mapping reductions from R to RE

Question

Let $L_1$ be some language in $R$. Let $L_2$ be some language in $RE$. Is it necessarily that $L_1 \leq_m L_2$ ? I know that for non trivial $L_1$,$L_1$ in $R$ it is right to say that $L_1 \leq_m L_2$. But I can't prove the first case.

and another question: I am almost certain that the following is true, though I have not found any reference to it on the Internet: The identity function is a mapping reduction from $\emptyset$ to $\emptyset$.

Answer 1

If $L_2 \neq \Sigma^*$ and $L_2 \neq \emptyset$ then $L_1 \in R$ and $L_2 \in RE$ implies $L_1 \le_m L_2$.

Let $T$ be a Turing machine that decides $L_1$. Let $a,b \in \Sigma^*$ such that $a \in L_2$ and $b \not\in L_2$. For $x \in \Sigma^*$, define $\phi(x) = \begin{cases} a & \text{ if $T(x)$ accepts }\\ b & \text{ if $T(x)$ rejects } \end{cases}$. It is easy to check that $\phi$ is a mapping reduction from $L_1$ to $L_2$.

If $L_2 = \Sigma^*$ then $L_1 \in R$ and $L_2 \in RE$ does not imply $L_1 \le_m L_2$.

This can be seen, e.g., by choosing $L_1 = \emptyset$.

If $L_2 = \emptyset$ then $L_1 \in R$ and $L_2 \in RE$ does not imply $L_1 \le_m L_2$.

This can be seen, e.g., by choosing $L_1 = \Sigma^*$.