Suppose you have a program p and you want to write another program q so that q(1)↓ ⇔ p(0)↓. Well, make q(1) do exactly the same thing as p(0). Keep in mind that you have p in hand when you write q, and you can build p into q.

The following function f is a mapping reduction from A to B.

  f(p)
    Build and return the following program q.

    q(x)
       Return p(0)

It should be clear that f is computable and that, for every p, p ∈ A ⇔ q ∈ B.