Subset equals Image of Preimage implies Surjection/Proof 2

Proof
Suppose $f$ is not a surjection.

$T$ must have at least two elements for this to be the case.

Let one of these two elements not be the image of any element of $S$.

That is, let $b_1, b_2 \in T$ such that:
 * $\exists a \in S: f \left({a}\right) = b_1$
 * $\nexists x \in S: f \left({x}\right) = b_2$

Let $B = \left\{{b_1, b_2}\right\}$.

Then:

So by the Rule of Transposition:
 * $\forall B \subseteq S: B = \left({f \circ f^{-1} }\right) \left[{B}\right]$

implies that $g$ is an surjection.