Composition of Direct Image Mappings of Mappings
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Theorem
Let $A, B, C$ be non-empty sets.
Let $f: A \to B$ and $g: B \to C$ be mappings.
Let:
- $f^\to: \powerset A \to \powerset B$
and
- $g^\to: \powerset B \to \powerset C$
be the direct image mappings of $f$ and $g$.
Then:
- $\paren {g \circ f}^\to = g^\to \circ f^\to$
Proof 1
Let $S \subseteq A$ such that $S \ne \O$.
Then:
\(\ds \map {\paren {g^\to \circ f^\to} } S\) | \(=\) | \(\ds \map {g^\to} {\map {f^\to} S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map g x: x \in \map {f^\to} S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map g x: \exists y \in S: x = \map f y}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map g {\map f y}: y \in S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\map {\paren {g \circ f} } y: y \in S}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {g \circ f}^\to} S\) |
Now we treat the case where $S = \O$:
\(\ds \map {\paren {g^\to \circ f^\to} } \O\) | \(=\) | \(\ds \O\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {g \circ f}^\to} \O\) |
and the result is complete.
$\blacksquare$
Proof 2
We have that a mapping is a relation.
Hence Composition of Direct Image Mappings of Relations applies.
$\blacksquare$