Take $x_1,x_2,x_3,x_4\in\phi^{-1}(W)$ with $x_1+x_2=x_3+x_4$. Write $x_3=x_1+a$ and $x_4=x_1+b$; then $x_2=x_1+a+b$. Hence
$$
D=\phi(x_1)-\phi(x_3)-\phi(x_4)+\phi(x_2)
$$
belongs to $X$. All four values of $\phi$ lie in the same affine subspace $W=V+w$, and their coefficients in $D$ sum to zero, so $D\in V$. The hypothesis $V\cap X=\{0\}$ gives $D=0$, precisely the additive-quadruple identity required of a <Freiman homomorphism>. This is the <second-difference obstruction to a Freiman homomorphism>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 129 / 2 / i / Solution

Let $X$ be the image of all additive second differences
$$
\phi(x)-\phi(x+a)-\phi(x+b)+\phi(x+a+b).
$$
If a subspace $V$ satisfies $V\cap X=\{0\}$, then the restriction of $\phi$ to the inverse image of every coset of $V$ is a Freiman homomorphism.


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