Let $M,N$ be transitive structures of the language of set theory. Let $A\subseteq M$ be amenable to $M$, that is, for all $x\in M$, we have $x\cap A\in M$. Let $\sigma:M\to N$ be a cofinal $\Sigma_0$-preserving embedding ($\Sigma_0$ here refers to formulas in the language of set theory alone). Define $A'\subseteq N$ by:
$A'=\bigcup\{\sigma(A\cap x)\mid x\in M\}$.
Is it the case that $A'$ is amenable to $N$?
I'm thinking to show that, given $x\in N$, $x\cap A'\in N$ by showing that $x\cap A'$ is in the range of $\sigma$ or there is some $B\in M$ such that $(x\cap A')\in \sigma(B)$. But I've not been getting far with these. Any help appreciated!
We will additionally need that $M$ models union EDIT: and that $N$ is closed under intersecions. Let $y\in N$. Since $\sigma$ is cofinal, there is $x^\prime\in M$ so that $y\in\sigma(x^\prime)$. If we put $x=\bigcup x^\prime$ then by $\Sigma_0$-elementarity, $\sigma(x)=\bigcup\sigma(x^\prime)$. Hence $y\subseteq\sigma(x)$. Now I claim that $$y\cap A^\prime= y\cap (\sigma(x)\cap A^\prime)=y\cap \sigma(x\cap A)\in N$$ We only need to argue that the last equality holds true, for which it is enough to show $\sigma(x)\cap A^\prime=\sigma(x\cap A)$. $\supseteq$ is trivial, so suppose $z\in \sigma(x)\cap A^\prime$. As $z\in A^\prime$, there is $x_0\in M$ with $z\in\sigma(x_0\cap A)$. Thus $$z\in\sigma(x_0\cap A)\cap \sigma(x)=\sigma(x\cap x_0\cap A)\subseteq\sigma(x\cap A)$$ by $\Sigma_0$-elementarity.