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I read the following in a paper:

Let $G\subset J$ a subgroup of the Jacobian $J$ which is a countable union of Zariski closed subsets in the abelian variety $J$, so the irredundant decomposition of $G$ contains a unique irreducible component $A$ passing through $0$ which is an abelian subvariety of $J$.

Let \begin{equation*} i:A\hookrightarrow J \end{equation*} be the closed embedding of $A$ into $J$. Let \begin{equation*} H^1(A,\mathbb{Z})\rightarrow H^1(J,\mathbb{Z}) \end{equation*} the homomorphism in cohomology groups induced by $i$.

My question is: how $H^1(A,\mathbb{Z})\rightarrow H^1(J,\mathbb{Z})$ is obtained? It seems to be just the pushforward... If so, does it mean that the Jacobian $J$ and the irreducible component passing through $0$ of a subgroup of the Jacobian have the same dimension?

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    $\begingroup$ Could you please give a link to the paper? $\endgroup$
    – KBS
    Sep 21, 2022 at 18:54
  • $\begingroup$ Why do you think something here should imply that $J$ and $A$ have the same dimension? $\endgroup$
    – KReiser
    Sep 21, 2022 at 19:09
  • $\begingroup$ Dear @KBS the paper is a more general setting: arxiv.org/abs/1405.6430v2, pag 18. In this paper instead of working with the Jacobian $J$ they work with an Abelian variety denoted by $A$, and to which I am denoting by $A$ in my question they denote by $A_0$. $\endgroup$
    – Roxana
    Sep 21, 2022 at 19:16
  • $\begingroup$ Dear @KReiser, because the pushforward, by definition, should be from the $1$-th cohomology of $A$ to the $(1+2r)$-th cohomology of $J$, where $r=\dim(J)-\dim(A)$. $\endgroup$
    – Roxana
    Sep 21, 2022 at 19:20

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