Why $\textrm{Hom}(\Lambda^2H_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z)$? On page 21 of the first volume of Geometry of Algebraic Curves, there is stated the isomorphism
\begin{equation}
\textrm{Hom}(\Lambda^2H_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z),
\end{equation}
where $A$ is a complex torus defined by a lattice $\Lambda=H_1(A,\mathbb Z)\subset \mathbb C^g$. It is not explained, so it should be obvious, but I find no way to prove it.
Can anyone give me a clue to see why this isomorphism holds? Is it a special case of something more general (that I do not know)?
Thanks in advance.
 A: For a complex torus one can calculate explicitly cohomology as exterior algebra (over integers) on the lattice $\Lambda\subset \mathbb C^g$. You can find this result, for example, in Griffths and Harris Principles of Algebraic geometry. Than your isomorphism is a connection between homology and cohomology in this situation.
A: Just to add a little to Alex's answer:
Topologically, a $g$-dimensional abelian variety is the product of $2g$ circles.
Using the Kunneth Theorem, you can prove that the cohomology ring of a product of circles is generated by the $H^1$, and indeed is isomorphic to the exterior algebra on $H^1$ (with any coefficients).  
A: 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} 
