Identity Mapping is Relation Isomorphism

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Theorem

Let $\left({S, \mathcal R}\right)$ be a relational structure.


Then the identity mapping $I_S: S \to S$ is a relation isomorphism from $\left({S, \mathcal R}\right)$ to itself.


Proof

By definition of identity mapping:

$\forall x \in S: I_S \left({x}\right) = x$

So:

$x \mathrel {\mathcal R} y \implies I_S \left({x}\right) \mathrel {\mathcal R} I_S \left({y}\right)$


From Identity Mapping is Bijection, $I_S$ is a bijection.

Hence:

$I_S^{-1} \left({x}\right) = x$

So:

$x \mathrel {\mathcal R} y \implies I_S^{-1} \left({x}\right) \mathrel {\mathcal R} I_S^{-1} \left({y}\right)$

$\blacksquare$


Sources