If $A \leq_m B$ and $A$ is not mapping reducible to $co\text{-}B$, then $A \leq_T co\text{-}B$.Is this true?My intuition is false even if we can find some special case to make it true such as $A=B=co\text{-}A_{TM}$. However, I still can't find a counterexample. Could anyone give me a little hint?   
If A is mapping-reducible to B and is not mapping-reducible to co-B, is A Turing-reducible to co-B?
computability;reductions