Subset equals Preimage of Image iff Mapping is Injection

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Theorem

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

Let $f^{-1}$ denote the inverse of $f$.


Then:

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ if and only if $f$ is an injection

where:

$f \sqbrk A$ denotes the image of $A$ under $f$
$f^{-1} \circ f$ denotes the composition of $f^{-1}$ and $f$.


This can be expressed in the language and notation of direct image mappings and inverse image mappings as:

$\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ if and only if $f$ is an injection


Proof

Sufficient Condition

Let $g$ be such that:

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

Then by Subset equals Preimage of Image implies Injection, $f$ is an injection.

$\Box$


Necessary Condition

Let $f$ be an injection.

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

$\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$

$\blacksquare$


Sources

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