If A is mapping-reducible to B and is not mapping-reducible to co-B, is A Turing-reducible to co-B? 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?
Hint: If $A \leq_m B$ then there is a computable function $f$ such that $x \in A$ iff $f(x) \in B$ iff it is not the case that $f(x) \in \text{co-}B$. Can you use this to Turing-reduce $A$ to $\text{co-}B$?