I would think of torus as follows: in the complex plane $\mathbb{C}$, consider two $\mathbb{R}$-independent vectors $\{v_1,v_2\}$. Then $v_1,v_2$ together with $0$ will determine a parallelogram, and this parallelogram (or vectors $v_1,v_2$) generate a lattice in $\mathbb{C}$ as below:
The quotient of $\mathbb{C}$ by this lattice is a torus, which is a Riemann surface.
If $f\colon \mathbb{C}\rightarrow\mathbb{C}$ is a biholomorphic map which takes this lattice to itself, then it will induce a biholomorphic map from the torus to itself.
I want to know whether every biholomorphic map from torus to itself comes in this way? More precisely, I want to understand automorphisms of torus, and how they come from automorphisms of $\mathbb{C}$?
(I am not well familiar with this area, I am trying to understand these points in ground level; please help me.)
Summary of comments:
A biholomorphism $f:\mathbb C/\Gamma_1 \to \mathbb C/\Gamma_2$ will induces a holomorphism map $F$ between their universal covers, which are both $\mathbb C$. -- John Ma
The above map $F:\mathbb{C}\to\mathbb{C}$ is also a biholomorphism, since $f^{-1}$ can also be lifted in the above way to some $G$ such that $F\circ G=\operatorname{id}$ on an open subset of $\mathbb{C}$.
Every biholomorphism of $\mathbb{C}$ is $z\mapsto az+b$.
Some of the maps listed in 3 descend to the torus: namely, the automorphisms of the lattice, and translations (since all translations commute).
If two biholomorphisms of $\mathbb{C}$ differ by a translation by a lattice element, they induce the same map of the torus. And apart from the lattice-preserving biholomorphisms of $\mathbb{C}$, any translation induces an automorphism of the torus. -- Daniel Fischer