Subset equals Image of Preimage implies Surjection

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

Let:
 * $\forall B \subseteq T: B = \paren {f \circ f^{-1} } \sqbrk B$

where $f \sqbrk B$ denotes the image of $B$ under $f$.

Then $f$ is a surjection.

Also see

 * Image of Preimage of Subset under Surjection equals Subset
 * Subset equals Image of Preimage iff Mapping is Surjection