Relation Isomorphism Preserves Reflexivity

Theorem
Let $\struct {S, \mathcal R_1}$ and $\struct {T, \mathcal R_2}$ be relational structures.

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

Then $\mathcal R_1$ is a reflexive relation $\mathcal R_2$ is also a reflexive relation.

Proof
it is necessary to prove only that if $\mathcal R_1$ is reflexive then $\mathcal R_2$ 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 $\mathcal R_1$ is a reflexive relation it follows that:
 * $x \mathrel {\mathcal R_1} x$

As $\phi$ is a relation isomorphism it follows that:
 * $\map \phi x \mathrel {\mathcal R_2} \map \phi x$

Hence the result.