Induced Structure from Doubleton is Isomorphic to External Direct Product with Self

Theorem
Let $A$ be a set.

Let $\struct {S, \odot}$ be an algebraic structure.

Let $S^A$ denote the set of mappings from $A$ to $S$.

Let $\struct {S^A, \odot}$ denote the algebraic structure on $S^A$ induced by $\odot$.

Then $\struct {S^A, \odot}$ is isomorphic to the external direct product of $\struct {S, \odot}$ with itself.

Proof
Let $A = \set {a, b}$ where $a$ and $b$ are arbitrary.

Let $\phi: S^A \to S^2$ be defined as:


 * $\forall f \in S^A: \map \phi f = \tuple {\map f a, \map f b}$

We are to show that $\phi$ is an isomorphism.

First we demonstrate that $\phi$ is a homomorphism.

So, let $f, g \in S^A$.

Recall that:
 * by definition of pointwise operation:
 * $\forall f, g \in S^A: \map {\paren {f \odot g} } x = \map f x \odot \map g x$


 * by definition of external direct product:
 * $\forall \tuple {x_1, x_2}, \tuple {y_1, y_2} \in S \times S: \tuple {x_1, x_2} \odot \tuple {y_1, y_2} = \tuple {x_1 \odot y_1, x_2 \odot y_2}$

We have:

and so $\phi$ has been shown to be a homomorphism.

It remains to be shown that $\phi$ is a bijection.

Let $\map \phi f = \map \phi g$.

We have:

Thus by definition $\phi$ is an injection.

Let $\tuple {s_1, s_2} \in S$ be arbitrary.

By the nature of $S^A$, there exists a mapping $h: A \to S$ such that:
 * $\map h a = s_1$
 * $\map h b = s_2$

Hence:
 * $\map \phi h = \tuple {s_1, s_2}$

As $\tuple {s_1, s_2}$ is arbitrary, it follows that $\phi$ is a surjection.

Thus $\phi$ is both an injection and a surjection, hence by definition a bijection.

We have that $\phi$ is a homomorphism which is a bijection.

That is, $\phi$ is an isomorphism.

The result follows.