Inverse Relational Structures of Isomorphic Structures are Isomorphic
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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 $14.26$