Let $\mathbf{a}$ and $\mathbf{b}$ be two Turing degrees such that $\mathbf{a'} = \mathbf{a} \oplus \mathbf{b}$. Must it be the case that $\mathbf{a'} \leq \mathbf{b'}$? What if in addition, we know that $\mathbf{b'} \leq \mathbf{a'}$?
This seems to be a basic fact if true, but I cannot see how to easily prove it in general, which leads me to think it should be false. Is there a known counterexample then?
As an extension of Noah's answer, jump-inversion allows you to take any degree ${\bf d} \ge {\bf 0}'$ and obtain a 1-generic ${\bf a}$ with ${\bf a}' = {\bf a}\oplus {\bf 0}' = {\bf d}$. In particular, with ${\bf d} = {\bf 0}^{(3)}$ we have a counterexample.
It's certainly not true in general. If ${\bf a}$ is sufficiently Cohen generic then ${\bf a}\oplus{\bf 0'}\equiv_T{\bf a}'$, so a fortiori we get ${\bf a}\oplus {\bf b}'\ge_T{\bf a}'$ for any ${\bf b}$ whatsoever.
Here is an extreme example. Suppose that $0^{\sharp}$ exists. Then for any nonrecursive real $b\in L$, by Posner-Robinson theorem, there is a 1-generic real $a$ so that $0^{\sharp}\equiv_T a\oplus b\equiv_T a'$. Then for whatever "simply defined'' jump operator $J$, $J(b)$ would be computed by $a'$ but not computes even $a$. This can be generalized to arbitrary defined "jump operator'' under certain set theory axioms.
