Subset equals Image of Preimage implies Surjection/Proof 2

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

Let $f_g: \mathcal P \left({S}\right) \to \mathcal P \left({T}\right)$ be the mapping induced by $g$.

Similarly, let $f_{g^{-1}}: \mathcal P \left({T}\right) \to \mathcal P \left({S}\right)$ be the mapping induced by the inverse $g^{-1}$.

Let:
 * $\forall B \in \mathcal P \left({T}\right): B = \left({f_g \circ f_{g^{-1}}}\right) \left({B}\right)$

Then $g$ is a surjection.

Proof
Suppose $g$ 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:
 * $\exists a \in S: g \left({a}\right) = b_1$
 * $\not \exists x \in S: g \left({x}\right) = b_2$

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

Then:

So by the Rule of Transposition:
 * $\forall B \in \mathcal P \left({S}\right): B = \left({f_g \circ f_{g^{-1} } }\right) \left({B}\right)$

implies that $g$ is an surjection.