Relation Isomorphism Preserves Equivalence Relations

Theorem

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

Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.

Then $\RR_1$ is an equivalence relation if and only if $\RR_2$ is also an equivalence relation.

Proof

Let $\phi: S \to T$ be a relation isomorphism.

By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.

Without loss of generality, it thus is necessary to prove only that if $\RR_1$ is an equivalence relation then $\RR_2$ is an equivalence relation.

So, suppose $\RR_1$ is an equivalence relation.

By definition:

$(1): \quad \RR_1$ is reflexive

$(2): \quad \RR_1$ is symmetric

$(3): \quad \RR_1$ is transitive.

It follows that:

From Relation Isomorphism Preserves Reflexivity, $\RR_2$ is reflexive.

From Relation Isomorphism Preserves Symmetry, $\RR_2$ is symmetric.

From Relation Isomorphism Preserves Transitivity, $\RR_2$ is transitive.

So by definition $\RR_2$ is an equivalence relation.

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$