For each language $L \in L(RE)$ there are a homomorphism $h$ and two context-free languages $L_1$ and $L_2$ such that $L = h(L_1 \cap L_2)$.
I understand that this is because context-free languages are not closed under intersection so $L_1 \cap L_2$ will produce non-context-free languages. Since context-sensitive grammars doesn't allow erasing rules which context-free grammars allow to, then the language must be recursively enumerable language. Then recursively enumerable languages are closed under homomorphism.
If $L_1$ and $L_2$ are context-sensitive languages, does $L = h(L_1 \cap L_2)$ still result in $L \in L(RE)$, as context-sensitive languages do closed under intersection but not morphisms?