Subset equals Image of Preimage implies Surjection/Proof 1

Proof
Let $f$ be such that:
 * $\forall B \in \mathcal P \left({T}\right): B = \left({f^\to \circ f^\gets}\right) \left({B}\right)$

In particular, it holds for $T$ itself.

Hence:

So:
 * $T \subseteq \operatorname{Im} \left({f}\right) \subseteq T$

and so by definition of set equality:
 * $\operatorname{Im} \left({f}\right) = T$

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