Subset equals Image of Preimage implies Surjection/Proof 1

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

In particular, it holds for $T$ itself.

Hence:

So:
 * $T \subseteq \Img f \subseteq T$

and so by definition of set equality:
 * $\Img f = T$

So, by definition, $f$ is a surjection.