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Say you have a curve $X$ of genus $g \geq 2$, and a surjective map $\phi:X \to E$, where $E$ is an elliptic curve. Denote by $J_X$, $J_E$ the Jacobians of $X$, $E$ respectively. Then we get an induced pushout map ${\phi}_\ast : J_X \to J_E$ and an induced pullback map $\phi^\ast: J_E \to J_X$. My questions are: what do these maps do, explicitly? Also, what are $\phi^\ast \phi_{\ast}$ and $\phi_{\ast} \phi^\ast$?

As an aside, is there a good reference where I can learn more about pushout and pullback maps in more generality (with a view towards algebraic geometry and concrete uses)?

Thanks!

anon1234
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    These have very little to do with pullbacks in the sense of category theory and nothing at all to do with pushouts. For one thing, $\phi_*$ is called the pushforward. – Zhen Lin Nov 06 '13 at 19:46
  • There are only a few articles on the pushouts of morphisms of schemes, if this is your question. They exist only in rather special situations. – Cantlog Nov 06 '13 at 21:57
  • These comments actually clarify a lot - thanks! – anon1234 Nov 06 '13 at 23:06

2 Answers2

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A partial answer: You can learn more about pullbacks & pushouts in a much more abstract setting with Category Theory.

Shaun
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    Thanks! But I am trying to avoid too abstract settings -- what I'm aiming for is a working knowledge (in the context of algebraic geometry). – anon1234 Nov 06 '13 at 17:59
  • It might be a good place to start looking for that, those, since there's a very strong connection between the two. – Shaun Nov 06 '13 at 19:11
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First of all $J_X = \text{Div}_0(X) / \text{P}(X)$, the divisors of $X$ of degree 0 modulo the principle divisors of $X$.

So $\phi \colon X \to E$ induces a map $\phi_\# \colon \text{Div}_0(X) \to \text{Div}_0(E)$ given by $\phi_\#(\sum n_i P_i) = \sum n_i \phi(P_i)$. This $\phi_\#$ maps principle divisors to principle divisors, so in turn induces the map $\phi_* \colon J_X \to J_E$.

For the other map, note that the preimage $\phi^{-1}(Q)$ of a point $Q$ of $E$ consists of finitely many points, the number of which does not depend on $Q$ (provided that you count the points in the preimage with the appropriate multiplicity). So $\phi^{-1}(Q)$ can be seen as a divisor of $X$ and so $\phi$ induces a map $\phi^\# \colon \text{Div}(E) \to \text{Div}(X)$ given by $\phi^\#(\sum_i n_i Q_i) = \sum_i n_i \phi^{-1}(Q_i)$. Divisors of degree 0 map to divisiors of degree 0 under this map, so this $\phi^\#$ restricts to a map $\phi^\# \colon \text{Div}_0(E) \to \text{Div}_0(X)$. This map $\phi^\#$ also maps principle divisors to principle divisors, so induces the map $\phi^* \colon J_E \to J_X$.

Magdiragdag
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  • A few typos: $J_X=Div_0(X)/P(X)$ and $\phi_{#}: Div_0(X)\to Div_0(E)$. One can notice that $\phi_\phi^$ is the multiplication by the degree of $\phi$. While not much can be said on $\phi^\phi_$ (consider for instance what it does on a point of $X$), even if you are expert in category theory. – Cantlog Nov 06 '13 at 21:54
  • Thanks both of you, Cantlog and Magdiragdag - this is exactly what I was looking for! – anon1234 Nov 06 '13 at 23:08
  • Fixed the typos; thanks Cantlog - and a good point to remark that $\phi_* \phi^*$ is multiplication by the degree of $\phi$. – Magdiragdag Nov 06 '13 at 23:14
  • An interesting point also is that $\phi^*$ is injective if and only if $\phi$ doesn't factor over a non-trivial isogeny from an elliptic curve $E'$ to $E$. – rfauffar Nov 07 '13 at 01:29