Jacobi inversion

If $g=0$, [F9] gives the one-point Jacobian and zero dual space; the empty tuple and empty sum represent its unique element, with $N=1$, and $X^0$ is a point. Thus all statements hold. For the remainder assume $g\ge 1$. Fix points $a_1,\dots ,a_g$ and simply connected coordinate disks $V_j$ with coordinate $z_j$ centred at $a_j$ as in [F1]. For each $i,j$ let $f_{ij}$ be the holomorphic primitive of the coefficient of ${\omega }_i$ on $V_j$ with $f_{ij}(a_j)=0$, which exists by [F2], and define $F:V_1\times \cdots \times V_g\to {\mathbb{C}}^g$ by $F(x{)}_i:={\sum }_{j=1}^g{f}_{ij}(x_j)$. Each component is a sum of holomorphic functions of one coordinate, hence $F$ is holomorphic by [F3], and $F(a)=0$.

The complex Jacobian of $F$ at $a$ is the matrix $(\mathrm{\partial }{F}_i\mathrm{/}\mathrm{\partial }{z}_j(a))=(h_{ij}(a_j))$, where $h_{ij}$ is the coefficient of ${\omega }_i$ in the chart $z_j$, i.e. the evaluation ${\omega }_i(a_j)$ of the differential on ${\mathrm{\partial }}_{z_j}$. By [F1] the evaluation $\omega \mapsto (\omega (a_1),\dots ,\omega (a_g))$ is an isomorphism, so this matrix is invertible.

By the inverse function theorem [F4] applied at $a$, there are open neighbourhoods $U_0\subseteq V_1\times \cdots \times V_g$ of $a$ and $V_0$ of $0$ such that $F{\mathrm{\mid }}_{U_0}:U_0\to V_0$ is biholomorphic; in particular there is $\epsilon >0$ with the ball $B(0,\epsilon )\subseteq F(V_1\times \cdots \times V_g)$.

For $x\in V_1\times \cdots \times V_g$ and each $j$ choose a path ${\gamma }_j$ in the simply connected disk $V_j$ from $a_j$ to $x_j$, and put $c_x:={\sum }_j{\gamma }_j$ and $D_x:={\sum }_j(x_j-a_j)=\mathrm{\partial }{c}_x$. By [F2], ${\int }_{c_x}{\omega }_i={\sum }_j{f}_{ij}(x_j)=F(x{)}_i$ for every $i$, so under the identification [F7] the vector $F(x)$ is the functional $\omega \mapsto {\int }_{c_x}\omega$; by [F5] the Abel-Jacobi class of $D_x=\mathrm{\partial }{c}_x$ is $u(D_x)=[F(x)]$.

Let $\xi \in \mathrm{\Omega }(X{)}^{\ast }$, identified with a vector of ${\mathbb{C}}^g$ by [F7]. Choose $N\ge 1$ with $\xi \mathrm{/}N\in B(0,\epsilon )$ and write $\xi \mathrm{/}N=F(x)$ for some $x\in V_1\times \cdots \times V_g$, using step 2.1. Then $\xi =NF(x)$ is the functional $\omega \mapsto {\int }_{N{c}_x}\omega$ of the chain $N{c}_x$, whose boundary is the degree-zero divisor $D:=N{D}_x=N{\sum }_j(x_j-a_j)$; by [F5] and step 2.2, $u(D)=[NF(x)]=[\xi ]$. Hence $u$ is surjective and the displayed representation of $\xi$ holds.

For any class in the Jacobian, step 3.1 supplies a degree-zero divisor $D$ representing it. Set $D^{\mathrm{\prime }}=D+{\sum }_{j=1}^g[a_j]$, of degree $g$. Riemann–Roch [F6] gives $\mathrm{\ell }(D^{\mathrm{\prime }})-\mathrm{\ell }(K-D^{\mathrm{\prime }})=1$, so choose nonzero $f\in L(D^{\mathrm{\prime }})$. The divisor $D^{\mathrm{\prime }\mathrm{\prime }}=(f)+D^{\mathrm{\prime }}$ is effective of degree $g$, since principal divisors have degree zero, and can be written $D^{\mathrm{\prime }\mathrm{\prime }}={\sum }_{j=1}^g[y_j]$ with multiplicities. By [F9] and additivity [F5], $u(D)=u(D+(f))=u({\sum }_j([y_j]-[a_j]))$. Hence the class is the image of $(y_1,\dots ,y_g)$ under the displayed map, proving its surjectivity.

Steps 3.1 and 4.1 prove the surjectivity of $u$ with the explicit division-by-$N$ form and the surjectivity of $X^g\to \mathrm{Jac}(X)$ under the inherited AC of [F8]. The displayed map on $X^g$ depends only on the multiset $\{y_1,\dots ,y_g\}$ because $u$ is additive, so it factors through the quotient by coordinate permutations, and that factor is surjective.