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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 ?

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  • $\begingroup$ For each $D$, you can find a such $r$. But you wouldn't be able to prove $F$ is additve. So you need a uniform integer $r$ as in answer_bot's answer. Then you lose the condition on the dimension of $L(D+rP_0)$, this is reason why you need extra works to show the independence on $E$ as again in the proof of answer_bot. $\endgroup$ – user143488 Apr 25 '14 at 6:00
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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.

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  • $\begingroup$ Thank you! I'd have the following question: Why did you pick $n>2g+99$ ? Wouldn't $n=g$ suffice to get $L(D+rP_0)$ nonzero ? $\endgroup$ – Klaus Apr 25 '14 at 0:28
  • $\begingroup$ Dear answer_bot: in my opinion the beautiful result in your last line is too difficult to be left as an exercise. $\endgroup$ – Georges Elencwajg Apr 25 '14 at 21:44

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