Recursive Mapping

Theorem

Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $\left({S^*, \circ, \preceq}\right)$ be $\left({S, \circ, \preceq}\right) \setminus \left\{{0}\right\}$ as defined in Zero Complement.

Let $\left({T, *}\right)$ be an algebraic structure, and let $a \in T$.

Then there is one and only one mapping $f_a: \left({S^*, \circ, \preceq}\right) \to \left({T, *}\right)$ such that:

$\forall n \in S \setminus \left\{{0}\right\}: f_a \left({n}\right) = \begin{cases} a & : n = 1 \\ f_a \left({r}\right) * a & : n = r \circ 1 \end{cases}$

Proof

Let $s: T \to T$ be the mapping defined as:

$\forall x \in T: s \left({x}\right) = x * a$

The result follows by a direct application of the Principle of Recursive Definition to $s$, $1 \in S$ and $a \in T$:

$f_a \left({n}\right) = \begin{cases} a & : n = 1 \\ f_a \left({r}\right) * a & : n = r \circ 1 \end{cases}$

Because $g_a$ as defined in Recursive Mapping with Identity is unique, the restriction of $g_a$ to $S^*$ is $f_a$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 16$: Theorem $16.7$