Relation Isomorphism Preserves Reflexivity

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

Proof
WLOG 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 $x = \phi^{-1} \left({y}\right)$.

As $\phi$ is a bijection it follows from Inverse Element of Bijection that:
 * $y = \phi \left({x}\right)$

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:
 * $\phi \left({x}\right) \mathrel {\mathcal R_2} \phi \left({x}\right)$

Hence the result.