If$A \leq_T B$ is given, can you reduce $\overline{A}$ to $B$ and vice-versa

Question

If you are given two languages $A$, $B$ and $$A \leq_T B.$$ Is it possible to $\overline{A} \leq_T B$ or $A \leq_T \overline{B}$?

Here is my shot.

Case 1: $\overline{A} \leq_T B$

This is only possible if $A \leq_m B$ exists and $B=\overline{B}$. As you can transform any many-one reduction to its complement, we can show that if $A \leq_m B$, then $\overline{A} \leq_m \overline{B}=B$. Thus $\overline{A} \leq_m B$.

Case 2: $A \leq_T \overline{B}$

This is the same as above but we need to change the role of $A$ and $B$.

Answer 1

You say:

This is only possible if $A \le_m B$ exists and $B=\overline{B}$

But that condition simplifies to "false" since it can never the case that $B=\overline{B}$.

Anyway the answer to both cases is "yes". If $A \le_T B$ then there is a Turing machine $T$ with oracle $B$ that decides $A$. By complementing $T$'s output we obtain a Turing machine with oracle $B$ that decides $\overline{A}$ showing that $\overline{A} \le_T B$. Moreover, consider a Turing machine $T'$ with oracle $\overline{B}$ that simulates $T$ except for the following: whenever $T$ invokes its oracle for $B$ on some input $w$, $T'$ invokes its oracle for $\overline{B}$ on $w$ and then complements the result. Clearly $T'$ still decides $A$, thus showing that $A \le_T \overline{B}$.