[proofplan] We construct a map from $A+A$ to $B+B$ by sending a represented sum $a_1+a_2$ to $\phi(a_1)+\phi(a_2)$. The Freiman $2$-isomorphism condition is exactly what is needed to show that this definition is independent of the chosen representation of the sum. We then prove that the resulting map is bijective: surjectivity follows from the surjectivity of $\phi$, and injectivity follows from the reverse implication in the Freiman relation. [/proofplan] [step:Define the induced map on sumsets and prove it is well-defined] Define \begin{align*} \Phi: A+A &\to B+B \end{align*} as follows. For $x \in A+A$, choose $a_1,a_2 \in A$ such that $x=a_1+a_2$, and set \begin{align*} \Phi(x):=\phi(a_1)+\phi(a_2). \end{align*} The value lies in $B+B$ because $\phi(a_1),\phi(a_2) \in B$. We verify that $\Phi$ is independent of the chosen representation of $x$. Suppose $a_1,a_2,a_3,a_4 \in A$ satisfy \begin{align*} a_1+a_2=a_3+a_4. \end{align*} Since $\phi$ is a Freiman $2$-isomorphism, the forward implication gives \begin{align*} \phi(a_1)+\phi(a_2)=\phi(a_3)+\phi(a_4). \end{align*} Thus every two representations of the same element of $A+A$ give the same value in $B+B$, so $\Phi$ is a well-defined map. [/step] [step:Use surjectivity of $\phi$ to prove surjectivity of the induced map] Let $y \in B+B$. By definition of $B+B$, there exist $b_1,b_2 \in B$ such that \begin{align*} y=b_1+b_2. \end{align*} Since $\phi: A \to B$ is surjective, there exist $a_1,a_2 \in A$ with \begin{align*} \phi(a_1)=b_1, \qquad \phi(a_2)=b_2. \end{align*} Set $x:=a_1+a_2 \in A+A$. By the definition of $\Phi$, \begin{align*} \Phi(x)=\phi(a_1)+\phi(a_2)=b_1+b_2=y. \end{align*} Hence $\Phi$ is surjective. [/step] [step:Use the reverse Freiman implication to prove injectivity of the induced map] Let $x,z \in A+A$ satisfy $\Phi(x)=\Phi(z)$. Choose $a_1,a_2,c_1,c_2 \in A$ such that \begin{align*} x=a_1+a_2, \qquad z=c_1+c_2. \end{align*} By the definition of $\Phi$, \begin{align*} \phi(a_1)+\phi(a_2)=\Phi(x)=\Phi(z)=\phi(c_1)+\phi(c_2). \end{align*} Since $\phi$ is a Freiman $2$-isomorphism, the reverse implication gives \begin{align*} a_1+a_2=c_1+c_2. \end{align*} Therefore $x=z$, so $\Phi$ is injective. [/step] [step:Conclude equality of doubling cardinalities] The map $\Phi: A+A \to B+B$ is both injective and surjective, hence it is a bijection. Since $A$ and $B$ are finite, the sumsets $A+A$ and $B+B$ are finite. A bijection between finite sets preserves cardinality, so \begin{align*} |A+A|=|B+B|. \end{align*} This proves the theorem. [/step]