Relation Isomorphism Preserves Reflexivity

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 a reflexive relation if and only if $\RR_2$ is also a reflexive relation.

Proof

Without loss of generality it is necessary to prove only that if $\RR_1$ is reflexive then $\RR_2$ is reflexive.

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

Let $y \in T$.

Let $x = \map {\phi^{-1} } y$.

As $\phi$ is a bijection it follows from Inverse Element of Bijection that:

$y = \map \phi x$

As $\RR_1$ is a reflexive relation it follows that:

$x \mathrel {\RR_1} x$

As $\phi$ is a relation isomorphism it follows that:

$\map \phi x \mathrel {\RR_2} \map \phi x$

Hence the result.

$\blacksquare$

Sources

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