Subset equals Image of Preimage iff Mapping is Surjection

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.

Similarly, let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.


Then:

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

if and only if $f$ is a surjection.


Proof

Sufficient Condition

Let $f$ be such that:

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

From Subset equals Image of Preimage implies Surjection, $f$ is a surjection.

$\Box$


Necessary Condition

Let $f$ be a surjection.

Then by Image of Preimage of Subset under Surjection equals Subset:

$B = \map {f^\to} {\map {f^\gets} B}$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Subset_equals_Image_of_Preimage_iff_Mapping_is_Surjection&oldid=417550"