General Morphism Property for Semigroups

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.

Let $\phi: S \to T$ be a homomorphism.

Then:
 * $\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$

Hence it follows that:
 * $\forall n \in \N_{>0}: \forall s \in S: \map \phi {s^n} = \paren {\map \phi s}^n$

Proof
$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ can be proved by induction.

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$

$P(1)$ is true, as this just says:
 * $\map \phi {s_1} = \map \phi {s_1}$.

Basis for the Induction
$P(2)$ is the case:
 * $\map \phi {s_1 \circ s_2} = \map \phi {s_1} * \map \phi {s_2}$

This follows from the fact that $\phi$ is a homomorphism.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k}$

Then we need to show:


 * $\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k + 1} } = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k} * \map \phi {s_{k + 1} }$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$

The result for $n \in \N_{>0}$ follows directly from the above, by replacing each occurrence of $s_k$ with $s$.