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 $$\textrm{Hom}(\Lambda^2H_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z),$$ 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)?

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By the characteric property of the exterior product, there is a canonical isomorphism of $\mathbb Z$- modules $$\textrm{Hom}_\mathbb Z(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= Alt^2 _\mathbb Z(H_1(A,\mathbb Z),\mathbb Z) \quad (1)$$ 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 $$\textrm{Hom}_\mathbb Z(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= Alt^2_\mathbb Z (\Lambda,\mathbb Z) \quad (2)$$

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 $$\textrm{Hom}(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)= \wedge ^2 \check{\Lambda} \quad (4)$$
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 $$\textrm{Hom}(\Lambda^2_\mathbb ZH_1(A,\mathbb Z),\mathbb Z)\cong H^2(A,\mathbb Z)\quad \text {(FINAL)}$$

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I have given the proof in such tedious detail in order that it might be easier to localize and discuss a possibly obscure point, but also to try and convince myself that all isomorphisms are canonical . Of course this says the same as the other, more pleasantly concise answers – Georges Elencwajg Dec 20 '12 at 17:45
Dear Georges, thank you very much for your effort in showing all the details. I still have a doubt about (1) and (6). To be completely honest, I know what the alternating algebra $\textrm{Alt}(V)$ is (for us, $V=\Lambda$), but I do not know the notation $\textrm{Alt}(V,\mathbb Z)$. Do you mean "work with $\mathbb Z$-modules instead of vector spaces"?; I know there is an isomorphism $\textrm{Alt}(V)\cong \Lambda(V)$, so in (1) I have problems with the "Hom" in the lhs. As for (6), I miss the application of Kunneth formula, but again it's my fault: I am not familiar with it yet. – Brenin Dec 22 '12 at 18:12
Dear atricolf: for a $\mathbb Z$-module $A$ the notation $Alt^2_\mathbb Z (A,\mathbb Z)$ denotes the $\mathbb Z$-module consisting of bilinear maps $b:A\times A\to \mathbb Z$ which are alternating, i.e. such that $b(a,a)=0$ for all $a\in A$. This $\mathbb Z$-module $Alt^2_\mathbb Z (A,\mathbb Z)$ is canonically isomorphic not to the $\mathbb Z$-module $\wedge^2_\mathbb Z A$ but to its dual $Hom_\mathbb Z(\wedge_\mathbb Z ^2A,\mathbb Z)$. This is because of the perfect pairing $Alt^2_\mathbb Z (A,\mathbb Z)\times \wedge^2 A\to \mathbb Z: (b,a\wedge a')\mapsto b(a,a')$. – Georges Elencwajg Dec 22 '12 at 18:49
Dear Georges, thank you very much! – Brenin Dec 24 '12 at 9:59
My pleasure, atricolf! – Georges Elencwajg Dec 24 '12 at 10:10

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.

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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).