S)$ (Teichmüller's space), by: $$[h]\cdot[X,f]:=[X,f\circ h^{1}],$$ for all $[h]\in\text{Mod}(S)$.

The orbit space $\mathcal{T}(S)/\text{Mod}(S)$ is then naturally identified with $\mathcal{M}_g$, the moduli space of Riemann surfaces of genus $g$.

Why this is the case? Why would an orbit be identified with a biholomorphism class of Riemann surfaces?

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Loosely speaking, in $T(S)$ you identify two marked Riemann surfaces when the markings $f_1$ and $f_2$ are homotopic to the identity in some sense (more precisely, when they are homotopic to a biholomorphism between $X_1$ and $X_2$).

So the same underlying (biholomorphism class of) Riemann surface $X$ can give you different points in $T(S)$ for different markings. But when you mod out by $Mod(S)$, those points are now the same in $T(S)/Mod(S)$. In other words, with your notations, in $T(S)/Mod(S)$ two points $[X_1,f_1]$ and $[X_2,f_2]$ are the same if and only if $X_1$ is biholomorphic to $X_2$.

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  • $\begingroup$ Whether this explanation is satisfying might depend on the OP's definition of Teichmüller space. If they are defining it using hyperbolic structures, the connection to complex structures is less obvious. $\endgroup$ – Cheerful Parsnip Apr 9 at 16:09
  • $\begingroup$ This is exactly the definition I'm used to. Now it all makes sense! Thank you $\endgroup$ – Gabriel Apr 9 at 16:28

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