Relation Isomorphism Preserves Transitivity

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 transitive relation iff $\mathcal R_2$ is also a transitive relation.

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

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

Thus WLOG it suffices to prove only that if $\mathcal R_1$ is transitive, then $\mathcal R_2$ is also transitive.

So, suppose $\mathcal R_1$ is a transitive relation.

Let $y_1, y_2, y_3 \in T$ such that $y_1 \,\mathcal R_2\, y_2$ and $y_2 \,\mathcal R_2\, y_3$.

Let $x_1 = \phi^{-1} \left({y_1}\right)$, $x_2 = \phi^{-1} \left({y_2}\right)$ and $x_3 = \phi^{-1} \left({y_3}\right)$.

As $\phi$ is a bijection it follows from Inverse Element of Bijection that $y_1 = \phi \left({x_1}\right)$, $y_2 = \phi \left({x_2}\right)$ and $y_3 = \phi \left({x_3}\right)$.

As $\phi^{-1}$ is a relation isomorphism it follows that:
 * $x_1 = \phi^{-1} \left({y_1}\right) \,\mathcal R_1\, \phi^{-1} \left({y_2}\right) = x_2$
 * $x_2 = \phi^{-1} \left({y_2}\right) \,\mathcal R_1\, \phi^{-1} \left({y_3}\right) = x_3$

As $\mathcal R_1$ is a transitive relation it follows that $x_1 \,\mathcal R_1\, x_3$.

As $\phi$ is a relation isomorphism it follows that $y_1 = \phi \left({x_1}\right) \,\mathcal R_2\, \phi \left({x_3}\right) = y_3$.

Hence if $y_1 \,\mathcal R_2\, y_2$ and $y_2 \,\mathcal R_2\, y_3$, then also $y_1 \,\mathcal R_2\, y_3$.

Hence, $\mathcal R_2$ is a transitive relation, by definition.

Hence the result.