Identity Mapping is Relation Isomorphism

Theorem

Let $\struct {S, \RR}$ be a relational structure.


Then the identity mapping $I_S: S \to S$ is a relation isomorphism from $\struct {S, \RR}$ to itself.


Proof

By definition of identity mapping:

$\forall x \in S: \map {I_S} x = x$

So:

$x \mathrel \RR y \implies \map {I_S} x \mathrel \RR \map {I_S} y$


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

Hence:

$\map {I_S^{-1} } x = x$

So:

$x \mathrel \RR y \implies \map {I_S^{-1} } x \mathrel \RR \map {I_S^{-1} } y$

$\blacksquare$


Sources

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