Why is it biholomorphic? The proof of Theorem 13.5 in "Lectures on Riemann Surfaces" by Otto Forster begins by saying  

Set $U_1:={\mathbb P}^1 \backslash \infty$ and $U_2:={\mathbb P}^1 \backslash 0$.
  Since $U_1 = {\mathbb C}$ and $U_2$ is biholomorphic to ${\mathbb C}$, it follows from (13.4) that $H^1(U_i, {\cal O})=0$. 

In "1.5 Examples of Riemann Surfaces" in the book the maps $\phi_i:U_i \rightarrow {\mathbb C}, i=1,2$ are defined as follows:

$\phi_1$ is the identity map and
  $$
\phi_2(z) := 
\left\{
\begin{array}{ll}
1/z & \mbox{for} \; z \in {\mathbb C}^*\\
0   & \mbox{for} \; z = \infty
\end{array}
\right.
$$

But, It seems to me that $\phi_2$ cannot be biholomorphic at $\infty$, because since
$\phi_2'(z)=-1/z^2$, 
$$
\lim_{z\rightarrow \infty} \phi_2'(z) = 0.
$$
Could someone point out where I made a mistake ?  
 A: The "paradox" you are asking about is extremely interesting and I can only congratulate you on the dynamic  way you are studyng mathematics. 
1) The first confusing point is that  for a holomorphic function $\phi$  on an open subset $U$ of a  manifold, in your case $\phi_2$ and $U_2$ , the naïve notion of derivative $\phi'(a)$ at a point $a\in U$ as a number does not work: you would get different  numbers according to the chart you use.
The correct notion is that of a linear form on the tangent space
$$ d_a\phi:T_a (U) \to   T_{\phi(a)} \mathbb R  = \mathbb R        $$
The recipe for computing $d_a\phi$  is to choose a chart $w$ in  a  neighbourhood of $a$, to consider the composed function $\phi_w=\phi \circ w^{-1}$ and to decree that we have 
$$ d_a\phi (t\cdot \frac {\partial}{\partial w}) =t\cdot \phi_w'(w(a))        \quad (t\in \mathbb R) $$
If you do that in your situation with $U=U_2, a=\infty, \phi=\phi_2=w$, you will find completely tautologically that $d_\infty (\phi_2):T_\infty (\mathbb P^1)\to \mathbb R $ is given by $d_a\phi_2 (t\cdot \frac {\partial}{\partial w})= t$, since $(\phi_2)_w=\phi_2 \circ w^{-1}$ is the identity.  
2) The second confusing point is that you are not allowed to calculate $d_\infty\phi_2$ by means of the chart $\phi_1=z$  since its domain does not contain infinity: $\infty\notin U_1=dom(\phi_1)=\mathbb C$.  
3) In the language of divisors (introduced on page 127 of your  book) the divisor of the global meromorphic differential form $dw\in \Gamma ( \mathbb P^1, \Omega_X ^1 \otimes_{\mathcal  O_X} \mathcal M_X)$ is $div(w)=-2\cdot (0)$ and for $dz\in \Gamma ( \mathbb P^1, \Omega_X ^1 \otimes_{\mathcal  O_X} \mathcal M_X)$ it is $div(z)=-2\cdot (\infty)$.
Both  results confirm that the line bundle of holomorphic $1$-forms on $\mathbb P^1$, a Riemann surface of genus $g=0$, has degree $2g-2=2\cdot0-2=-2$.
