Actually, I just want to understand first sentence of Here. It says that
Let $X$ be the $\mathbb{CP}_{1}$ with $n$ points deleted. Let $n \geq 3$. If I understand correctly, the universal covering of $X$ is isomorphic to the upper half plane as a complex analytic space.
And the answer is related to Fuchsian group, but I still do not find a introductory paper about that. Could you explain this statement, or just give some reference about why Fuchsian group is related to universal covering of $X$?
First, you need to know the Uniformization Theorem:
Theorem. Every simply-connected Riemann surface $X$ is biholomorphic to a model surface which is either $S^2$ or ${\mathbb C}$ of the hyperbolic plane $H^2$ (the upper half-plane if you like).
Depending on which of the three models occur, one calls $X$ spherical, euclidean or hyperbolic. (Sadly, this terminology is inconsistent with the AG terminology, where "elliptic" means "euclidean" while "rational" means "spherical" and "general type" is "hyperbolic".) The same applies to (connected) surfaces $Y$ which are covered by $X$: Such a surface is biholomorphic to the quotient of a model surface by a discrete group $G$, of biholomorphic automorphisms, $G\cong \pi_1(Y)$, acting freely (without fixed points) on $X$. Now, the question becomes which groups $G$ can act in such a way on the model surfaces. The group $Aut({\mathbb C})$ of conformal automorphisms of ${\mathbb C}$ consists of complex affine transformations $$ z\mapsto az+b. $$ (One usually proves this in an undergraduate CA class.) Since $G<Aut({\mathbb C})$ is supposed to act freely, it follows that $a=1$ for all elements of $G$. Thus, $G$ is free abelian of rank $\le 2$. This implies that all surfaces of euclidean type have abelian fundamental group. Since $\pi_1$ of $n$ times punctured sphere is free of rank $n-1$, if $n\ge 3$ then this group is nonabelian, hence, the surface cannot have parabolic type. It obviously cannot have elliptic type. Thus, we conclude that such a surface is the quotient of the hyperbolic plane by a group of automorphisms (hyperbolic isometric) acting properly discontinuously and freely. The same argument shows that every connected RS of euclidean type is either $T^2$ (aka an "ellipic curve"), or is conformal to ${\mathbb C}^\times$ or to ${\mathbb C}$.