We know that the halting problem $A_{TM}$ and the diagonal language K are mapping reducible to each other. Furthermore both are complete with respect to the mapping reduce relation.
I would like to prove that any r.e. language is mapping reducible to K directly without reducing to $A_{TM}$. How can this be done?
Let $A=L(M)$ be any r.e. language. We reduce $A$ to $K$ using the following reduction function $f(x)$.
Given any word $x$, construct a TM $N_x$ such that, when run on input string $y$, it ignores it, and simulates $M$ on $x$. If $M$ accepts, we make $N_x$ accept, otherwise we make $N_x$ diverge. We let $f(x) = \langle N_x \rangle$.
$f$ is clearly a total computable function. It is also simple to check that $x\in A \iff f(x)\in K$.
Note that $f(x)\in K$ means that the TM encoded by $f(x)$ halts on input $f(x)$. This means that $N_x$ halts on input $y = \langle N_x \rangle$. The value of $y$ is actually irrelevant, since $N_x$ will ignore it, and halt only when $M$ accepts $x$, i.e. when $x\in A$.
In a sense, the "trick" here was to ignore the diagonal property which is used in the definition of $K$, and just make $N_x$ halt on all inputs $y$ or diverge on all inputs $y$. In such way, we surely include the diagonal, and we simplify the reduction function $f$.
If we really wanted, it is possible to define a function $f(x)$ which produces the encoding of $N_x$, which always halts (or always diverges, if we prefer that) on any $y\neq f(x)$, and instead halts on the the diagonal $y=f(x)$ if and only if $x\in A$. Defining such $f$, however, is harder, and requires the second recursion theorem. Ignoring $y$ makes $f$ much simpler to construct.
