In the book Algebraic Geometry I edited by Safarevich, the following universal property of the Jacobian variety of an algebraic curve is given page 158 (with no more details):
The Abel mapping $a: C\to J(C)$ is universal: for any regular map $f: C\to A$ into an abelian variety $A$, there exists a unique regular homomorphism $F: J(C)\to A$ such that $F\circ a=f$.
Suppose that $C$ is a compact Riemann surface. Then $J(C)$ is isomorphic as a group to $\text{Pic}^0(C)$. Pick a base point $P_0\in C$. The Abel mapping is $$P\to [P-P_0].$$ We can see what $F$ should be:
- Assuming the existence of $F$, we would have $$F([E-dP_0])=\sum n_if(P_i)$$ if $E=\sum n_iP_i$ is an effective divisor of degree $d$.
- Pick an integer $r\ge g$. By Riemann-Roch, any divisor class $[D]$ of degree zero contains an effective divisor $E$ of degree $g$: $[D]=[E-rP_0]$. Thus we would finally have $$F([D])=F([E-rP_0])$$ which defines $F$ everywhere.
However, I can't see how this could be well-defined.
In the second step, we would like to be able to pick $r$ so that $\dim L(D+rP_0)=1$, but is this even possible ?
This is kind of a fun question. To prove you thing is well defined consider the morphism $$ C^n = C \times \ldots \times C \longrightarrow A,\quad (p_1, \ldots, p_n) \longmapsto \sum f(p_i) $$ This clearly factors through the symmetric power $S^n(C) = C^n/S_n$ giving a morphism $f^n : S^n(C) \to A$. Since C is a smooth curve, this symmetric power is a smooth projective variety too. In fact $S^n(C)$ is the moduli space of effective divisors of degree $n$ on $C$. If $n > 2g + 99$, then through every point $D$ of $S^n(C)$ there is a projective space passing through, namely the linear system $|D|$. To see that your map is well defined we have to show that $f^n$ maps each of these $|D|$ to a point.
To do this it certainly suffices to show that any morphism $f : \mathbf{P}^1 \to A$ is constant. Now taking $C = \mathbf{P}^1$ in the discussion above, we see that if $f$ is not constant, then we get for every $n$ a nonconstant map $\mathbf{P}^n = S^n(\mathbf{P}^1) \to A$. To finish the proof by contradiction one has just to do the following
Exercise: If $X$ is a variety of dimension $< n$, then every morphism $\mathbf{P}^n \to X$ is constant.