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

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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:

\(\ds \map \phi {f \odot g}\) \(=\) \(\ds \tuple {\map {\paren {f \odot g} } a, \map {\paren {f \odot g} } b}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \tuple {\map f a \odot \map g a, \map f b \odot \map g b}\) Definition of Pointwise Operation
\(\ds \) \(=\) \(\ds \tuple {\map f a, \map f b} \odot {\map g a, \map g b}\) Definition of External Direct Product
\(\ds \) \(=\) \(\ds \map \phi f \odot \map \phi g\) Definition of $\phi$

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

$\Box$


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

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


We have:

\(\ds \map \phi f\) \(=\) \(\ds \map \phi g\)
\(\ds \leadsto \ \ \) \(\ds \tuple {\map f a, \map f b}\) \(=\) \(\ds \tuple {\map g a, \map g b}\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds \map f a\) \(=\) \(\ds \map g a\) Equality of Ordered Pairs
\(\, \ds \land \, \) \(\ds \map f b\) \(=\) \(\ds \map g b\)
\(\ds \leadsto \ \ \) \(\ds f\) \(=\) \(\ds g\) Equality of Mappings

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.

$\Box$


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

That is, $\phi$ is an isomorphism.

The result follows.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.11 \ \text{(d)}$
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