Specifically, is it the case that (for $a,b\in\omega^\omega$) $a$ $\leq_T$ $b$ implies $a$ $\leq_c$ $b$? I suspect it might be trivial, but not knowing much Recursion Theory, it's hard to see how it could.
Thank you in advance.
Specifically, is it the case that (for $a,b\in\omega^\omega$) $a$ $\leq_T$ $b$ implies $a$ $\leq_c$ $b$? I suspect it might be trivial, but not knowing much Recursion Theory, it's hard to see how it could. Thank you in advance. 


Yes, if $a\le_T b$ then $a$ is firstorder definable from $b$; in particular $a\in L(b)$ so $a\le_c b$. 


As Bjorn pointed out, $a\leq_Tb$ implies $a\leq_cb$. But it should also be mentioned that constructible degrees are much coarser then Turing degrees: Suppose $\aleph_1^L$ is the real $\aleph_1$. Then every constructible degree (set of $\leq_c$equivalent reals) is uncountable (take your real $a\in\omega^\omega$ and consider all coordinate wise sums of $a$ and a constructible real. These sums are all of the same constructible degree as $a$). Every Turing degree is countable since there are only countably many Turing machines. 

