Lower Adjoint at Element is Minimum of Preimage of Singleton of Element implies Composition is Identity
Jump to navigation
Jump to search
Theorem
Let $L = \struct {S, \preceq}, R = \struct {T, \precsim}$ be ordered sets.
Let $g: S \to T, d: T \to S$ be mappings such that
- $\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {\set t} }$
Then $g \circ d = I_T$
where $I_T$ denotes the identity mapping of $T$.
Proof
Let $t \in T$.
By assumption:
- $\map d t = \min \set {g^{-1} \sqbrk {\set t} }$
By definition of min operation:
- $\map d t = \map \inf {g^{-1} \sqbrk {\set t} }$ and $\map d t \in g^{-1} \sqbrk {\set t}$
By definition of image of set:
- $\map g {\map d t} \in \set t$
By definition of singleton:
- $\map g {\map d t} = t$
Thus by definition of composition of mappings:
- $\map {\paren {g \circ d} } t = t$
$\blacksquare$
Sources
- Mizar article WAYBEL_1:23