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
$\blacksquare$