# Isomorphism between Algebraic Structures induces Isomorphism between Induced Structures

## Theorem

Let $A$ be a set.

Let $\struct {S, \odot}$ and $\struct {T, \otimes}$ be algebraic structures.

Let:

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

Let:

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

Let $\phi$ be an isomorphism from $S$ to $T$.

Let $\chi: S^A \to T^A$ be the mapping defined as:

$\forall f \in S^A: \map \chi f = \phi \circ f$

where:

$\phi \circ f$ denotes the composition of $\phi$ with $f$.

Then:

$\chi$ is an isomorphism from $\struct {S^A, \odot}$ to $\struct {T^A, \otimes}$

## Proof

Let $f: A \to S$ and $g: A \to S$ be arbitrary elements of $S^A$.

We have that $\phi$ is an isomorphism.

Hence $\phi$ is a fortiori a homomorphism which is a bijection.

Hence, from Bijection iff exists Mapping which is Left and Right Inverse, $\phi$ has an inverse mapping $\phi^{-1}: T \to S$ such that:

$\phi \circ \phi^{-1} = I_T$
$\phi^{-1} \circ \phi = I_S$

where $I_S$ and $I_T$ denote the identity mappings on $S$ and $T$ respectively.

From Inverse of Bijection is Bijection, $\phi^{-1}$ is also a bijection.

We have that:

 $\ds \map \chi {f \odot g}$ $=$ $\ds \phi \circ \paren {f \odot g}$ Definition of $\chi$ $\ds$ $=$ $\ds \paren {\phi \circ f} \otimes \paren {\phi \circ g}$ Composition of Mappings is Left Distributive over Homomorphism of Pointwise Operation $\ds$ $=$ $\ds \map \chi f \otimes \map \chi g$ Definition of $\chi$

demonstrating that $\chi$ is a homomorphism.

$\Box$

Then:

 $\ds \map \chi f$ $=$ $\ds \map \chi g$ by assumption $\ds \leadsto \ \$ $\ds \phi \circ f$ $=$ $\ds \phi \circ g$ Definition of $\chi$ $\ds \leadsto \ \$ $\ds \phi^{-1} \circ \phi \circ f$ $=$ $\ds \phi^{-1} \circ \phi \circ g$ as $\phi$ is a bijection $\ds \leadsto \ \$ $\ds f$ $=$ $\ds g$

Hence $\chi$ is an injection.

$\Box$

Let $g: A \to T$ be an arbitrary element of $T^A$.

As $\phi^{-1}: T \to S$ is a bijection:

$\exists h \in S^A: h = \map {\phi^{-1} } g = \phi^{-1} \circ g$

That is:

$\exists h \in S^A: \map \phi h = g$

As $g$ is arbitrary:

$\forall g \in T^A: \exists h \in S^A: g = \phi \circ h = \map \chi h$

That is, $\chi$ is a surjection.

$\Box$

We have shown that $\chi$ is:

an injection
a surjection
a homomorphism

and so by definition $\chi$ is an isomorphism.

$\blacksquare$

## 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 $13.11 \ \text{(c)}$