How to show that if there's a mapping reduction from L to its complement, it doesn't imply that L∈R?

Question

I have the following prove/disprove claim:

if $$L\leq_m L^{c}$$ then $$L\in R$$

I figured out that I can theoretically provide a counter-example where both $$L,L^{c}\not\in(RE\cup co-RE)$$ but couldn't find a mapping reduction between two sets with this property. I tried it with EQ and its complement, but couldn't find a mapping reduction.

Answer 1

Start with any non-recursive set $A$ of natural numbers. Then $L = \{ 2n \mid n \in A \} \cup \{ 2n+1 \mid n \notin A \}$ is a counterexample, because for all natural numbers $x, x\in L$ iff $x+(-1)^x \notin L.$ ($L$ is clearly not recursive, since $A$ is reducible to it.)

This actually shows that $L$ and $L^c$ are recursively isomorphic, which is stronger than what you were asking for.