Subset_equals_Preimage_of_Image_iff_Mapping_is_Injection

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