Exercise 2.4.2 from Tao-Vu book

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Let $A$ be an additive set in a group $Z$, and let $\phi:Z\to Z'$ be a group homomorphism. Establish the inequalities $$|A|\leq |\phi(A)|\sup \limits_{x\in Z'}|A\cap \phi^{-1}(x)|\leq |2A|.$$ Hint: use the Ruzsa covering lemma to cover $A$ by translates of a subset of $\phi^{-1}(0).$

Proof: since $A$ is an additive set, then $A=\{a_1,\dots,a_n\}$ and let $A'=\{a'_1,\dots,a'_n\},$ where $a'_k:=\phi(a_k)$ for $1\leq k\leq n$. After some deep thoughts I was able to understand the hint and came to the concluison that we need to cover $A$ with the following sets: $(A-a_k)\cap \phi^{-1}(0)+a_k \equiv A\cap \phi^{-1}(a'_k)$ for $1\leq k\leq n$.

Applying Ruzsa's covering lemma to the pair of additive sets $A\cap \phi^{-1}(a'_k)$ and $A$ we obtain that there exists additive set $X_k$ such that $$A\subseteq A\cap \phi^{-1}(a'_k)-A\cap \phi^{-1}(a'_k)+X_k \quad \text{and} \quad |X_k|\leq \dfrac{|A\pm A\cap \phi^{-1}(a'_k)|}{|A\cap \phi^{-1}(a'_k)|}.$$

Then I stucked and don't know what to do further.

EDIT: If $A\subseteq A\cap \phi^{-1}(a'_k)-A\cap \phi^{-1}(a'_k)+X_k$, then $\phi(A)\subseteq \phi(A\cap \phi^{-1}(a'_k))-\phi(A\cap \phi^{-1}(a'_k))+\phi(X_k)$. However, $\phi(A\cap \phi^{-1}(a'_k))=\{a'_k\};$ Hence $\phi(A)\subseteq a'_k-a'_k+\phi(X_k)=\phi(X_k);$ Hence $$|\phi(A)|\leq |\phi(X_k)|\leq |X_k|\leq \dfrac{|A+A\cap \phi^{-1}(a'_k)|}{|A\cap \phi^{-1}(a'_k)|}\leq \dfrac{|2A|}{|A\cap \phi^{-1}(a'_k)|};$$ We were able to show that: $|\phi(A)||A\cap \phi^{-1}(a'_k)|\leq |2A|$ for all $1\leq k\leq n$. But $\max\limits_{1\leq k\leq n}|A\cap \phi^{-1}(a'_k)|=\sup \limits_{x\in Z'}|A\cap \phi^{-1}(x)|$ and we obtain the RHS inequality which is: $|\phi(A)|\sup \limits_{x\in Z'}|A\cap \phi^{-1}(x)|\leq |2A|.$

Hmm it seems nice! I was wondering how we can obtain the LHS inequality?