Let $A/\mathbb{Q}$ be an abelian variety with good reduction at a prime $p$. Assume $\mathcal{A}/\mathbb{Z}\_{(p)}$ is an integral model at $p$(hence proper smooth).


For any number field $K$ and any prime ideal $\mathfrak{p}$ of $K$ over $p$ with residue field $\kappa$, by valuation criterion on $\mathcal{O}\_{K,\mathfrak{p}}$, any $K$-point of $A$ extends uniquely to an $\kappa$-point of $\mathcal{A}$'s special fiber.


By fixing a prime ideal of $\overline{\mathbb{Z}}$ lying over $p$ and taking injective limit, we get a well defined reduction map\begin{equation} A(\overline{\mathbb{Q}})\rightarrow \mathcal{A}\_{\mathbb{F}\_p}(\overline{\mathbb{F}\_p}). \end{equation}


Does the above map induce isomprhisms on $\ell$-primary parts of both groups? Notice that both of their $\ell$-primary subgroups are isomorphic to $(\mathbb{Q}\_{\ell}/\mathbb{Z}\_{\ell})^{\oplus 2g}$.



The reduction map on the $\ell$-primary torsion of abelian variety
This is true. One reference for this is Theorem C.1.4 in Hindry-Silverman's "Diophantine Geometry: An Introduction" which says that for an abelian variety $A$, number field $K$ and any prime $\mathfrak p$ above $p$, the reduction map $A(K)\to A(O\_K/\mathfrak p)$ is injective on prime-to-$p$-torsion. The result you ask for follows since $\overline{\mathbb Q}$ is a union of number fields.


One way to prove the required injectivity follows from the description of the kernel of the reduction map over the completion $K\_{\mathfrak p}$, which can be identified with the formal group associated to $A$. Formal groups over $p$-adic fields have no prime-to-$p$ torsion.


