If the halting set is Turing-reducible to a c.e. set B, is B m-equivalent to the halting set? Say you have a c.e set $B$. Then $B$ is $m$-reducible to the halting set $K$. If I know additionally that $K \leq_T B$, can I infer $K \equiv_m B$?
Intuitively, I would say yes, as the Turing degree of $K$ i.e. $0’$ is the largest c.e. Turing degree and all c.e. sets are $m$-comparable to $K$. So the the only thing that could happen is $B \lneq_m K$. But then $K \leq_T B \lneq_m K$, which seems implausible. 
 A: Your intuition is reasonable, but it turns out to be wrong: there are indeed such sets!
A simple set is a co-infinite c.e. set with no infinite c.e. set in its complement. Simple sets need not be $\equiv_TK$ (indeed, there are low simple sets), but there are simple sets which are Turing-equivalent to $K$ (see the theorem at the bottom of page $304$ of Post's original paper.). $K$ itself is not simple - indeed, it is as far from simple as possible, namely creative - but no creative set is $m$-reducible to a simple set.
(I wrote earlier - and I blame my lack of sleep and caffeine - that no non-simple set could be $m$-reducible to a simple set. This is utter nonsense: let $S$ be simple and let $\hat{S}=\{2x: x\in S\}$. $\hat{S}$ is not simple - the set of odd numbers is an infinite c.e. set in the complement of $\hat{S}$ - but fixing some $n\not\in S$ we have that $\hat{S}\le_mS$ via the map sending $2k+1$ to $n$ and $2k$ to $k$ for all $k$. In my defence, it didn't take long for me to realize how bonkers my claim was.)
Incidentally, these sorts of considerations led Post and others to suspect that Post's problem ("are there Turing-incomparable c.e. sets?") would be answered by such "combinatorial" constructions. This turned out not to pan out, however - for example, there is a low simple set. Instead, the priority method was needed.
