# Intuition behind $\mathcal{M}_g\cong\mathcal{T}(S)/\text{Mod}(S)$.

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?

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 $$X_2$$.