Inverse Relational Structures of Isomorphic Structures are Isomorphic

Theorem

Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.

Let ${\RR_1}^{-1}$ and ${\RR_2}^{-1}$ be the inverses of $\RR_1$ and $\RR_2$ respectively.

Let $f: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a relation isomorphism.

Then $f: \struct {S, {\RR_1}^{-1} } \to \struct {T, {\RR_2}^{-1} } $ is also a relation isomorphism.

Proof

\(\ds \forall x, y \in S: \, \)

\(\ds x\)

\({\RR_1}^{-1}\)

\(\ds y\)

\(\ds \leadstoandfrom \ \ \)

\(\ds y\)

\(\RR_1\)

\(\ds x\)

Definition of Inverse Relation

\(\ds \leadstoandfrom \ \ \)

\(\ds \map f y\)

\(\RR_2\)

\(\ds \map f x\)

Definition of Relation Isomorphism

\(\ds \leadstoandfrom \ \ \)

\(\ds \map f x\)

\({\RR_2}^{-1}\)

\(\ds \map f y\)

Definition of Inverse Relation

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.26$