If $L_1\in R$ and $L_2$ is non-trivial language then $L_1\leq L_2$

Question

Language $L$ is trivial if $L=\varnothing$ or $L=\Sigma^*$. I'm trying to prove the following theorem:

If $L_1\in R$ and $L_2$ is non-trivial language then $L_1\leq L_2$.

If $L_2$ is non-trivial the $L_2\neq\varnothing$ and $L_2\neq\Sigma^*$. If I want to prove that $L_1\leq L_2$, then I need to show that there exists $f:\Sigma^*\to\Sigma^*$ which is total, can be calculated by a Turing machine and follows: $\forall x\in\Sigma^*$, $x\in L_1$ iff $f(x)\in L_2$. I also know that $L_1\in R$ then there is a turing machine $M_1$ that decides language $L_1$. But how do I continue from here? How $f$ should look like?

Answer 1

We know there are two $x_1,x_2\in \Sigma^*$ such that $x_1\in L_2$ and $x_2\notin L_2$, since $L_2$ is not trivial. Now, we will build the following reduction:

$f(x)=\cases{x_1 & $x\in L(M_1)=L_1$ \\ x_2 & otherwise}$

Notice that $f$ is computable: calculate $M_1(x)$, and output either $x_1$ or $x_2$ according to the answer you got.