Identity Mapping is Frame Homomorphism

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Theorem

Let $L = \struct{S, \preceq}$ be a frame.

Let $\operatorname{id}_S$ denote the identity mapping on $S$.

Then:

$\operatorname{id}_S$ is a frame homomorphism of $L$ to $L$

Proof

$\operatorname{id}_S$ is Finite Meet Preserving

Let $F \subseteq S_1$ be a finite subset.

We have:

\(\ds \inf \operatorname{id}_S \sqbrk F\)

\(=\)

\(\ds \inf F\)

Definition of Identity Mapping

\(\ds \)

\(=\)

\(\ds \map {\operatorname{id}_S} {\inf F}\)

Definition of Identity Mapping

Since $F$ was arbitrary, it follows that $\operatorname{id}_S$ is finite meet preserving by definition.

$\Box$

$\operatorname{id}_S$ is Arbitrary Join Preserving

Let $A \subseteq S_1$ be any subset of $S$.

We have:

\(\ds \sup \operatorname{id}_S \sqbrk F\)

\(=\)

\(\ds \sup F\)

Definition of Identity Mapping

\(\ds \)

\(=\)

\(\ds \map {\operatorname{id}_S} {\sup F}\)

Definition of Identity Mapping

Since $A$ was arbitrary, it follows that $\operatorname{id}_S$ is arbitrary join preserving by definition.

$\Box$

By definition, $\operatorname{id}_S$ is a frame homomorphism

$\blacksquare$