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When people consider a map $\varphi: C \rightarrow C$ between algebraic curves and they mention the "associated map" on the Jacobian of $C$. Which map do they mean? Do they mean $\varphi^{\*}$ or $\varphi_*$ (or some other map)?

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Does it matter? The principal polarization implies that either one of $\phi^\ast$ and $\phi_\ast$ determines the other... –  Dan Petersen Jun 14 '12 at 18:40
    
Typically, one extends the map linearly to divisors, i.e. $\sum n_iP_i \mapsto \sum n_i\phi(P_i)$, so this is $\phi_*$. But, as Dan pointed out, usually it doesn't matter. –  Felipe Voloch Jun 14 '12 at 19:22

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We have to remind ourselves who the Jacobian $J(C)$ is. One quick and dirty answer is that the Jacobian of a curve $C$--seen as a riemann surface--is just the quotient $H_1(C, \mathbb{R}) /H_1(C, \mathbb{Z})$ -- this is an insufficient definition since I have not specified either a metric or a Kahler form. But it's sufficient to convince us that the induced map $J(\phi): J(C) \to J(C')$ ought to be $H_1(\phi)$ factored through the quotient. Again, the above description of $J(C)$ is incomplete but sufficient our purpose.

Any riemannian manifold has a meaningful associated Jacobi variety. In the case of complex curves, there's that alternative description as the quotient $H^0(C, \Omega^1)^* / jH_0(C, \mathbb{Z})$ where $j: C \to H_1(C, \mathbb{Z})$ is the Abel-Jacobi period mapping (defined up to a choice of base point on $C$--but all homotopy equivalent). Here we find that the induced map $J'(C)$ between jacobians is $\phi^*$ where one deserves to convince themselves that map actually factors through the quotient.

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As Bill Clinton once said, it depends on the meaning of "is". The jacobian is what you said only over $\mathbb{C}$. –  Felipe Voloch Jun 14 '12 at 20:28
    
Ah, quite right. What i've described is almost exclusively over $\mathbb{C}$. Over arbitrary fields I have no idea. –  J. Martel Jun 15 '12 at 0:11

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