By the characteric property of the exterior product, there is a canonical isomorphism of $\mathbb Z$- modules \begin{equation} \textrm{Hom}_\mathbb Z(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= Alt^2 _\mathbb Z(H_1(A,\mathbb Z),\mathbb Z) \quad (1)\end{equation} For a complex torus $A= V/\Lambda$ (where $V$ is a complex vector space and $\Lambda \subset V$ a full lattice) there is a canonical isomorphism $H_1(A,\mathbb Z)=\Lambda \quad (0)$ so that $(1)$ becomes \begin{equation} \textrm{Hom}_\mathbb Z(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= Alt^2_\mathbb Z (\Lambda,\mathbb Z) \quad (2)\end{equation}
and multilinear algebra furnishes an isomorphism $Alt^2 (\Lambda ,\mathbb Z)=\wedge ^2 \check{\Lambda} \quad (3)$ where we have used the notation $\check {\Lambda }=Hom_\mathbb Z(\Lambda,\mathbb Z)$
Hence $(2)$ becomes \begin{equation} \textrm{Hom}(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= \wedge ^2 \check{\Lambda} \quad (4)\end{equation}
From algebraic topology we have an isomorphism $ H^1(A,\mathbb Z)\stackrel {algtop} {=}Hom_\mathbb Z(H_1(A,\mathbb Z),\mathbb Z)\stackrel {(0)} {=} Hom_\mathbb Z(\Lambda,\mathbb Z)\stackrel {def} {=}\check {\Lambda } \quad (5)$ which joined to Künneth's theorem permits to prove that the cup product $\Lambda ^2_\mathbb Z \check {\Lambda } \stackrel {(5)}{=}\Lambda ^2_\mathbb Z H^1(A,\mathbb Z) \stackrel {cup}{\to }H^2(A,\mathbb Z) \quad (6)$ is an isomorphism.
From $(4)$ and $(6)$ we obtain the required canonical isomorphism of $\mathbb Z$- modules \begin{equation} \textrm{Hom}(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z)\quad \text {(FINAL)}\end{equation}