Let $L_1, L_2 \subseteq \Sigma^*$ be Turing recognizable languages. Suppose there is a computable function $f : \Sigma^* \to \Sigma^*$ such that $x \notin L_1 \to f(x) \in L_2$. Can we prove that $L_1$ is Turing decidable?
I tried constructing $L_3 = \{x : f(x) \in L_2\}$ and proved that $\overline{L_1} \subseteq L_3$. However, I realized that if $L_3 = \Sigma^*$, there is no way to prove whether $L_1$ is decidable or not.
Did I mess up somewhere?
Suppose $f$ is a constant function such that $\forall x f(x) \in L_2$. Then $f$ is computable and satisfies your condition whether or not $L_1$ is decidable. It seems like you need more conditions on $f$.