# The cohomology of the Jacobian and its subvariety

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?

• Why do you think something here should imply that $J$ and $A$ have the same dimension? Sep 21, 2022 at 19:09
• 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 $A_0$. Sep 21, 2022 at 19:16
• 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 $r=\dim(J)-\dim(A)$. Sep 21, 2022 at 19:20