Let $A$ be a finite abelian group. I have read that there is an isomorphism of abelian groups between the group $H^2(A;\mathbb{C}^{\times})$, and the group of skew-symmetric bicharacters on $A$, i.e. bilinear maps $b:A\times A\rightarrow \mathbb{C}^{\times}$ such that $b(x,x)=1$ for every $x\in A$. How does one prove this result?
In particular, is it true that this isomorphism is not induced by the canonical map: $$\{bicharacters\}\rightarrow H^2(A;\mathbb{C}^{\times})$$
The answer can be found in proposition 2.6 of this article by D. Tambara.
Namely, let $X^2_a(A)$ denote the group of skew-symmetric bicharacters. There is a canonical map $Alt:Z^2(A;\mathbb{C}^{\times})\rightarrow X^2_a(A)$, given by $\gamma\mapsto \{(a,b)\mapsto \gamma(a,b)/\gamma(b,a)\}$. It is not hard to check that this map factors through $Alt:H^2(A;\mathbb{C}^{\times})\rightarrow X^2_a(A)$.
If $\gamma\in Z^2(A;\mathbb{C}^{\times})$ is such that $Alt(\gamma)=1$, then the central extension of $A$ by $\mathbb{C}^{\times}$ specified by $\gamma$ is abelian. But $\mathbb{C}^{\times}$ is divisible, whence injective as a $\mathbb{Z}$-module. Thus, this extension is trivial, and $\gamma$ is a 2-coboundary.
Surjectivity can be checked by inspection.