$\DeclareMathOperator\Hom{Hom}$Let $A$ be a finite abelian group and $\text{Sym}(A)$ the (abelian) group of symmetric bilinear forms over $A$ valued in $\mathbb{R}/\mathbb{Z}$.
A quadratic function on $A$ is $q:A\rightarrow \mathbb{R}/\mathbb{Z}$ such that $q(-a)=q(a)$ and $$ \chi_q(a,b)=q(a+b)-q(a)-q(b) $$ is bilinear. Hence if $q$ is quadratic $\chi_q\in \text{Sym}(A)$. The map $q\rightarrow \chi_q$ is clearly linear and thus defines a homomorphism from the group $Q(A)$ of quadratic functions to $\text{Sym}(A)$.
Is this map surjective? In other words, is there always a quadratic refinement of a symmetric bilinear form valued in $\mathbb{R}/\mathbb{Z}$? I think the answer is yes but I am not sure about a concrete proof.
Assuming the answer to the first question is yes, what is the kernel of the map? I think it should be isomorphic to $\Hom(A,\mathbb{Z}_2)$ by looking at examples (and from other abstract arguments, see Is there a clear pattern for the degree $2n$ cohomology group of the $n$'th Eilenberg-MacLane space?), but It is not clear to me how to prove this, and even how to see $\Hom(A,\mathbb{Z}_2)$ as a subgroup of $Q(A)$.
