If $X \leq\_T Y + 0'$ does there always exist $Z \leq\_T Y$, $Z \leq\_T X$ with $X \leq\_T Z'$?


Obviously, we can find $Z \leq Y$ where the $y$-th column of $Z$ has a limit equal to $X(y)$. Just let $\langle y, s\rangle$ be given by the computation of $X$ from $0'\_s + Y$. However, I realized I wasn't sure if it was possible for it to be the case that any such approximation includes information that $X$ can't compute. Probably, I'm overlooking something obvious.


If not, is there a natural class of $X$ for which this property is guaranteed?



If $X \leq_T Y + 0'$ does there exist $Z \leq_T Y$, $Z \leq_T X$ with $X \leq_T Z'$?
Ohh, I think I'm being dumb. The answer is no.


Given $X \not\leq\_T 0'$ we build $Y$ using the finite extension method and modify the usual minimal pair construction by coding in the bits of $X$ into $Y$ between the minimal pair requirements.


Now, since $0'$ can figure out how the minimal pair requirements are met it can decode the bits of $X$ in $Y$. Thus $0' +Y$ computes $X$ but any $Z$ must be computable and as $X \not\leq\_T 0'$ the claim fails.


