How can I prove the existence or not of a surjective morphism of Riemann surfaces $\Bbb C^\times \to \Bbb C$ ? and viceversa? In case that it exists, how can I prove if there's any that can be extended to a morphism $\Bbb P^1\to \Bbb P^1$?
$\Bbb C$ is identified with a chart $U_1 \subseteq \Bbb P^1$ and $\Bbb C^\times$ with $U_0 \cap U_1 \subseteq \Bbb P^1$.
I guess it will suffice to find and example in case it existed, but I have no clue how to proceed.
A morphism of Riemann surfaces is a holomorphic function from $X \to Y $ for $X$, $Y$ complex manifolds of dimension $n$ and $m$, and
Just to match domain and codomain I think I could use
$exp(1/z):\Bbb C^\times \to \Bbb C$ and
$exp(z):\Bbb C \to \Bbb C^\times $
but how do I prove or not that is a morphism of Riemann surfaces and if can be extended.
$$p(z)=z(z+1)$$ is an example of such morphism. Every non-constant polynomial is surjective. In our case we omit $0$ from the domain, but $z=-1$ also maps to $0$, therefore $p$ is surjective. We can easily expand it to the projective line by setting $p(0)=0$ and $p(\infty)=\infty$. From here on you can easily check that the expanded polynomial is surjective and holomorphic in $0$ and $\infty$.
Edit for clarification.
let $f(z)=z(z+1)$ be a polynomial from $\mathbb{C}$ to itself. Then $f|_{C^x}=p$. For expansion, define $$p'\colon\mathbb{P}^1 \rightarrow \mathbb{P}^1$$ as $$p'(z)=p'([z,1])=[f(z),1]$$ and $$p'([1,0]) = [1,0]$$.
Consider charts $$\phi_i \colon U_i \rightarrow \mathbb{C}$$
Where $U_i=\{[x_0,x_1];x_i\neq 0\} \subset \mathbb{P}^1$
$$\phi_0\colon [1,y] \mapsto y$$ $$\phi_0^{-1} \colon y \mapsto [1,y]$$
and
$$\phi_1\colon [x,1] \mapsto x$$ $$\phi_1^{-1} \colon x \mapsto [x,1]$$
then $p'$ is holomorphic if $\phi_i \circ p' \circ \phi_i^{-1}$ is holomorphic.
First let us check for $[z,1]\in U_1$ then $$\phi_1 \circ p' \circ \phi_1^{-1}(z)=\phi_1 \circ p'([z,1])=\phi_1([f(z),1])=f(z)$$ which is a holomorphic function. So we have checked that $p'$ is holomorphic for all elements in $U_1$. It only remains to show that $p'$ is holomorphic at $\infty=[1,0]\in U_0$. Then let $[1,z]\in U_0$ and $z\neq 0$. We get
\begin{align} &\phi_0 \circ p' \circ \phi_0^{-1}(z)=\phi_0 \circ p'([1,z])=\phi_0 \circ p'([1/z,1]) \\ &=\phi_0([f(1/z),1])=\phi_0([1,\frac{1}{f(1/z)}])=\frac{1}{f(1/z)}=\frac{z^2}{1+z} \end{align}
Which is holomorphic on some small neighborhood of $z=0$. but for $z=0$, we get
$$\phi_0 \circ p' \circ \phi_0^{-1}(0)=\phi_0 \circ p'([1,0])=\phi_0([1,0])=0$$
and we get a function $\phi_0 \circ p' \circ \phi_0^{-1}(z)=\frac{z^2}{1+z}$ for $z\neq0$ and $\phi_0 \circ p' \circ \phi_0^{-1}(0)=0$ which is obviously holomorphic on some small neighborhood of $z=0$.
And so we have checked that $p'$ is also holomorphic at $z=0$.