Inverse Image of Direct Image of Inverse Image equals Inverse Image Mapping

Theorem

Let $f: S \to T$ be a mapping.

Let:

$f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$

$f^\gets: \powerset T \to \powerset S$ denote the inverse image mapping of $f$

where $\powerset S$ denotes the power set of $S$.

Then:

$f^\gets \circ f^\to \circ f^\gets = f^\gets$

where $\circ$ denotes composition of mappings.

Proof

\(\ds \forall A \in \powerset S: \, \)

\(\ds A\)

\(\subseteq\)

\(\ds \map {\paren {f^\gets \circ f^\to} } A\)

Subset of Domain is Subset of Preimage of Image

\(\ds \leadsto \ \ \)

\(\ds \forall B \in \powerset T: \, \)

\(\ds \map {f^\gets} B\)

\(\subseteq\)

\(\ds \map {\paren {f^\gets \circ f^\to} } {\map {f^\gets} B}\)

Image of Subset under Mapping is Subset of Image, setting $A = \map {f^\gets} B \in \powerset S$

and:

\(\ds \forall B \in \powerset T: \, \)

\(\ds \map {\paren {f^\to \circ f^\gets} } B\)

\(\subseteq\)

\(\ds B\)

Subset of Codomain is Superset of Image of Preimage

\(\ds \leadsto \ \ \)

\(\ds \forall A \in \powerset S: \, \)

\(\ds \map {\paren {f^\gets \circ f^\to \circ f^\gets} } B\)

\(\subseteq\)

\(\ds \map {f^\gets} B\)

Image of Subset under Mapping is Subset of Image

Thus we have:

$\forall B \in \powerset T: \map {f^\gets} B \subseteq \map {\paren {f^\gets \circ f^\to \circ f^\gets} } B \subseteq \map {f^\gets} B$

and the result follows.

$\blacksquare$

Also see

• Direct Image of Inverse Image of Direct Image equals Direct Image Mapping

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $4$