# Relation Isomorphism Preserves Equivalence Relations

## Theorem

Let $\left({S, \mathcal R_1}\right)$ and $\left({T, \mathcal R_2}\right)$ be relational structures.

Let $\left({S, \mathcal R_1}\right)$ and $\left({T, \mathcal R_2}\right)$ be (relationally) isomorphic.

Then $\mathcal R_1$ is an equivalence relation if and only if $\mathcal R_2$ 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.

Thus WLOG it is necessary to prove only that if $\mathcal R_1$ is an equivalence relation then $\mathcal R_2$ is an equivalence relation.

So, suppose $\mathcal R_1$ is an equivalence relation.

By definition:

- $(1): \quad \mathcal R_1$ is reflexive
- $(2): \quad \mathcal R_1$ is symmetric
- $(3): \quad \mathcal R_1$ is transitive.

It follows that:

- From Relation Isomorphism Preserves Reflexivity, $\mathcal R_2$ is reflexive.
- From Relation Isomorphism Preserves Symmetry, $\mathcal R_2$ is symmetric.
- From Relation Isomorphism Preserves Transitivity, $\mathcal R_2$ is transitive.

So by definition $\mathcal R_2$ is an equivalence relation.

Hence the result.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $14.9 \ \text{(c)}$