Subset equals Preimage of Image iff Mapping is Injection

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

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

Then:
 * $\forall A \subseteq S: A = \paren {f^{-1} \circ f} \sqbrk A$ $f$ 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:
 * $\forall A \in \powerset S: A = \map {\paren {f^\gets \circ f^\to} } A$ $f$ is an injection

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.

Necessary Condition
Let $f$ be an injection.

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


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