I got stuck in an exercise from Tao and Vu's book Additive Combinatorics. It is ex. 5.3.4. on page 226.
In the following let (Z,+) and (W,+) be two abelian groups and let A $\subset$ Z and B $\subset$ W be two finite subsets.
The definition of a Freiman hom. goes as follows:
Let $k \in \mathbb{N}\setminus \{0\}$. We call a map $\varphi:A\rightarrow B$ a $k$-Freiman homomorphism : $\Leftrightarrow \; \sum_{i=1}^{k}{a_{i}}= \sum_{i=1}^{k}{\tilde{a}_{i}} \implies \sum_{i=1}^{k}{\varphi(a_{i})}= \sum_{i=1}^{k}{\varphi(\tilde{a}_{i})}$ for $(a_{i})_{i = 1}^{k}, (\tilde{a}_{i})_{i = 1}^{k}$ elements in A.
The exercise I'm stucked with is:
Let $\varphi: A \rightarrow B$ be a Freiman $k$-homomorphism for any $k\geq 2$. Suppose further that $A$ generates $Z$ as a group.
$\exists!$ group homomorphism $\psi: Z \rightarrow W$ and $\exists! c \in W$ such that $\varphi(x)=\psi(x)+c \;\; \forall x \in A.$
Here are my thoughts so far:
Clearly $c$ should be something like $\varphi(0)$ but in general $0 \notin A.$ I can define a $\tilde{\varphi}$ on $Z$ to be just $\widetilde{\varphi}(z):=\sum{\varphi(a_i)}$ for $z=\sum{a_i} \in Z.$ However, I don't see how I can show that this is well-defined:
If $z=\sum_{i=1}^{n}{a_i}=z=\sum_{i=1}^{m}{\tilde{a}_i}$ and (WLOG) $\; m-n \geq 0\,$ I could add just zeroes on the left side to get the same number of summands and then use the property of the Freiman homomorphism. But as I cannot make sense of $\varphi(0)$ this doesn't help for getting the well-definition.
I also cannot add any other elements from $A$, can I?
The remaining parts of the proof I will be able to do, but I am stuck with this problem. I am very thankful for any hints or comments.