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?